romanian mathematical magazine
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R M MROMANIAN MATHEMATICAL MAGAZINE
Founding EditorFounding EditorDANIEL SITARUDANIEL SITARU
Available onlineAvailable onlinewww.ssmrmh.rowww.ssmrmh.ro
ISSN-L 2501-0099ISSN-L 2501-0099
RMM - Calculus Marathon 1501 - RMM - Calculus Marathon 1501 - 16001600
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1 RMM-CALCULUS MARATHON 1501-1600
Proposed by
Daniel Sitaru – Romania,Vasile Mircea Popa-Romania
Asmat Qatea-Afghanistan,Kaushik Mahanta-Assam-India
Srinivasa Raghava-AIRMC-India,Sujeethan Balendran-SriLanka
Narendra Bhandari-Bajura-Nepal,Orxan Abasov-Azerbaijan
Abdul Mukhtar-Nigeria,Ty Halpen-Florida-SUA,Angad Singh-India,George
Apostolopoulos-Messolonghi-Greece,Amrit Awasthi-India,Surjeet Singhania-
India.Florică Anastase-Romania,Neculai Stanciu-Romania,Mohammad
Hamed Nasery-Afghanistan,Costel Florea-Romania,Mikael Bernardo-
Mozambique,Simon Peter-Madagascar,Durmuş Ogmen-Turkiye
Ajetunmobi Abdulqoyyum-Nigeria,Syed Shahabudeen-India
Probal Chakraborty-India,Tobi Joshua-Nigeria,Ose Favour-Nigeria
Onikoyi Adeboye-Nigeria,Marin Chirciu-Romania,Marian Ursărescu-Romania
Ruxandra Daniela Tonilă-Romania
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2 RMM-CALCULUS MARATHON 1501-1600
Solutions by
Daniel Sitaru – Romania,Rana Ranino-Setif-Algerie
Jose Ferreira Queiroz-Olinda-Brazil,Ty Halpen-Florida-USA,Amrit Awasthi-
Punjab-India,Syed Shahabudeen-Kerala-India,Akerele Olofin-Nigeria
Cornel Ioan Vălean-Romania,Ngulmun George Baite-India,Mohammad
Rostami-Afghanistan,Yen Tung Chung-Taichung-Taiwan,Orlando Irahola
Ortega-Bolivia,Ghuiam Naseri-Afghanistan,Marian Ursărescu-
Romania,Ajetunmobi Abdulquyyum-Nigeria,Muhammad Afzal-Pakistan
Adrian Popa-Romania,Angad Singh-India,Ruxandra Daniela Tonilă-Romania
Artan Ajredini-Presheva-Serbie,Naren Bhandari-Bajura-Nepal,Soumitra
Mandal-India,Ose Favour-Nigeria,Kartick Chandra Betal-India,Asmat Qatea-
Afghanistan,Ravi Prakash-New Delhi-India,Ahmed Yackoube Chach-
Mauritania,George Florin Șerban-Romania,Serlea Kabay-Liberia,Obaidullah
Jaihon-Afghanistan,Mikael Bernardo-Mozambique,Florică Anastase-Romania
Felix Marin-Romania,Kamel Gandouli Rezgui-Tunisia,Surjeet Singhania-India
Probal Chakraborty-Kolkata-India,Kaushik Mahanta-Assam-India
Hussain Reza Zadah-Afghanistan,Timson Azeez Folorunsho-Nigeria
Luca Paes Barreto-Pernambuco-Brazil,Abdul Mukhtar-Nigeria
Remus Florin Stanca-Romania,Mohammad Hamed Nasery-Afghanistan
Satyam Roy-India,Sujit Bhowmick-India,Sediqakbar Restheen-Afghanistan
Santiago Alvarez-Mexico,Almas Babirov-Azerbaijan,
Fayssal Abdelli-Bejaia-Algerie, Ajenikoko Gbolahan-Nigeria
Dawid Bialek-Poland
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3 RMM-CALCULUS MARATHON 1501-1600
1501. Prove that:
𝚿(𝟕
𝟖) − 𝚿(
𝟑
𝟖) = 𝝅√𝟐 − 𝟐√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)
where 𝚿(𝒙) is the digamma function.
Proposed by Vasile Mircea Popa-Romania
Solution 1 by Rana Ranino-Setif-Algerie
We know that:
∫𝒙𝒏
𝟏 + 𝒙𝒅𝒙
𝟏
𝟎
=𝟏
𝟐(𝝍(
𝒏
𝟐+ 𝟏) −𝝍 (
𝒏
𝟐+𝟏
𝟐))
𝛀 = 𝟐∫𝒙−𝟏𝟒
𝟏 + 𝒙𝒅𝒙
𝟏
𝟎
=𝒙=𝒙𝟒
𝟖∫𝒙𝟐
𝟏 + 𝒙𝟒𝒅𝒙
𝟏
𝟎
= 𝟒(∫𝒙𝟐 + 𝟏
𝟏 + 𝒙𝟒𝒅𝒙
𝟏
𝟎
+∫𝒙𝟐 − 𝟏
𝟏 + 𝒙𝟒𝒅𝒙
𝟏
𝟎
) =
= 𝟒(∫𝟏 +
𝟏𝒙𝟐
𝒙𝟐 +𝟏𝒙𝟐
𝒅𝒙𝟏
𝟎
+∫𝟏 −
𝟏𝒙𝟐
𝒙𝟐 +𝟏𝒙𝟐
𝒅𝒙𝟏
𝟎
)
𝒖 = 𝒙 −𝟏
𝒙; 𝒗 = 𝒙 +
𝟏
𝒙⇒ 𝛀 = 𝟒(∫
𝒅𝒖
𝒖𝟐 + 𝟐
𝟎
−∞
−∫𝒅𝒗
𝒗𝟐 − 𝟐
∞
𝟐
) =
= 𝟒(𝝅
𝟐√𝟐−𝟏
𝟐√𝟐𝐥𝐨𝐠 (
𝒗 − √𝟐
𝒗+ √𝟐)|𝟐
∞
= 𝝅√𝟐 − √𝟐 𝐥𝐨𝐠 (𝟐 + √𝟐
𝟐 − √𝟐) =
= 𝝅√𝟐 − √𝟐 𝐥𝐨𝐠 [(𝟐 + √𝟐
√𝟐)
𝟐
]
Therefore,
𝝍(𝟕
𝟖) −𝝍(
𝟑
𝟖) = 𝝅√𝟐 − 𝟐√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)
Solution 2 by Jose Ferreira Queiroz-Olinda-Brazil
𝝍(𝒓
𝒎) = −𝜸 − 𝐥𝐨𝐠𝟐𝒎 −
𝝅
𝟐⋅ 𝐜𝐨𝐭 (
𝝅𝒓
𝒎) + 𝟐 ∑ 𝐜𝐨𝐬 (
𝟐𝝅𝒏𝒗
𝒎)
[𝒎−𝟏𝟐]
𝒏=𝟏
𝐥𝐨𝐠 (𝐬𝐢𝐧 (𝝅𝒏
𝒎))
For 𝒓 = 𝟕,𝒎 = 𝟖, we have:
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4 RMM-CALCULUS MARATHON 1501-1600
𝝍(𝟕
𝟖) = −𝜸 − 𝐥𝐨𝐠 𝟏𝟔 −
𝝅
𝟐𝐜𝐨𝐭 (
𝟕𝝅
𝟖) + 𝟐∑𝐜𝐨𝐬 (
𝟒𝒏𝝅
𝟖) 𝐥𝐨𝐠 (𝐬𝐢𝐧 (
𝒏𝝅
𝟖))
𝟑
𝒏=𝟏
=
= −𝜸 − 𝟒 𝐥𝐨𝐠 𝟐 +𝝅
𝟐(√𝟐 + 𝟏) +
√𝟐
𝟐𝐥𝐨𝐠(𝟐 − √𝟐) −
√𝟐
𝟐𝐥𝐨𝐠(𝟐 + √𝟐)
For 𝒓 = 𝟑,𝒎 = 𝟖 we have:
𝝍(𝟑
𝟖) = −𝜸 − 𝐥𝐨𝐠 𝟏𝟔 −
𝝅
𝟐𝐜𝐨𝐭 (
𝟑𝝅
𝟖) + 𝟐∑𝐜𝐨𝐬 (
𝟔𝒏𝝅
𝟖) 𝐥𝐨𝐠 (𝐬𝐢𝐧 (
𝒏𝝅
𝟖))
𝟑
𝒏=𝟏
=
= −𝜸 − 𝟒 𝐥𝐨𝐠𝟐 −𝝅
𝟐(√𝟐 − 𝟏) −
√𝟐
𝟐𝐥𝐨𝐠(𝟐 − √𝟐) +
√𝟐
𝟐𝐥𝐨𝐠(𝟐 + √𝟐)
Now, 𝝍(𝟕
𝟖) − 𝝍(
𝟑
𝟖) = 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠(𝟐 − √𝟐) − √𝟐 𝐥𝐨𝐠(𝟐 + √𝟐) =
= 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠(𝟐 − √𝟐
𝟐 + √𝟐) = 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠(
𝟐
(𝟐 + √𝟐)𝟐) =
= 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠 (𝟏
𝟑 + 𝟐√𝟐) = 𝝅√𝟐 + √𝟐 𝐥𝐨𝐠 (
𝟏
(𝟏 + √𝟐)𝟐)
Therefore,
𝝍(𝟕
𝟖) −𝝍(
𝟑
𝟖) = 𝝅√𝟐 − 𝟐√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)
1502. Prove that:
∏𝟐𝒏+ (−𝟏)
𝒏𝟐+𝒏𝟐
𝟐𝒏 + 𝐜𝐨𝐬 (𝒏𝝅𝟐)
∞
𝒏=𝟏
=√𝟒 − 𝟐√𝟐
𝟐
Proposed by Asmat Qatea-Afghanistan
Solution 1 by Ty Halpen-Florida-USA
∏𝟐𝒏+ (−𝟏)
𝒏𝟐+𝒏𝟐
𝟐𝒏 + 𝐜𝐨𝐬 (𝒏𝝅𝟐 )
∞
𝒏=𝟏
=
= (∏𝟐(𝟐𝒏) + (−𝟏)
(𝟐𝒏)𝟐+𝟐𝒏𝟐
𝟐(𝟐𝒏) + 𝐜𝐨𝐬(𝒏𝝅)
∞
𝒏=𝟏
)(∏𝟐(𝟐𝒏 − 𝟏) + (−𝟏)
(𝟐𝒏−𝟏)𝟐+𝟐𝒏−𝟏𝟐
𝟐(𝟐𝒏 − 𝟏) + 𝐬𝐢𝐧(𝒏𝝅)
∞
𝒏=𝟏
)
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5 RMM-CALCULUS MARATHON 1501-1600
= (∏𝟒𝒏 + (−𝟏)𝒏
𝟒𝒏 + (−𝟏)𝒏
∞
𝒏=𝟏
)(∏𝟒𝒏− 𝟐 + (−𝟏)𝒏
𝟒𝒏 − 𝟐
∞
𝒏=𝟏
) =∏(𝟏 +(−𝟏)𝒏
𝟒𝒏 − 𝟐)
∞
𝒏=𝟏
Now, use the result:
∏(𝟏+(−𝟏)𝒏𝒄
𝒂𝒏 + 𝒃)
∞
𝒏=𝟏
=𝟐−𝒃𝒂√𝝅𝚪(
𝒂 + 𝒃𝟐 )
𝚪 (𝒂 + 𝒃 − 𝒄𝟐𝒂 )𝚪 (
𝟐𝒂+ 𝒃 + 𝒄𝟐𝒂 )
Fro 𝒂 = 𝟒, 𝒃 = −𝟐, 𝒄 = 𝟏:
∏(𝟏+(−𝟏)𝒏
𝟒𝒏 − 𝟐)
∞
𝒏=𝟏
=√𝟐𝝅𝚪 (
𝟏𝟐)
𝚪(𝟏𝟖) 𝚪(
𝟕𝟖)= √𝟐𝐬𝐢𝐧 (
𝝅
𝟖) =
√𝟒 − 𝟐√𝟐
𝟐
Solution 2 by Amrit Awasthi-Punjab-India
Consider the following cases:
I) 𝒏 = 𝟒𝒌 + 𝟏: (−𝟏)𝒏𝟐+𝒏
𝟐 = −𝟏, 𝐜𝐨𝐬 (𝒏𝝅
𝟐) = 𝟎
II) 𝒏 = 𝟒𝒌 + 𝟐: (−𝟏)𝒏𝟐+𝒏
𝟐 = −𝟏, 𝐜𝐨𝐬 (𝒏𝝅
𝟐) = −𝟏
III) 𝒏 = 𝟒𝒌 + 𝟑: (−𝟏)𝒏𝟐+𝒏
𝟐 = 𝟏, 𝐜𝐨𝐬 (𝒏𝝅
𝟐) = 𝟎
IV) 𝒏 = 𝟒𝒌: (−𝟏)𝒏𝟐+𝒏
𝟐 = 𝟏, 𝐜𝐨𝐬 (𝒏𝝅
𝟐) = 𝟏
Hence, rewriting the product, we have:
𝛀 =∏𝟐(𝟒𝒌 + 𝟏) − 𝟏
𝟐(𝟒𝒌 + 𝟏) + 𝟎⋅𝟐(𝟒𝒌 + 𝟐) − 𝟏
𝟐(𝟒𝒌 + 𝟐) − 𝟏⋅𝟐(𝟒𝒌 + 𝟑) + 𝟏
𝟐(𝟒𝒌 + 𝟑) + 𝟎⋅𝟐(𝟒(𝒌 + 𝟏)) + 𝟏
𝟐(𝟒(𝒌 + 𝟏)) + 𝟏
∞
𝒌=𝟎
=
= 𝐥𝐢𝐦𝒏→∞
∏𝟖𝒌+ 𝟏
𝟖𝒌 + 𝟐⋅𝟖𝒌 + 𝟕
𝟖𝒌 + 𝟔
𝒏
𝒌=𝟎
=𝟏
𝟐⋅𝟕
𝟔𝐥𝐢𝐦𝒏→∞
∏𝒌+
𝟏𝟖
𝒌+𝟐𝟖
⋅𝒌 +
𝟕𝟖
𝒌 +𝟔𝟖
𝒏
𝒌=𝟏
=
=𝟏
𝟐⋅𝟕
𝟔𝐥𝐢𝐦𝒏→∞
𝚪(𝒏 +𝟏𝟖 + 𝟏)
𝚪(𝟏𝟖 + 𝟏)
𝚪(𝒏 +𝟐𝟖 + 𝟏)
𝚪(𝟐𝟖+ 𝟏)
⋅
𝚪 (𝒏 +𝟕𝟖 + 𝟏)
𝚪 (𝟕𝟖 + 𝟏)
𝚪 (𝒏 +𝟔𝟖 + 𝟏)
𝚪 (𝟔𝟖+ 𝟏)
=
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6 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝟐⋅𝟕
𝟔𝐥𝐢𝐦𝒏→∞
𝒏𝟏𝟖+𝟏−
𝟐𝟖+𝟕𝟖+𝟏−
𝟔𝟖−𝟏 ⋅
𝚪 (𝟐𝟖 + 𝟏)𝚪 (
𝟔𝟖 + 𝟏)
𝚪 (𝟕𝟖 + 𝟏)𝚪 (
𝟏𝟖 + 𝟏)
=
=𝟏
𝟐⋅𝟕
𝟔⋅
𝟑𝟏𝟔𝚪 (
𝟏𝟒)𝚪 (𝟏 −
𝟏𝟒)
𝟕𝟔𝟒𝚪 (
𝟏𝟖)𝚪 (𝟏 −
𝟏𝟖)=𝟏
𝟐⋅𝟕
𝟔⋅𝟑
𝟏𝟔⋅𝟔𝟒
𝟕⋅
𝝅
𝐬𝐢𝐧𝝅𝟒
𝝅
𝐬𝐢𝐧𝝅𝟖
=𝟏
𝟐⋅𝟕
𝟔⋅𝟏𝟐 𝐬𝐢𝐧
𝝅𝟖
𝟕 𝐬𝐢𝐧𝝅𝟒
= √𝟐 𝐬𝐢𝐧𝝅
𝟖
=√𝟒 − 𝟐√𝟐
𝟐
1503. Prove that:
∫ ∫ ∫ …∫𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏
√𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)(𝟏 + 𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏)
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
=
= 𝝅𝒏√𝝅𝒏+𝟐 𝑭𝒏𝒏+𝟏 (
𝟏
𝟐,𝟏
𝟐,… ,
𝟏
𝟐⏟ (𝒏+𝟏)−𝒕𝒊𝒎𝒆𝒔
; 𝟏, 𝟏, . . . , 𝟏⏞ 𝒏−𝒕𝒊𝒎𝒆𝒔
; −𝟏)
Proposed by Kaushik Mahanta-Assam-India
Solution 1 by Syed Shahabudeen-Kerala-India
∫ ∫ ∫ …∫𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏
√𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)(𝟏 + 𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏)
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
=
=∑(−𝟏)𝒌
𝟒𝒌(𝟐𝒌
𝒌)
∞
𝒌=𝟎
∫ ∫ ∫ …∫(𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏)
𝒌−𝟏𝟐 𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏
√(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
=
= ∑(−𝟏)𝒌
𝟒𝒌(𝟐𝒌
𝒌)𝚪𝒏 (𝒌 +
𝟏𝟐) 𝚪
𝒏 (𝟏𝟐)
𝚪𝒏(𝒌 + 𝟏)
∞
𝒌=𝟎
=
(𝟐𝒌)!
𝟒𝒌𝒌!=(𝟏𝟐)𝒌
=∑(−𝟏)𝒌 (𝟏
𝟐)𝒌
𝚪𝒏 (𝒌 +𝟏𝟐) 𝚪
𝒏 (𝟏𝟐)
𝚪𝒏(𝒌 + 𝟏)𝒌!
∞
𝒌=𝟎
=∑(−𝟏)𝒌
𝒌!(𝟏
𝟐)𝒌
𝚪𝒏 (𝒌 +𝟏𝟐) 𝚪
𝒏 (𝟏𝟐)
𝚪𝒏(𝒌 + 𝟏)
∞
𝒌=𝟎
=
= 𝚪𝟐𝒏 (𝟏
𝟐)∑(−𝟏)𝒌 (
𝟏
𝟐)𝒌
(𝟏𝟐)𝒌
𝒏
(𝟏)𝒌𝒏𝒌!
∞
𝒌=𝟎
= 𝝅𝒏∑(𝟏𝟐)𝒌
𝒏+𝟏
(−𝟏)𝒌
(𝟏)𝒌𝒏𝒌!
∞
𝒌=𝟎
=
= 𝝅𝒏√𝝅𝒏+𝟐 𝑭𝒏𝒏+𝟏 (𝟏
𝟐,𝟏
𝟐,… ,
𝟏
𝟐⏟ (𝒏+𝟏)−𝒕𝒊𝒎𝒆𝒔
; 𝟏, 𝟏, . . . , 𝟏⏞ 𝒏−𝒕𝒊𝒎𝒆𝒔
; −𝟏)
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7 RMM-CALCULUS MARATHON 1501-1600
Solution 2 by Akerele Olofin-Nigeria
∫ ∫ ∫ …∫𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏
√𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)(𝟏 + 𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏)
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
=
= ∫ ∫ ∫ …∫𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏
√𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
𝟏
𝟏 + 𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏=
= ∫ ∫ ∫ …∫𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏
√𝒙𝟏𝒙𝟐𝒙𝟑…𝒙𝒏(𝟏 − 𝒙𝟏)(𝟏 − 𝒙𝟐)(𝟏 − 𝒙𝟑)… (𝟏 − 𝒙𝒏)
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
∑(
𝟏𝟐+ 𝒌 − 𝟏
𝒌) (−𝟏)𝒌 (∏𝒙𝒊
𝒏
𝒌=𝟏
)
𝒌∞
𝒌=𝟎
=∑(−𝟏)𝒌 (𝒌 −
𝟏𝟐
𝒌)
∞
𝒌=𝟎
∫ ∫ ∫ …∫(∏ 𝒙𝒊
𝒏𝒊=𝟏 )𝒌−
𝟏𝟐
√∏ (𝟏 − 𝒙𝒊)𝒏𝒊=𝟏
𝒅𝒙𝟏𝒅𝒙𝟐𝒅𝒙𝟑…𝒅𝒙𝒏
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
𝟏
𝟎
⇒ 𝑱𝒏 =∑(−𝟏)𝒌(𝒌 −
𝟏𝟐
𝒌)∏∫ 𝒙
𝒊
𝒌−𝟏𝟐(𝟏 − 𝒙𝒊)
−𝟏𝟐𝒅𝒙𝒊
𝟏
𝟎
𝒏
𝒊=𝟏
∞
𝒌=𝟎
∵ ∫ 𝒙𝒊
𝒌−𝟏𝟐(𝟏 − 𝒙𝒊)
−𝟏𝟐
𝟏
𝟎
𝒅𝒙𝒊 = 𝜷(𝒌 +𝟏
𝟐,𝟏
𝟐) =
𝚪 (𝒌 +𝟏𝟐)𝚪 (
𝟏𝟐)
𝚪(𝒌 + 𝟏)
Therefore,
𝑱𝒏 =∑𝚪(𝒌 +
𝟏𝟐)
𝚪 (𝟏𝟐)𝚪
(𝒌 + 𝟏)(𝚪 (𝒌 +
𝟏𝟐)𝚪 (
𝟏𝟐)
𝚪(𝒌 + 𝟏))
𝒏∞
𝒌=𝟎
= 𝝅𝒏∑𝚪𝒏+𝟏 (𝒌 +
𝟏𝟐)
𝚪𝒏+𝟏 (𝟏𝟐)𝚪
𝒏(𝒌 + 𝟏)
(−𝟏)𝒌
𝚪(𝒌 + 𝟏)
∞
𝒌=𝟎
=
= 𝝅𝒏√𝝅𝒏+𝟐 𝑭𝒏𝒏+𝟏 (
𝟏
𝟐,𝟏
𝟐,… ,
𝟏
𝟐⏟ (𝒏+𝟏)−𝒕𝒊𝒎𝒆𝒔
; 𝟏, 𝟏, . . . , 𝟏⏞ 𝒏−𝒕𝒊𝒎𝒆𝒔
; −𝟏)
1504. If we have:
∫ (𝐬𝐢𝐧𝝅𝒙
𝒙𝟑−𝐜𝐨𝐬𝝅𝒙
𝒙𝟐−𝐬𝐢𝐧𝐡 𝝅𝒙
𝒙𝟑+𝐜𝐨𝐬𝐡𝝅𝒙
𝒙𝟐)𝒆−𝝅𝒙
√𝒙
∞
𝟎
𝒅𝒙 = 𝝅𝟐𝜶 + 𝝅𝟑𝜷
then find the values of 𝜶 and 𝜷.
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Syed Shahabudeen-Kerala-India
𝛀 = ∫ (𝐬𝐢𝐧 𝝅𝒙
𝒙𝟑−𝐜𝐨𝐬 𝝅𝒙
𝒙𝟐−𝐬𝐢𝐧𝐡𝝅𝒙
𝒙𝟑+𝐜𝐨𝐬𝐡𝝅𝒙
𝒙𝟐)𝒆−𝝅𝒙
√𝒙
∞
𝟎
𝒅𝒙 =
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8 RMM-CALCULUS MARATHON 1501-1600
= ∫𝒆−𝝅𝒙 𝐬𝐢𝐧𝝅𝒙
𝒙𝟕𝟐
∞
𝟎
𝒅𝒙 +∫𝒆−𝝅𝒙 𝐜𝐨𝐬𝝅𝒙
𝒙𝟓𝟐
∞
𝟎
𝒅𝒙 −∫𝒆−𝝅𝒙 𝐬𝐢𝐧𝐡𝝅𝒙
𝒙𝟕𝟐
∞
𝟎
𝒅𝒙 +∫𝒆−𝝅𝒙 𝐜𝐨𝐬𝐡𝝅𝒙
𝒙𝟓𝟐
∞
𝟎
𝒅𝒙
= 𝑨 −𝑩− 𝑪+ 𝑫.
𝑨 = ∫𝒆−𝝅𝒙 𝐬𝐢𝐧𝝅𝒙
𝒙𝟕𝟐
∞
𝟎
𝒅𝒙 = 𝑰𝒎∫𝒆𝒊𝝅𝒙𝒆−𝒊𝝅𝒙
𝒙𝟕𝟐
∞
𝟎
𝒅𝒙 = 𝑰𝒎(𝓛 {𝒆𝒊𝝅𝒙𝒙−𝟕𝟐}) = 𝑰𝒎(
𝚪 (−𝟓𝟐)
(𝝅 − 𝒊𝝅)−𝟓𝟐
)
= −𝟖𝝅𝟑
𝟏𝟓𝑰𝒎(
𝟏
(𝟏 − 𝒊)−𝟓𝟐
) = −𝟖𝝅𝟑
𝟏𝟓𝑰𝒎(
(𝟏 + 𝒊)−𝟓𝟐
𝟐−𝟓𝟐
) = −𝟖𝝅𝟑
𝟏𝟓𝑰𝒎
(
(√𝟐𝒆
𝒊𝝅𝟒 )−𝟓𝟐
𝟐−𝟓𝟐
)
=
=𝟖𝝅𝟑𝟐
𝟓𝟒
𝟏𝟓𝐬𝐢𝐧𝟓𝝅
𝟖
𝑩 = ∫𝒆−𝝅𝒙 𝐜𝐨𝐬 𝝅𝒙
𝒙𝟓𝟐
∞
𝟎
𝒅𝒙 = 𝑹𝒆∫𝒆𝒊𝝅𝒙𝒆−𝒊𝝅𝒙
𝒙𝟓𝟐
∞
𝟎
𝒅𝒙 = 𝑹𝒆 (𝓛 {𝒆𝒊𝝅𝒙𝒙−𝟓𝟐}) = 𝑹𝒆(
𝚪 (−𝟑𝟐)
(𝝅 − 𝒊𝝅)−𝟑𝟐
)
=𝟒𝝅𝟐
𝟑𝑹𝒆
(
(√𝟐𝒆
𝒊𝝅𝟒 )
−𝟑𝟐
𝟐−𝟑𝟐
)
=𝟒𝝅𝟐𝟐
𝟑𝟒
𝟑𝐜𝐨𝐬
𝟑𝝅
𝟖
𝑪 = ∫𝒆−𝝅𝒙 𝐬𝐢𝐧𝐡𝝅𝒙
𝒙𝟕𝟐
∞
𝟎
𝒅𝒙 =𝟏
𝟐(𝓛 {𝒆𝝅𝒙𝒙−
𝟕𝟐} − 𝓛 {𝒆−𝝅𝒙𝒙−
𝟕𝟐}) =
=𝟏
𝟐𝐥𝐢𝐦𝒔→𝝅
(𝚪 (−
𝟓𝟐)
(𝒔 − 𝝅)−𝟓𝟐
−𝚪(−
𝟓𝟐)
(𝒔 + 𝝅)−𝟓𝟐
) =−𝚪(−
𝟓𝟐)
𝟐(𝟐𝝅)−𝟓𝟐
=𝟏𝟔𝝅𝟑√𝟐
𝟏𝟓
𝑫 = ∫𝒆−𝝅𝒙 𝐜𝐨𝐬𝐡𝝅𝒙
𝒙𝟓𝟐
∞
𝟎
𝒅𝒙 =𝟏
𝟐(𝓛{𝒆𝝅𝒙𝒙−
𝟓𝟐} + 𝓛 {𝒆−𝝅𝒙𝒙−
𝟓𝟐}) =
=𝟏
𝟐𝐥𝐢𝐦𝒔→𝝅
(𝚪 (−
𝟑𝟐)
(𝒔 − 𝝅)−𝟑𝟐
+𝚪 (−
𝟑𝟐)
(𝒔 + 𝝅)−𝟑𝟐
) =𝚪(−
𝟑𝟐)
𝟐(𝟐𝝅)−𝟑𝟐
=𝟒𝝅𝟐√𝟐
𝟑
𝛀 =𝟖𝝅𝟑𝟐
𝟓𝟒
𝟏𝟓𝐬𝐢𝐧𝟓𝝅
𝟖−𝟒𝝅𝟐𝟐
𝟑𝟒
𝟑𝐜𝐨𝐬
𝟑𝝅
𝟖−𝟏𝟔𝝅𝟑√𝟐
𝟏𝟓+𝟒𝝅𝟐√𝟐
𝟑=
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9 RMM-CALCULUS MARATHON 1501-1600
= 𝝅𝟐 (𝟒√𝟐
𝟑−𝟐𝟏𝟏𝟒
𝟑𝐜𝐨𝐬
𝟑𝝅
𝟖) + 𝝅𝟑 (
𝟐𝟏𝟕𝟒
𝟏𝟓𝐬𝐢𝐧𝟓𝝅
𝟖−𝟏𝟔√𝟐
𝟏𝟓)
Therefore, 𝜶 =𝟒√𝟐
𝟑−𝟐𝟏𝟏𝟒
𝟑𝐜𝐨𝐬
𝟑𝝅
𝟖, 𝜷 =
𝟐𝟏𝟕𝟒
𝟏𝟓𝐬𝐢𝐧
𝟓𝝅
𝟖−𝟏𝟔√𝟐
𝟏𝟓
1505. Evaluate the integral in a closed-form
𝛀 = ∫ (𝐭𝐚𝐧𝟐 𝒙
𝐜𝐨𝐬𝟑𝒙𝟐
+𝟗√𝟐 𝐜𝐨𝐬
𝟑𝒙𝟒
𝟐 𝐬𝐢𝐧𝟑𝒙𝟒+ 𝟏
+𝟏𝟔 𝐬𝐢𝐧 𝒙
𝟒 𝐜𝐨𝐬 𝒙 + 𝟏)
𝝅𝟑
𝟎
𝒅𝒙
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Rana Ranino-Setif-Algerie
𝛀 = ∫ (𝐭𝐚𝐧𝟐 𝒙
𝐜𝐨𝐬𝟑𝒙𝟐
+𝟗√𝟐𝐜𝐨𝐬
𝟑𝒙𝟒
𝟐 𝐬𝐢𝐧𝟑𝒙𝟒 + 𝟏
+𝟏𝟔𝐬𝐢𝐧𝒙
𝟒 𝐜𝐨𝐬 𝒙 + 𝟏)
𝝅𝟑
𝟎
𝒅𝒙 =
= ∫𝐭𝐚𝐧𝟐 𝒙
𝐜𝐨𝐬𝟑𝒙𝟐
𝝅𝟑
𝟎
𝒅𝒙 + ∫𝟗√𝟐𝐜𝐨𝐬
𝟑𝒙𝟒
𝟐 𝐬𝐢𝐧𝟑𝒙𝟒 + 𝟏
𝝅𝟑
𝟎
𝒅𝒙 +∫𝟏𝟔𝐬𝐢𝐧 𝒙
𝟒 𝐜𝐨𝐬 𝒙 + 𝟏
𝝅𝟑
𝟎
𝒅𝒙 = 𝑨 +𝑩 + 𝑪
𝑨 = ∫𝐭𝐚𝐧𝟐 𝒙
𝐜𝐨𝐬𝟑𝒙𝟐
𝝅𝟑
𝟎
𝒅𝒙 = 𝟐∫𝐭𝐚𝐧𝟐 𝟐𝒙
𝐜𝐨𝐬𝟑 𝒙
𝝅𝟔
𝟎
𝒅𝒙 = 𝟖∫𝐭𝐚𝐧𝟐 𝒂𝐬𝐞𝐜𝟑 𝒙
(𝟏 − 𝐭𝐚𝐧𝟐 𝒙)𝟐𝐜𝐨𝐭𝟒 𝒙
𝐜𝐨𝐭𝟒 𝒙
𝝅𝟔
𝟎
𝒅𝒙 =
= 𝟖∫𝐜𝐬𝐜𝟐 𝒙 𝐬𝐞𝐜𝒙
(𝐜𝐨𝐭𝟐 𝒙 − 𝟏)𝟐
𝝅𝟔
𝟎
𝒅𝒙 =𝒕=𝐜𝐬𝐜 𝒙
𝟖∫𝒕𝟐
(𝒕𝟐 − 𝟐)(𝒕𝟐 − 𝟏)
∞
𝟎
𝒅𝒕 =
= ∫ (𝟑√𝟐
𝒕 + √𝟐−𝟑√𝟐
𝒕 − √𝟐+
𝟒
𝒕 − 𝟏−
𝟒
𝒕 + 𝟏+
𝟐
(𝒕 + √𝟐)𝟐 +
𝟐
(𝒕 − √𝟐)𝟐)
∞
𝟐
𝒅𝒕
𝑨 = [𝟑√𝟐 𝐥𝐨𝐠 (𝒕 + √𝟐
𝒕 − √𝟐) + 𝟒 𝐥𝐨𝐠 (
𝒕 − 𝟏
𝒕 + 𝟏) −
𝟒𝒕
𝒕𝟐 − 𝟐]𝟐
∞
=
= 𝟒 + 𝟒 𝐥𝐨𝐠𝟑 − 𝟔√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)
𝑩 = ∫𝟗√𝟐𝐜𝐨𝐬
𝟑𝒙𝟒
𝟐 𝐬𝐢𝐧𝟑𝒙𝟒 + 𝟏
𝝅𝟑
𝟎
𝒅𝒙 =𝒕=𝟐 𝐬𝐢𝐧
𝟑𝒙𝟒𝟔√𝟐∫
𝒅𝒕
𝒕
𝟏+√𝟐
𝟏
= 𝟔√𝟐 𝐥𝐨𝐠(𝟏 + √𝟐)
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10 RMM-CALCULUS MARATHON 1501-1600
𝑪 = ∫𝟏𝟔𝐬𝐢𝐧𝒙
𝟒 𝐜𝐨𝐬 𝒙 + 𝟏
𝝅𝟑
𝟎
𝒅𝒙 =𝒕=𝟒 𝐜𝐨𝐬 𝒙+𝟏
𝟒∫𝒅𝒕
𝒕
𝟓
𝟑
= 𝟒 𝐥𝐨𝐠 𝟓 − 𝟒 𝐥𝐨𝐠𝟑
Therefore,
𝛀 = ∫ (𝐭𝐚𝐧𝟐 𝒙
𝐜𝐨𝐬𝟑𝒙𝟐
+𝟗√𝟐𝐜𝐨𝐬
𝟑𝒙𝟒
𝟐 𝐬𝐢𝐧𝟑𝒙𝟒 + 𝟏
+𝟏𝟔𝐬𝐢𝐧 𝒙
𝟒 𝐜𝐨𝐬 𝒙 + 𝟏)
𝝅𝟑
𝟎
𝒅𝒙 = 𝟒(𝟏 + 𝐥𝐨𝐠 𝟓)
1506. Find:
𝛀 = ∫𝐥𝐨𝐠𝟐 𝒙 𝐥𝐨𝐠 (
(𝟏 + 𝒙)𝟐
𝟒𝒙)
𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 − 𝟐 𝐥𝐨𝐠(𝟐)∫𝐥𝐨𝐠𝟐(𝒙)
𝟏 + 𝒙
𝟏
𝟎
𝒅𝒙 =𝝅𝟒
𝟕𝟐
without using harmonic series, beta function.
Proposed by Sujeethan Balendran-SriLanka
Solution by Cornel Ioan Vălean-Romania
It is known that:
(𝟏): (−𝟏)𝒎
(𝒎 − 𝟏)!∫𝐥𝐨𝐠𝒎−𝟏(𝒙) 𝐥𝐨𝐠 (
𝟏 + 𝒙𝟐 )
𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 =
=𝟏
𝟐(𝒎𝜻(𝒎+ 𝟏) − 𝟐 𝐥𝐨𝐠(𝟐) (𝟏 − 𝟐𝟏−𝒎)𝜻(𝒎)− ∑(𝟏− 𝟐−𝒌)(𝟏− 𝟐𝟏+𝒌−𝒎)𝜻(𝒌 + 𝟏)𝜻(𝒎− 𝒌))
𝒎−𝟐
𝒌=𝟏
Based on (𝟏), we get that:
∫𝐥𝐨𝐠𝟐 𝒙 𝐥𝐨𝐠 (
𝟏 + 𝒙𝟐 )
𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 =𝟑
𝟐𝐥𝐨𝐠(𝟐) 𝜻(𝟑) −
𝟏𝟗
𝟕𝟐𝟎𝝅𝟒,
which we need in our calculations below.
We also need that:
∫𝐥𝐨𝐠𝟐 𝒙
𝟏 + 𝒙
𝟏
𝟎
𝒅𝒙 = ∑(−𝟏)𝒏−𝟏∫ 𝒙𝒏−𝟏𝟏
𝟎
𝐥𝐨𝐠𝟐(𝒙)
∞
𝒏=𝟏
𝒅𝒙 = 𝟐∑(−𝟏)𝒏−𝟏𝟏
𝒏𝟑
∞
𝒏=𝟏
=𝟑
𝟐𝜻(𝟑) 𝐚𝐧𝐝
∫𝐥𝐨𝐠𝟑(𝒙)
𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 = ∑∫ 𝒙𝒏−𝟏 𝐥𝐨𝐠𝟑(𝒙)𝒅𝒙𝟏
𝟎
∞
𝒏=𝟏
= −𝟔∑𝟏
𝒏𝟒
∞
𝒏=𝟏
= −𝟔𝜻(𝟒).
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11 RMM-CALCULUS MARATHON 1501-1600
Returning to the main result, cleverly splitting and using the auxiliary results above, we
get:
𝛀 = 𝟐∫𝐥𝐨𝐠𝟐(𝒙) 𝐥𝐨𝐠 (
𝟏 + 𝒙𝟐 )
𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 −∫𝐥𝐨𝐠𝟑(𝒙)
𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 − 𝟐 𝐥𝐨𝐠(𝟐)∫𝐥𝐨𝐠𝟐(𝒙)
𝟏 + 𝒙
𝟏
𝟎
𝒅𝒙 =𝝅𝟒
𝟕𝟐
1507. Prove that:
∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐
𝟏
𝟎
(−𝒙 +𝒙𝟐
𝟐𝟐−𝒙𝟑
𝟑𝟐+⋯)𝒅𝒙 =
𝑮𝜻(𝟐)
𝟖
where 𝑮 − is Catalan’s constant.
Proposed by Narendra Bhandari-Bajura-Nepal
Solution 1 by Ngulmun George Baite-India
𝑰 = ∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐
𝟏
𝟎
(−𝒙 +𝒙𝟐
𝟐𝟐−𝒙𝟑
𝟑𝟐+⋯)𝒅𝒙
𝐋𝐞𝐭 𝛇 = −𝒙 +𝒙𝟐
𝟐𝟐−𝒙𝟑
𝟑𝟐+⋯ =∑(−𝟏)𝒏
𝒙𝒏
𝒏𝟐
∞
𝒏=𝟏
= ∑(−𝒙)𝒏
𝒏𝟐
∞
𝒏=𝟏
⇒ 𝜻 = 𝑳𝒊𝟐(−𝒙)
𝑰 = ∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐𝑳𝒊𝟐(−𝒙)
𝟏
𝟎
𝒅𝒙
∵ 𝑳𝒊𝟐(−𝟐) = ∫𝟐 𝐥𝐨𝐠 𝒕
𝟏 + 𝒕𝟐
𝟏
𝟎
𝒅𝒕
𝑰 = ∫ ∫𝒙 𝐥𝐨𝐠𝒙 𝐥𝐨𝐠 𝒚
(𝟏 + 𝒙𝟐)(𝟏 + 𝒙𝒚)𝒅𝒚
𝟏
𝟎
𝒅𝒙𝟏
𝟎
=
=𝟏
𝟐[∫ (∫
𝒙 𝐥𝐨𝐠𝒙 𝐥𝐨𝐠𝒚
(𝟏 + 𝒙𝟐)(𝟏 + 𝒙𝒚)𝒅𝒚
𝟏
𝟎
)𝒅𝒙𝟏
𝟎
+∫ (∫𝒚 𝐥𝐨𝐠 𝒙 𝐥𝐨𝐠𝒚
(𝟏 + 𝒚𝟐)(𝟏 + 𝒙𝒚)𝒅𝒚
𝟏
𝟎
)𝒅𝒙𝟏
𝟎
] =
=𝟏
𝟐∫ ∫ (
𝒙
(𝟏 + 𝒙𝟐)(𝟏 + 𝒙𝒚)+
𝒚
(𝟏 + 𝒚𝟐)(𝟏 + 𝒙𝒚)) 𝐥𝐨𝐠 𝒙 𝐥𝐨𝐠 𝒚𝒅𝒚
𝟏
𝟎
𝒅𝒙𝟏
𝟎
=
=𝟏
𝟐∫ ∫
𝒙 + 𝒚 + 𝒙𝟐𝒚 + 𝒙𝒚𝟐
(𝟏 + 𝒙𝟐)(𝟏 + 𝒚𝟐)(𝟏 + 𝒙𝒚)𝐥𝐨𝐠𝒙 𝐥𝐨𝐠 𝒚𝒅𝒚
𝟏
𝟎
𝟏
𝟎
𝒅𝒙
=𝟏
𝟐∫ ∫
(𝒙 + 𝒚)(𝟏 + 𝒙𝒚)
(𝟏 + 𝒙𝟐)(𝟏 + 𝒚𝟐)(𝟏 + 𝒙𝒚)
𝟏
𝟎
𝐥𝐨𝐠 𝒙 𝐥𝐨𝐠𝒚 𝒅𝒚𝟏
𝟎
𝒅𝒙 =
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12 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝟐∫ ∫
(𝒙 + 𝒚)
(𝟏 + 𝒙𝟐)(𝟏 + 𝒚𝟐)𝐥𝐨𝐠 𝒙 𝐥𝐨𝐠𝒚
𝟏
𝟎
𝒅𝒚𝟏
𝟎
𝒅𝒙 = ∫𝒙 𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
∫𝐥𝐨𝐠 𝒚
𝟏 + 𝒚𝟐𝒅𝒚
𝟏
𝟎
=
= (∫ ∑(−𝟏)𝒌−𝟏𝒙𝟐𝒌−𝟏 𝐥𝐨𝐠 𝒙𝒅𝒙
∞
𝒌=𝟏
𝟏
𝟎
)(∫ ∑(−𝟏)𝒌−𝟏𝒚𝟐𝒌−𝟐 𝐥𝐨𝐠 𝒚𝒅𝒚
∞
𝒌=𝟏
=𝟏
𝟎
)
= (∑(−𝟏)𝒌−𝟏∫ 𝒙𝟐𝒌−𝟏 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
∞
𝒌=𝟏
)(∑(−𝟏)𝒌−𝟏∫ 𝒚𝟐𝒌−𝟐 𝐥𝐨𝐠 𝒚𝒅𝒚𝟏
𝟎
∞
𝒌=𝟏
) =
= (∑(−𝟏)𝒌−𝟏 [𝒙𝟐𝒌
𝟐𝒌]𝟎
𝟏
− ∫𝒙𝟐𝒌
𝟐𝒌
𝟏
𝒙𝒅𝒙
𝟏
𝟎
∞
𝒌=𝟏
)(∑(−𝟏)𝒌−𝟏 [𝒚𝟐𝒌−𝟏
𝟐𝒌 − 𝟏]𝟎
𝟏
−∫𝒚𝟐𝒌−𝟏
𝟐𝒌 − 𝟏
𝟏
𝒚𝒅𝒚
𝟏
𝟎
∞
𝒌=𝟏
) =
= (∑(−𝟏)𝒌−𝟏 ⋅−𝟏
(𝟐𝒌)𝟐
∞
𝒌=𝟏
)(∑(−𝟏)𝒌−𝟏 ⋅−𝟏
(𝟐𝒌 − 𝟏)𝟐
∞
𝒌=𝟏
) =
= (𝟏
𝟒𝜼(𝟐))𝑮 =
𝟏
𝟒𝜼(𝟐)𝑮 =
𝟏
𝟒⋅𝜻(𝟐)
𝟐⋅ 𝑮 =
𝟏
𝟖𝜻(𝟐)𝑮
Therefore,
∫𝐥𝐨𝐠𝒙
𝟏 + 𝒙𝟐
𝟏
𝟎
(−𝒙 +𝒙𝟐
𝟐𝟐−𝒙𝟑
𝟑𝟐+⋯)𝒅𝒙 =
𝑮𝜻(𝟐)
𝟖
Solution 2 by Mohammad Rostami-Afghanistan
𝛀 = ∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐∑(−𝒙)𝒏
𝒏𝟐
∞
𝒏=𝟏
𝒅𝒙𝟏
𝟎
= ∫(−𝟏)𝒏𝒙𝒏
𝒏𝟐∑(−𝒙𝟐)𝒌
𝝏
𝝏𝒂|𝒂=𝟎
∞
𝒌=𝟎
𝟏
𝟎
𝒙𝒂𝒅𝒙 =
= ∑(−𝟏)𝒏
𝒏𝟐
∞
𝒏=𝟏
∑(−𝟏)𝒌 𝝏
𝝏𝒂|𝒂=𝟎
∞
𝒌=𝟎
∫ 𝒙𝒏+𝟐𝒌+𝒂𝟏
𝟎
𝒅𝒙 =
= ∑(−𝟏)𝒏
𝒏𝟐∑(−𝟏)𝒌 [
𝟏
𝒏 + 𝟐𝒌 + 𝒂 + 𝒂]𝒂=𝟎
′∞
𝒌=𝟎
=
∞
𝒏=𝟏
= ∑(−𝟏)𝒏
𝒏𝟐∑(−𝟏)𝒌 ⋅
−𝟏
(𝟐𝒌 + 𝒏 + 𝟏)𝟐
∞
𝒌=𝟎
∞
𝒏=𝟏
; {𝒌 → 𝒌 −𝒏
𝟐}
𝛀 = ∑(−𝟏)𝒏
𝒏𝟐∑
(−𝟏)𝒌−𝒏𝟐
(−𝟏) [𝟐 (𝒌 −𝒏𝟐) + 𝒏 + 𝟏
]𝟐
∞+𝒏𝟐
𝒌=𝒏𝟐
∞
𝒏=𝟏
=
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13 RMM-CALCULUS MARATHON 1501-1600
= ∑(−𝟏)
𝒏𝟐−𝟏
𝒏𝟐
∞
𝒏=𝟏
∑(−𝟏)𝒌
(𝟐𝒌 + 𝟏)𝟐
∞
𝒌=𝒏𝟐
{𝒌 =𝒏
𝟐= 𝒕 ∈ 𝑷 = {𝟎, 𝟏, 𝟐, 𝟑, 𝟒… } ⇒ 𝒏 = 𝟐𝒕}
𝛀 =∑(−𝟏)𝒕−𝟏
𝟒𝒕𝟐
∞𝟐
𝒕=𝟏𝟐
∑(−𝟏)𝒕
(𝟐𝒕 + 𝟏)𝟐
∞
𝒕=𝟎
𝒕∈𝑷,𝒕≥𝟏⇒ 𝛀 =
𝟏
𝟒∑(−𝟏)𝒕−𝟏
𝒕𝟐⋅ 𝑮
∞
𝒕=𝟏
⇒
𝛀 =𝟏
𝟒⋅ 𝜼(𝟐) ⋅ 𝑮 =
𝟏
𝟒(𝟏 − 𝟐𝟏−𝟐)𝜻(𝟐)𝑮
Therefore,
∫𝐥𝐨𝐠𝒙
𝟏 + 𝒙𝟐
𝟏
𝟎
(−𝒙 +𝒙𝟐
𝟐𝟐−𝒙𝟑
𝟑𝟐+⋯)𝒅𝒙 =
𝑮𝜻(𝟐)
𝟖
1508.
𝑰𝒏 = ∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠(𝚪(𝒙))𝟏
𝟎
𝒅𝒙, 𝑷𝒏 = ∫𝒙𝒏
𝟏 + 𝒙𝒅𝒙
𝟏
𝟎
Prove:
𝟐√𝝅 ⋅ 𝑰𝒏 ⋅ 𝚪 (𝒏 + 𝟐
𝟐) = 𝚪 (
𝒏 + 𝟏
𝟐) (𝐥𝐨𝐠𝝅 + 𝑷𝒏)
Proposed by Asmat Qatea-Afghanistan
Solution by Rana Ranino-Setif-Algerie
𝑰𝒏 = ∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠(𝚪(𝒙))𝟏
𝟎
𝒅𝒙 =𝒙→𝟏−𝒙
∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠 𝚪(𝟏 − 𝒙)𝟏
𝟎
𝒅𝒙 =
=𝟏
𝟐∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠(𝚪(𝒙)𝚪(𝟏 − 𝒙))𝟏
𝟎
𝒅𝒙 =
=𝟏
𝟐𝐥𝐨𝐠𝝅∫ 𝐬𝐢𝐧𝒏(𝝅𝒙)𝒅𝒙
𝟏
𝟎
−𝟏
𝟐∫ 𝐬𝐢𝐧𝒏(𝝅𝒙) 𝐥𝐨𝐠(𝐬𝐢𝐧(𝝅𝒙))𝟏
𝟎
𝒅𝒙 =𝒕=𝝅𝒙 𝐥𝐨𝐠 𝝅
𝟐𝝅∫ 𝐬𝐢𝐧𝒏 𝒕 𝒅𝒕𝝅
𝟎
−
−𝟏
𝟐𝝅∫ 𝐬𝐢𝐧𝒏 𝒕 𝐥𝐨𝐠(𝐬𝐢𝐧 𝒕)𝝅
𝟎
=𝐥𝐨𝐠𝝅
𝝅∫ 𝐬𝐢𝐧𝒏 𝒕 𝒅𝒕
𝝅𝟐
𝟎
−𝟏
𝝅∫ 𝐬𝐢𝐧𝒏 𝒕
𝝅𝟐
𝟎
𝐥𝐨𝐠(𝐬𝐢𝐧 𝒕) 𝒅𝒕 =
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14 RMM-CALCULUS MARATHON 1501-1600
=𝐥𝐨𝐠𝝅
𝝅∫ 𝐬𝐢𝐧𝒏 𝒕 𝒅𝒕
𝝅𝟐
𝟎
−𝟏
𝝅
𝝏
𝝏𝒔|𝒔=𝟎
∫ 𝐬𝐢𝐧𝒏+𝒔 𝒕 𝒅𝒕
𝝅𝟐
𝟎
∫ 𝐬𝐢𝐧𝒏 𝒕 𝒅𝒕
𝝅𝟐
𝟎
=𝟏
𝟐𝜷 (𝒏 + 𝟏
𝟐,𝟏
𝟐) =
𝚪(𝒏 + 𝟏𝟐 ) 𝚪(
𝟏𝟐)
𝚪(𝒏 + 𝟐𝟐 )
= √𝝅𝚪(𝒏 + 𝟏𝟐 )
𝚪(𝒏 + 𝟐𝟐 )
𝑰𝒏 =𝐥𝐨𝐠𝝅
𝟐√𝝅
𝚪(𝒏 + 𝟏𝟐 )
𝚪(𝒏 + 𝟐𝟐)−
𝟏
𝟐√𝝅
𝝏
𝝏𝒔|𝒔=𝟎
𝚪(𝒏 + 𝒔 + 𝟏
𝟐 )
𝚪(𝒏 + 𝒔 + 𝟐
𝟐)=
=𝐥𝐨𝐠𝝅
𝟐√𝝅
𝚪 (𝒏 + 𝟏𝟐 )
𝚪 (𝒏 + 𝟐𝟐 )
−𝟏
𝟒√𝝅
𝐥𝐨𝐠𝝅
𝟐√𝝅
𝚪 (𝒏 + 𝟏𝟐 )
𝚪 (𝒏 + 𝟐𝟐 )
{𝝍(𝒏 + 𝟏
𝟐) −𝝍(
𝒏 + 𝟐
𝟐)}
𝟐√𝝅𝚪(𝒏 + 𝟐
𝟐) 𝑰𝒏 = 𝚪(
𝒏 + 𝟏
𝟐) {𝐥𝐨𝐠𝝅 +
𝟏
𝟐(𝝍(
𝒏 + 𝟐
𝟐) − 𝝍(
𝒏 + 𝟏
𝟐))
Therefore,
𝟐√𝝅 ⋅ 𝑰𝒏 ⋅ 𝚪 (𝒏 + 𝟐
𝟐) = 𝚪 (
𝒏 + 𝟏
𝟐) (𝐥𝐨𝐠𝝅 + 𝑷𝒏)
1509. Find:
𝛀 = ∫ 𝒙 𝐜𝐨𝐭 𝒙 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)
𝝅𝟐
𝟎
𝒅𝒙
Proposed by Sujeethan Balendran-SriLanka
Solution by Cornel Ioan Vălean-Romania
∵ ∑(∫ 𝒕𝟐𝒏−𝟏𝟏 − 𝒕
𝟏 + 𝒕𝒅𝒕
𝟏
𝟎
)𝐬𝐢𝐧𝟐(𝟐𝒏𝒙)
𝒏
∞
𝒏=𝟏
= 𝐥𝐨𝐠(𝐬𝐢𝐧𝒙) 𝐥𝐨𝐠(𝐜𝐨𝐬 𝒙) , 𝟎 < 𝑥 <𝝅
𝟐
We also have the trivial results,
𝒂𝒏 = ∫ 𝒙 𝐭𝐚𝐧𝒙 𝐬𝐢𝐧𝟐(𝟐𝒏𝒙)𝒅𝒙
𝝅𝟐
𝟎
=𝝅
𝟒𝑯𝟐𝒏 −
𝝅
𝟏𝟔⋅𝟏
𝒏 𝐚𝐧𝐝
𝒃𝒏 = ∫ 𝐥𝐨𝐠(𝐜𝐨𝐬 𝒙) 𝐬𝐢𝐧𝟐(𝟐𝒏𝒙)𝒅𝒙
𝝅𝟐
𝟎
=𝝅
𝟏𝟔⋅𝟏
𝒏−𝝅
𝟒𝐥𝐨𝐠 𝟐
where both results are easily derived by exploiting the differences 𝒂𝒏+𝟏 − 𝒂𝒏 and
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15 RMM-CALCULUS MARATHON 1501-1600
𝒃𝒏+𝟏 − 𝒃𝒏 or using Fourrier series.
Returning to the main integrals where we use integration by parts and then exploit the
auxiliary results above, we have:
𝑰 = 𝟐∫ 𝒙 𝐥𝐨𝐠(𝐬𝐢𝐧 𝒙) 𝐥𝐨𝐠(𝐜𝐨𝐬 𝒙) 𝐭𝐚𝐧 𝒙𝒅𝒙
𝝅𝟐
𝟎
−∫ 𝐥𝐨𝐠(𝐬𝐢𝐧 𝒙) 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)
𝝅𝟐
𝟎
𝒅𝒙 =
= 𝟐∑𝟏
𝒏(∫ 𝒕𝟐𝒏−𝟏
𝟏 − 𝒕
𝟏 + 𝒕𝒅𝒕
𝟏
𝟎
)𝒂𝒏
∞
𝒏=𝟏
−∑𝟏
𝒏(∫ 𝒕𝟐𝒏−𝟏
𝟏 − 𝒕
𝟏 + 𝒕𝒅𝒕
𝟏
𝟎
)𝒃𝒏
∞
𝒏=𝟏
=
= −𝝅𝟑
𝟒𝟖𝐥𝐨𝐠 𝟐 −
𝝅
𝟒𝐥𝐨𝐠 𝟐∫
𝐥𝐨𝐠(𝟏 − 𝒕𝟐)
𝒕𝒅𝒕
𝟏
𝟎⏟ −𝝅𝟐/𝟏𝟐
+𝝅
𝟒∫𝐥𝐨𝐠𝟐(𝟏 − 𝒕)
𝒕𝒅𝒕
𝟏
𝟎⏟ 𝟐𝜻(𝟑)
+𝝅
𝟏𝟔∫𝑳𝒊𝟐(𝒕
𝟐)
𝒕𝒅𝒕
𝟏
𝟎⏟ 𝜻(𝟑)/𝟐
+
+𝝅
𝟐𝐥𝐨𝐠 𝟐∫
𝐥𝐨𝐠(𝟏 − 𝒕𝟐)
𝟏 + 𝒕𝒅𝒕
𝟏
𝟎⏟
𝐥𝐨𝐠𝟐 𝟐−𝝅𝟐
𝟏𝟐
−𝝅
𝟐∫𝐥𝐨𝐠𝟐(𝟏 + 𝒕)
𝟏 + 𝒕𝒅𝒕
𝟏
𝟎⏟ 𝟏
𝟑 𝐥𝐨𝐠𝟑 𝟐
−𝝅
𝟐∫𝐥𝐨𝐠𝟐(𝟏 − 𝒕)
𝟏 + 𝒕𝒅𝒕
𝟏
𝟎⏟
𝟐𝑳𝒊𝟑(𝟏𝟐)
−
−𝝅
𝟒∫𝐥𝐨𝐠(𝟏 − 𝒕) 𝐥𝐨𝐠(𝟏 + 𝒕)
𝒕𝒅𝒕
𝟏
𝟎⏟ −𝟓/𝟖𝜻(𝟑)
=𝝅𝟑
𝟐𝟒𝐥𝐨𝐠 𝟐 +
𝝅
𝟔𝐥𝐨𝐠𝟑 𝟐 −
𝟑
𝟏𝟔𝝅𝜻(𝟑)
1510. For 𝒏 > 1, we have:
∫ ∫𝒙 𝐬𝐢𝐧 𝒕𝒙
𝐜𝐨𝐬𝐡𝝅𝒙𝒏
∞
𝟎
𝒅𝒕∞
𝟎
𝒅𝒙 =𝒏
𝟐
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Ngulmun George Baite-India
𝑰 = ∫ ∫𝒙𝐬𝐢𝐧 𝒕𝒙
𝐜𝐨𝐬𝐡𝝅𝒙𝒏
∞
𝟎
𝒅𝒕∞
𝟎
𝒅𝒙 =𝐜𝐨𝐬𝐡 𝒙=
𝟏𝟐(𝒆𝒙+𝒆−𝒙)
∫ ∫𝒙 𝐬𝐢𝐧 𝒕𝒙
𝟏𝟐 (𝒆
𝝅𝒙𝒏 + 𝒆−
𝝅𝒙𝒏 )
∞
𝟎
𝒅𝒕∞
𝟎
𝒅𝒙 =
= 𝟐∫ ∫𝒙𝐬𝐢𝐧 𝒕𝒙
𝒆𝝅𝒙𝒏 + 𝒆−
𝝅𝒙𝒏
∞
𝟎
∞
𝟎
𝒅𝒕𝒅𝒙 = 𝟐∫ ∫𝒙𝐬𝐢𝐧 𝒕𝒙
𝒆𝝅𝒙𝒏 (𝟏 + 𝒆−
𝟐𝝅𝒙𝒏 )
∞
𝟎
𝒅𝒕∞
𝟎
𝒅𝒙 =
= 𝟐∫ ∫𝒆−𝝅𝒙𝒏 𝒙 𝐬𝐢𝐧 𝒕𝒙
𝟏 + 𝒆−𝟐𝝅𝒙𝒏
∞
𝟎
𝒅𝒕∞
𝟎
𝒅𝒙 = 𝟐∑(−𝟏)𝒌∫ [∫ 𝒙𝒆−𝝅𝒏(𝟐𝒌+𝟏)𝒙 𝐬𝐢𝐧 𝒕𝒙𝒅𝒙
∞
𝟎
] 𝒅𝒕∞
𝟎
∞
𝒌=𝟎
=
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16 RMM-CALCULUS MARATHON 1501-1600
𝐍𝐨𝐰, 𝓛{𝐬𝐢𝐧𝝎𝒙} = ∫ 𝒆−𝒔𝒙 𝐬𝐢𝐧𝝎𝒙∞
𝟎
𝒅𝒙 =𝝎
𝝎𝟐 + 𝒔𝟐
Differentiating respect to 𝒔, we have:
−∫ 𝒙𝒆−𝒔𝒙 𝐬𝐢𝐧𝝎𝒙∞
𝟎
𝒅𝒙 =−𝟐𝒔𝝎
(𝒔𝟐 + 𝝎𝟐)𝟐⇒ ∫ 𝒙𝒆−𝒔𝒙 𝐬𝐢𝐧𝝎𝒙
∞
𝟎
𝒅𝒙 =𝟐𝒔𝝎
(𝒔𝟐 +𝝎𝟐)𝟐
Put 𝒔 =𝝅
𝒏(𝟐𝒌 + 𝟏) and 𝝎 = 𝒕 ⇒
∫ 𝒙𝒆−𝝅𝒏(𝟐𝒌+𝟏)𝒙
∞
𝟎
𝐬𝐢𝐧 𝒕𝒙𝒅𝒙 =
𝟐𝝅𝒏(𝟐𝒌 + 𝟏)𝒕
(𝝅𝟐
𝒏𝟐(𝟐𝒌 + 𝟏)𝟐 + 𝒕𝟐)
𝟐
𝑰 = 𝟐∑(−𝟏)𝒌∫
𝟐𝝅𝒏(𝟐𝒌 + 𝟏)𝒕
(𝝅𝟐
𝒏𝟐(𝟐𝒌 + 𝟏)𝟐 + 𝒕𝟐)
𝟐
∞
𝟎
𝒅𝒕
∞
𝒌=𝟎
=
=𝟒𝝅
𝒏∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)
∞
𝒌=𝟎
∫𝒕
(𝝅𝟐
𝒏𝟐(𝟐𝒌 + 𝟏)𝟐+ 𝒕𝟐)
𝟐
∞
𝟎
𝒅𝒕 =𝒕=𝝅𝒙𝒏
=𝟒𝝅
𝒏∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)
∞
𝒌=𝟎
∫
𝝅𝒙𝒏
𝝅𝟒
𝒏𝟒((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐
∞
𝟎
⋅𝝅
𝒏𝒅𝒙 =
=𝟒𝝅
𝒏⋅𝒏𝟐
𝝅𝟐∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)∫
𝒙
((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐
∞
𝟎
𝒅𝒙
∞
𝒌=𝟎
=
=𝟒𝒏
𝝅∫ ∑(−𝟏)𝒌 ⋅
𝒙(𝟐𝒌 + 𝟏)
((𝟐𝒌 + 𝟏) + 𝒙𝟐)𝟐 𝒅𝒙
∞
𝒌=𝟎
∞
𝟎
; (𝟏)
∵ 𝐬𝐞𝐜𝐡𝒙 = 𝝅∑(−𝟏)𝒌 ⋅𝟐𝒌 + 𝟏
(𝟏𝟐 + 𝒌)
𝟐
𝝅𝟐 + 𝒙𝟐
∞
𝒌=𝟎
𝐩𝐮𝐭 (𝒙 →𝝅𝒙
𝟐) ⇒
𝐬𝐞𝐜𝐡 (𝝅𝒙
𝟐) = 𝝅∑(−𝟏)𝒌 ⋅
𝟐𝒌 + 𝟏
(𝟐𝒌 + 𝟏)𝟐𝝅𝟐
𝟒 + (𝝅𝒙𝟒 )
𝟐
∞
𝒌=𝟎
=𝟒
𝝅∑(−𝟏)𝒌
𝟐𝒌 + 𝟏
(𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐
∞
𝒌=𝟎
Differentiating respect to 𝒙, we have:
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17 RMM-CALCULUS MARATHON 1501-1600
𝒅
𝒅𝒙𝐬𝐞𝐜𝐡 (
𝝅𝒙
𝟐) =
𝟒
𝝅∑(−𝟏)𝒌
−𝟐𝒙(𝟐𝒌 + 𝟏)
((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐
∞
𝒌=𝟎
= −𝟖
𝝅∑(−𝟏)𝒌
𝒙(𝟐𝒌 + 𝟏)
((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐
∞
𝒌=𝟎
∑(−𝟏)𝒌𝒙(𝟐𝒌 + 𝟏)
((𝟐𝒌 + 𝟏)𝟐 + 𝒙𝟐)𝟐
∞
𝒌=𝟎
= −𝝅
𝟖
𝒅
𝒅𝒙𝐬𝐞𝐜𝐡 (
𝝅𝒙
𝟐)
From (𝟏) we have:
𝑰 =𝟒𝒏
𝝅∫ −
𝝅
𝟖
𝒅
𝒅𝒙𝐬𝐞𝐜𝐡 (
𝝅𝒙
𝟐)
∞
𝟎
𝒅𝒙 = −𝒏
𝟐∫
𝒅
𝒅𝒙𝐬𝐞𝐜𝐡
𝝅𝒙
𝟐
∞
𝟎
𝒅𝒙 =
= −𝒏
𝟐|𝐬𝐞𝐜𝐡 (
𝝅𝒙
𝟐)|𝟎
∞
= −𝒏
𝟐[𝐥𝐢𝐦𝒙→∞
𝐬𝐞𝐜𝐡 (𝝅𝒙
𝟐) − 𝐥𝐢𝐦
𝒙→𝟎𝐬𝐞𝐜𝐡 (
𝝅𝒙
𝟐)] =
𝒏
𝟐
Therefore,
∫ ∫𝒙 𝐬𝐢𝐧 𝒕𝒙
𝐜𝐨𝐬𝐡𝝅𝒙𝒏
∞
𝟎
𝒅𝒕∞
𝟎
𝒅𝒙 =𝒏
𝟐
1511. Find:
𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛
Proposed by Asmat Qatea-Afghanistan
Solution 1 by Yen Tung Chung-Taichung-Taiwan
𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛
𝟑𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛 +
+∫ ∫ ∫𝒚𝟐 − 𝒛𝒙
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛 +
+∫ ∫ ∫𝒛𝟐 − 𝒙𝒚
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛 =
= ∫ ∫ ∫𝒙𝟐 + 𝒚𝟐 + 𝒛𝟐 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛 = ∫ ∫ ∫𝟏
𝒙 + 𝒚 + 𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛 =
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18 RMM-CALCULUS MARATHON 1501-1600
= ∫ ∫ 𝐥𝐨𝐠(𝒙 + 𝒚 + 𝒛)|𝟏𝟐
𝟐
𝟏
𝒅𝒚𝒅𝒙𝟐
𝟏
=
= ∫ ∫ (𝐥𝐨𝐠(𝟐 + 𝒚 + 𝒛)𝟐
𝟏
− 𝐥𝐨𝐠(𝟏 + 𝒚 + 𝒛)) 𝒅𝒚𝟐
𝟏
𝒅𝒛 =
= ∫ [(𝟐 + 𝒚 + 𝒛)(𝐥𝐨𝐠(𝟐 + 𝒚 + 𝒛) − 𝟏) − (𝟏 + 𝒚 + 𝒛)(𝐥𝐨𝐠(𝟏 + 𝒚 + 𝒛) − 𝟏)]𝟏𝟐𝒅𝒛
𝟐
𝟏
=
= ∫ ((𝟒 + 𝒛)(𝐥𝐨𝐠(𝟒 + 𝒛) − 𝟏) − (𝟑 + 𝒛)(𝐥𝐨𝐠(𝟑 + 𝒛) − 𝟏) − (𝟑 + 𝒛)(𝐥𝐨𝐠(𝟑 + 𝒛) − 𝟏)𝟐
𝟏
+ (𝟐 + 𝒛)(𝐥𝐨𝐠(𝟐 + 𝒛) − 𝟏))𝒅𝒛 =
= ∫ (𝟒 + 𝒛) 𝐥𝐨𝐠(𝟒 + 𝒛)𝒅𝒙 −𝟐
𝟏
𝟐∫ (𝟑 + 𝒛) 𝐥𝐨𝐠(𝟑 + 𝒛)𝒅𝒛𝟐
𝟏
+∫ (𝟐 + 𝒛) 𝐥𝐨𝐠(𝟐 + 𝒛)𝒅𝒛𝟐
𝟏
=
= ∫ 𝒖 𝐥𝐨𝐠 𝒖𝒅𝒖𝟔
𝟓
− 𝟐∫ 𝒖 𝐥𝐨𝐠𝒖𝒅𝒖𝟓
𝟒
+∫ 𝒖𝐥𝐨𝐠 𝒖𝒅𝒖𝟒
𝟑
=
= (𝟏
𝟐𝒖𝟐 𝐥𝐨𝐠𝒖 −
𝟏
𝟒𝒖𝟐)|
𝟓
𝟔
− 𝟐(𝟏
𝟐𝒖𝟐 𝐥𝐨𝐠𝒖 −
𝟏
𝟒𝒖𝟐)|
𝟒
𝟓
+ (𝟏
𝟐𝒖𝟐 𝐥𝐨𝐠 𝒖 −
𝟏
𝟒𝒖𝟐)|
𝟑
𝟒
=
= (𝟏𝟖 𝐥𝐨𝐠 𝟔 −𝟐𝟓
𝟐𝐥𝐨𝐠𝟓 −
𝟏𝟏
𝟒) − 𝟐(
𝟐𝟓
𝟐𝐥𝐨𝐠 𝟓 − 𝟏𝟔 𝐥𝐨𝐠 𝟐 −
𝟗
𝟒) + (𝟏𝟔 𝐥𝐨𝐠 𝟐 −
𝟗
𝟐𝐥𝐨𝐠𝟑 −
𝟗
𝟒)
= 𝟔𝟔 𝐥𝐨𝐠𝟐 +𝟐𝟕
𝟐𝐥𝐨𝐠𝟑 −
𝟕𝟓
𝟐𝐥𝐨𝐠 𝟓
Therefore,
𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛 = 𝟐𝟐 𝐥𝐨𝐠 𝟐 +𝟗
𝟐𝐥𝐨𝐠 𝟑 −
𝟐𝟓
𝟐𝐥𝐨𝐠𝟓
Solution 2 by Syed Shahabudeen-Kerala-india
𝛀 = ∫ ∫ ∫𝒙𝟐 − 𝒚𝒛
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛
𝟑𝛀 = ∫ ∫ ∫ ∑𝒙𝟐 − 𝒚𝒛
𝒙𝟑 + 𝒚𝟑 + 𝒛𝟑 − 𝟑𝒙𝒚𝒛𝒄𝒚𝒄
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛 = ∫ ∫ ∫𝟏
𝒙 + 𝒚 + 𝒛
𝟐
𝟏
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛 =
= ∫ ∫ ∫ ∫ 𝒕𝒙+𝒚+𝒛−𝟏𝟏
𝟎
𝒅𝒙𝟐
𝟏
𝒅𝒚𝟐
𝟏
𝒅𝒛𝟐
𝟏
𝒅𝒕 = ∫ 𝒕−𝟏𝟏
𝟎
∫ 𝒕𝒙𝟐
𝟏
𝒅𝒙∫ 𝒕𝒚𝒅𝒚𝟐
𝟏
∫ 𝒕𝒛𝟐
𝟏
𝒅𝒛𝒅𝒕 =
= ∫ 𝒕−𝟏 (𝒕𝟑 − 𝒕
𝐥𝐨𝐠𝒚)
𝟑𝟏
𝟎
𝒅𝒕 = ∫𝒕𝟐(𝒕 − 𝟏)𝟑
𝐥𝐨𝐠𝟑 𝒕
𝟏
𝟎
𝒅𝒕
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19 RMM-CALCULUS MARATHON 1501-1600
𝛀 = −𝟏
𝟑∫𝒕𝟐(𝟏 − 𝒕)𝟑
𝐥𝐨𝐠𝟑 𝒕
𝟏
𝟎
𝒅𝒕
𝐋𝐞𝐭: 𝛀(𝒂) = −𝟏
𝟑∫𝒕𝒂(𝟏 − 𝒕)𝟑
𝐥𝐨𝐠𝟑 𝒕
𝟏
𝟎
𝒅𝒕 ⇒𝝏𝟑𝛀
𝝏𝒂𝟑= −
𝟏
𝟑∫ 𝒕𝒂(𝟏 − 𝒕)𝟑𝟏
𝟎
𝒅𝒕 =
= −𝟏
𝟑(𝚪(𝒂 + 𝟏)𝚪(𝟒)
𝚪(𝒂 + 𝟓)) = −𝟐 ⋅
𝟏
(𝒂 + 𝟏)(𝒂 + 𝟐)(𝒂 + 𝟑)(𝒂 + 𝟒)
𝛀(𝒂) = −𝟐∫𝟏
(𝒂 + 𝟏)(𝒂 + 𝟐)(𝒂 + 𝟑)(𝒂 + 𝟒)𝒅𝟑𝒂 =
= −𝟐∫(𝟏
𝟔(𝒂 + 𝟏)−
𝟏
𝟐(𝒂 + 𝟐)+
𝟏
𝟐(𝒂 + 𝟑)−
𝟏
𝟔(𝒂 + 𝟒))𝒅𝟑𝒂
∫𝟏
𝒂 + 𝒏𝒅𝟑𝒂 =
𝟏
𝟐(𝒂 + 𝒏)𝟐 𝐥𝐨𝐠(𝒂 + 𝒏) −
𝟑𝒂
𝟒(𝒂 + 𝟐)
𝑨(𝒂) =𝟏
𝟔∫(
𝟏
𝒂 + 𝟏−
𝟏
𝒂 + 𝟒)𝒅𝟑𝒂 =
𝟏
𝟏𝟐((𝒂 + 𝟏)𝟐 𝐥𝐨𝐠(𝒂 + 𝟏) − (𝒂 + 𝟒)𝟐 𝐥𝐨𝐠(𝒂 + 𝟒)
𝑩(𝒂) =𝟏
𝟐∫(
𝟏
𝒂 + 𝟑−
𝟏
𝒂 + 𝟐)𝒅𝟑𝒂 =
𝟏
𝟒((𝒂 + 𝟑)𝟐 𝐥𝐨𝐠(𝒂 + 𝟑) − (𝒂 + 𝟐)𝟐 𝐥𝐨𝐠(𝒂 + 𝟐))
𝑨(𝟐) =𝟏
𝟏𝟐(𝟗 𝐥𝐨𝐠 𝟑 − 𝟑𝟔 𝐥𝐨𝐠 𝟔),𝑩(𝟐) =
𝟏
𝟒(𝟐𝟓 𝐥𝐨𝐠 𝟓 − 𝟏𝟔 𝐥𝐨𝐠 𝟒)
𝛀(𝟐) = −𝟐(𝑨(𝟐) + 𝑩(𝟐)) =
= −𝟐(𝟏
𝟏𝟐(𝟗 𝐥𝐨𝐠𝟑 − 𝟑𝟔 𝐥𝐨𝐠 𝟔) +
𝟏
𝟒(𝟐𝟓 𝐥𝐨𝐠 𝟓 − 𝟏𝟔 𝐥𝐨𝐠 𝟒)
Therefore,
𝛀(𝟐) = 𝟐𝟐 𝐥𝐨𝐠𝟐 +𝟗
𝟐𝐥𝐨𝐠 𝟑 +
𝟐𝟓
𝟐𝐥𝐨𝐠 𝟓
1512. Find:
𝛀 = ∫𝟏 − 𝐬𝐢𝐧𝟒 𝒙
(𝟏 + 𝐬𝐢𝐧𝟒 𝒙)√𝟏 + 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙
𝝅𝟔
𝟎
Proposed by Sujeethan Balendran-SriLanka
Solution 1 by Cornel Ioan Vălean-Romania
All we need is a clever variable change and a well-known established integral result,
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20 RMM-CALCULUS MARATHON 1501-1600
∫√𝐭𝐚𝐧𝒙𝒅𝒙 =𝟏
√𝟐(𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧 𝒙 − 𝐜𝐨𝐬 𝒙) − 𝐜𝐨𝐬𝐡−𝟏(𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙) + 𝑪
-one way to evaluate it is by calculating ∫(√𝐭𝐚𝐧 𝒙 + √𝐜𝐨𝐭 𝒙)𝒅𝒙 and ∫(√𝐭𝐚𝐧 𝒙 −
√𝐜𝐨𝐭 𝒙)𝒅𝒙
By letting the variable change 𝟏−𝐬𝐢𝐧𝟒 𝒙
𝟏+𝐬𝐢𝐧𝟒 𝒙→ 𝐬𝐢𝐧 𝒙, we get
𝛀 =𝟏
𝟐√𝟐∫ √𝐭𝐚𝐧𝒙
𝝅𝟐
𝐬𝐢𝐧−𝟏(𝟏𝟓𝟏𝟕)
𝒅𝒙 =𝝅
𝟖−𝟏
𝟒𝐬𝐢𝐧−𝟏 (
𝟕
𝟏𝟕) +
𝟏
𝟒𝐜𝐨𝐬𝐡−𝟏 (
𝟐𝟑
𝟏𝟕) =
=𝟏
𝟒𝐥𝐨𝐠 (
𝟐𝟑 + 𝟒√𝟏𝟓
𝟏𝟕) +
𝟏
𝟐𝐜𝐨𝐭−𝟏 (𝟐√
𝟑
𝟓)
Solution 2 by Cornel Ioan Vălean-Romania
𝛀 = ∫𝟏 − 𝐬𝐢𝐧𝟒 𝒙
(𝟏 + 𝐬𝐢𝐧𝟒 𝒙)√𝟏 + 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙
𝝅𝟔
𝟎
= ∫
𝟏𝒕𝟑− 𝒕
(𝒕𝟐 +𝟏𝒕𝟐)√𝟏𝒕𝟐− 𝒕𝟐
𝟏𝟐
𝟎
𝒅𝒕 =√𝟏
𝒕𝟐−𝒕𝟐→𝒕
= ∫𝒕𝟐
𝟒 + 𝒕𝟒
∞
√𝟏𝟓𝟐
𝒅𝒕 =𝟏
𝟒𝐥𝐨𝐠 (
𝟐𝟑 + 𝟒√𝟏𝟓
𝟏𝟕) +
𝟏
𝟐𝐜𝐨𝐭−𝟏 (𝟐√
𝟑
𝟓)
1513. Find without any software:
𝛀 = ∫√𝐬𝐢𝐧 𝒙 ⋅ 𝐭𝐚𝐧 𝒙 𝒅𝒙
Proposed by Orxan Abasov-Azerbaijan
Solution 1 by Yen Tung Chung-Taichung-Taiwan
𝛀 = ∫√𝐬𝐢𝐧𝒙 ⋅ 𝐭𝐚𝐧 𝒙𝒅𝒙 = ∫(√𝐬𝐢𝐧𝒙)
𝟑
𝟏 − 𝐬𝐢𝐧𝟐 𝒙⋅ 𝐜𝐨𝐬 𝒙 𝒅𝒙 =
𝒚=√𝐬𝐢𝐧𝒙,𝟐𝒚𝒅𝒚=𝐜𝐨𝐬 𝒙𝒅𝒙
= ∫𝒚𝟑
𝟏 − 𝒚𝟒⋅ 𝟐𝒚𝒅𝒚 = 𝟐∫
𝒚𝟒
𝟏 − 𝒚𝟒𝒅𝒚 = 𝟐∫(−𝟏 +
𝟏
𝟏 − 𝒚𝟒)𝒅𝒚 =
= −𝟐𝒚 +∫(𝟏
𝟏− 𝒚𝟐+
𝟏
𝟏 + 𝒚𝟐)𝒅𝒚 = −𝟐𝒚 + 𝐭𝐚𝐧𝐡−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝑪 =
= −𝟐√𝐬𝐢𝐧𝒙 + 𝐭𝐚𝐧𝐡−𝟏(√𝐬𝐢𝐧𝒙) + 𝐭𝐚𝐧−𝟏(√𝐬𝐢𝐧 𝒙) + 𝑪 =
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21 RMM-CALCULUS MARATHON 1501-1600
= −𝟐√𝐬𝐢𝐧𝒙 +𝟏
𝟐𝐥𝐨𝐠 (
𝟏 + √𝐬𝐢𝐧𝒙
𝟏 − √𝐬𝐢𝐧𝒙) + 𝐭𝐚𝐧−𝟏(√𝐬𝐢𝐧 𝒙) + 𝑪
Solution 2 by Orlando Irahola Ortega-Bolivia
𝛀 = ∫√𝐬𝐢𝐧 𝒙 ⋅ 𝐭𝐚𝐧𝒙 𝒅𝒙 = ∫(√𝐬𝐢𝐧𝒙)
𝟑
𝟏 − 𝐬𝐢𝐧𝟐 𝒙⋅ 𝐜𝐨𝐬 𝒙𝒅𝒙 =
𝒕𝟐=𝐬𝐢𝐧𝒙 ,𝟐𝒕𝒅𝒕=𝐜𝐨𝐬 𝒙𝒅𝒙
= ∫𝟐𝒕𝟒
𝟏 − 𝒕𝟒𝒅𝒕 = −𝟐∫
(𝒕𝟒 − 𝟏) + 𝟏
𝒕𝟒 − 𝟏𝒅𝒕 ⇒
−𝟏
𝟐𝛀 = 𝒕 −∫
𝒅𝒕
𝒕𝟒 − 𝟏= 𝒕 − 𝑰,
𝑰 = ∫𝒅𝒕
𝒕𝟒 − 𝟏=𝒕=𝟏𝒚− ∫
𝒚𝟐
𝟏 − 𝒚𝟒𝒅𝒚 = −
𝟏
𝟐∫𝟐𝒚𝟐
𝟏 − 𝒚𝟒𝒅𝒚
=𝟏
𝟐∫(𝒚𝟐 − 𝟏) + (𝒚𝟐 + 𝟏)
(𝒚𝟐 − 𝟏)(𝒚𝟐 + 𝟏)𝒅𝒚 =
𝟏
𝟐∫
𝒅𝒚
𝒚𝟐 + 𝟏+𝟏
𝟐∫
𝒅𝒚
𝒚𝟐 − 𝟏=
=𝟏
𝟐𝐭𝐚𝐧−𝟏 𝒚 +
𝟏
𝟒𝐥𝐨𝐠 (
𝒚 − 𝟏
𝒚+ 𝟏) =
𝟏
𝟐𝐭𝐚𝐧−𝟏 (
𝟏
𝒕) +
𝟏
𝟒𝐥𝐨𝐠 (
𝟏 − 𝒕
𝟏 + 𝒕) =
=𝟏
𝟐𝐭𝐚𝐧−𝟏 𝒕 +
𝟏
𝟒𝐥𝐨𝐠 (
𝟏 − 𝒕
𝟏 + 𝒕) ⇒
−𝟏
𝟐𝛀 = 𝒕 −∫
𝒅𝒕
𝒕𝟒 − 𝟏= 𝒕 − 𝑰 = 𝒕 −
𝟏
𝟐𝐭𝐚𝐧−𝟏 𝒕 −
𝟏
𝟒𝐥𝐨𝐠 (
𝟏 − 𝒕
𝟏 + 𝒕)
Therefore,
𝛀 = −𝟐√𝐬𝐢𝐧𝒙 +𝟏
𝟐𝐥𝐨𝐠(
𝟏 + √𝐬𝐢𝐧 𝒙
𝟏 − √𝐬𝐢𝐧 𝒙) + 𝐭𝐚𝐧−𝟏(√𝐬𝐢𝐧𝒙) + 𝑪
Solution 3 by Ghuiam Naseri-Afghanistan
𝛀 = ∫√𝐬𝐢𝐧𝒙 ⋅ 𝐭𝐚𝐧 𝒙𝒅𝒙 = ∫𝐬𝐢𝐧𝒙
𝐜𝐨𝐬 𝒙√𝐬𝐢𝐧𝒙𝒅𝒙 = ∫
(𝐬𝐢𝐧𝒙)𝟑𝟐
𝐜𝐨𝐬 𝒙𝒅𝒙 =
= ∫𝐜𝐨𝐬𝒙 ⋅ (𝐬𝐢𝐧 𝒙)
𝟑𝟐
𝐜𝐨𝐬𝟐 𝒙𝒅𝒙 = ∫
𝐜𝐨𝐬𝒙 ⋅ (𝐬𝐢𝐧 𝒙)𝟑𝟐
𝟏 − 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙 =
𝒖=√𝐬𝐢𝐧𝒙∫
𝟐𝒖𝟒
−𝒖𝟒 + 𝟏𝒅𝒖 = 𝟐∫
𝒖𝟒
−𝒖𝟒 + 𝟏𝒅𝒖
= 𝟐∫(−𝟏
𝟐⋅
𝟏
𝒖𝟐 + 𝟏−𝟏
𝟒⋅𝟏
𝒖 + 𝟏+𝟏
𝟒⋅𝟏
𝒖 − 𝟏+ 𝟏)𝒅𝒖 =
= −∫𝟏
𝒖𝟐 + 𝟏𝒅𝒖 −
𝟏
𝟐∫
𝟏
𝒖 + 𝟏𝒅𝒖 +
𝟏
𝟐∫
𝟏
𝒖 − 𝟏𝒅𝒖 + 𝟐𝒖 =
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22 RMM-CALCULUS MARATHON 1501-1600
= − 𝐭𝐚𝐧−𝟏 𝒖 −𝟏
𝟐𝐥𝐨𝐠(𝒖 + 𝟏) +
𝟏
𝟐𝐥𝐨𝐠(𝒖 − 𝟏) + 𝟐𝒖 + 𝑪
Therefore,
𝛀 = −𝟐√𝐬𝐢𝐧𝒙 +𝟏
𝟐𝐥𝐨𝐠(
𝟏 + √𝐬𝐢𝐧 𝒙
𝟏 − √𝐬𝐢𝐧 𝒙) + 𝐭𝐚𝐧−𝟏(√𝐬𝐢𝐧𝒙) + 𝑪
1514. Find without any software:
𝛀 = ∫𝐥𝐨𝐠(𝟗𝒙 − 𝟒)
𝟑𝒙𝟐 + 𝟐
𝟐
𝟏
𝒅𝒙
Proposed by Daniel Sitaru-Romania
Solution by Marian Ursărescu-Romania
Put 𝟗𝒙 − 𝟒 = 𝒕 ⇒ 𝒙 =𝒕+𝟒
𝟗, 𝒅𝒙 =
𝟏
𝟗𝒅𝒕
𝛀 = ∫𝐥𝐨𝐠(𝟗𝒙 − 𝟒)
𝟑𝒙𝟐 + 𝟐
𝟐
𝟏
𝒅𝒙 =𝟏
𝟗∫
𝐥𝐨𝐠 𝒕
𝟑 (𝒕 + 𝟒𝟗 )
𝟐
+ 𝟐
𝟏𝟒
𝟓
𝒅𝒕 =𝟏
𝟗∫
𝐥𝐨𝐠 𝒕
𝟑(𝒕𝟐 + 𝟖𝒕 + 𝟏𝟔)𝟖𝟏 + 𝟐
𝟏𝟒
𝟓
𝒅𝒕 =
= 𝟑∫𝐥𝐨𝐠 𝒕
𝒕𝟐 + 𝟖𝒕 + 𝟕𝟎
𝟏𝟒
𝟓
𝒅𝒕; (𝟏)
𝐋𝐞𝐭: 𝑰 = ∫𝐥𝐨𝐠 𝒕
𝒕𝟐 + 𝟖𝒕 + 𝟕𝟎
𝟏𝟒
𝟓
𝒅𝒕 =𝒕=𝟕𝟎𝒚− 𝟕𝟎∫
𝐥𝐨𝐠 (𝟕𝟎𝒚 )
𝟕𝟎𝟐
𝒚𝟐+ 𝟖 ⋅
𝟕𝟎𝒚 + 𝟕𝟎
𝟓
𝟏𝟒
𝒅𝒚 =
= ∫𝐥𝐨𝐠𝟕𝟎 − 𝐥𝐨𝐠𝒚
𝒚𝟐 + 𝟖𝒚 + 𝟕𝟎
𝟒
𝟓
𝒅𝒚 = 𝐥𝐨𝐠𝟕𝟎∫𝟏
𝒚𝟐 + 𝟖𝒚 + 𝟕𝟎𝒅𝒚
𝟏𝟒
𝟓
− ∫𝐥𝐨𝐠 𝒚
𝒚𝟐 + 𝟖𝒚 + 𝟕𝟎
𝟏𝟒
𝟓
𝒅𝒚 =
=𝟏
𝟐𝐥𝐨𝐠𝟕𝟎∫
𝟏
(𝒚 + 𝒚)𝟒 + (√𝟓𝟒)𝟐 𝒅𝒚
𝟏𝟒
𝟓
=𝟏
𝟐√𝟓𝟒𝐥𝐨𝐠 𝟕𝟎 𝐭𝐚𝐧−𝟏 (
𝒚 + 𝟒
√𝟓𝟒)|𝟓
𝟏𝟒
=
=𝟏
𝟐√𝟓𝟒𝐥𝐨𝐠 𝟕𝟎 (𝐭𝐚𝐧−𝟏 (
𝟏𝟖
√𝟓𝟒) − 𝐭𝐚𝐧−𝟏 (
𝟗
√𝟓𝟒)) =
=𝟏
𝟐√𝟓𝟒𝐥𝐨𝐠𝟕𝟎 (𝐭𝐚𝐧−𝟏 √𝟔 − 𝐭𝐚𝐧−𝟏 (
𝟑
√𝟔)) ; (𝟐)
From (1),(2) it follows that:
𝛀 = ∫𝐥𝐨𝐠(𝟗𝒙 − 𝟒)
𝟑𝒙𝟐 + 𝟐
𝟐
𝟏
𝒅𝒙 =𝟏
𝟐√𝟔𝐥𝐨𝐠 𝟕𝟎 (𝐭𝐚𝐧−𝟏 √𝟔 − 𝐭𝐚𝐧−𝟏 (
𝟑
√𝟔))
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23 RMM-CALCULUS MARATHON 1501-1600
1515. Find a closed form:
𝛀 = ∫𝒙 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑
𝟏
𝟎
𝒅𝒙
Proposed by Abdul Mukhtar-Nigeria
Solution 1 by Mohammad Rostami-Afghanistan
𝛀 = ∫𝒙 𝐥𝐨𝐠𝒙
𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑
𝟏
𝟎
𝒅𝒙 = ∫𝒙 𝐥𝐨𝐠 𝒙
(𝟏 − 𝒙)(𝟏 + 𝒙𝟐)
𝟏
𝟎
𝒅𝒙 = −∫(𝟏 − 𝒙 − 𝟏) 𝐥𝐨𝐠 𝒙
(𝟏 − 𝒙)(𝟏 + 𝒙𝟐)𝒅𝒙
𝟏
𝟎
=
= −∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐
𝟏
𝟎
𝒅𝒙 +∫𝐥𝐨𝐠 𝒙
(𝟏 − 𝒙)(𝟏 + 𝒙𝟐)𝒅𝒙
𝟏
𝟎
=
= −∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
+𝟏
𝟐∫𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝒅𝒙
𝟏
𝟎
+𝟏
𝟐∫
𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
+𝟏
𝟐∫𝒙 𝐥𝐨𝐠𝒙
𝟏 + 𝒙𝟐
𝟏
𝟎
𝒅𝒙 =
= −𝟏
𝟐∫
𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐
𝟏
𝟎
𝒅𝒙 +𝟏
𝟐∫𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝒅𝒙
𝟏
𝟎
+𝟏
𝟐∫𝒙 𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
=
= −𝟏
𝟐∫ ∑(−𝒙𝟐)
𝝏
𝝏𝒂|𝒂=𝟎𝒙𝒂𝒅𝒙
∞
𝒌=𝟎
𝟏
𝟎
+𝟏
𝟐∫ ∑𝒙𝒏
𝝏
𝝏𝒃|𝒃=𝟎𝒙𝒃𝒅𝒙
∞
𝒏=𝟎
𝟏
𝟎
+𝟏
𝟐∫ 𝒙 ∑(−𝒙𝟐)𝒎
𝝏
𝝏𝒄|𝒄=𝟎𝒙𝒄
∞
𝒎=𝟎
𝒅𝒙𝟏
𝟎
=
= −𝟏
𝟐∑(−𝟏)𝒌
𝝏
𝝏𝒂|𝒂=𝟎
∞
𝒌=𝟎
∫ 𝒙𝟐𝒌+𝒂𝒅𝒙𝟏
𝟎
+𝟏
𝟐∑
𝝏
𝝏𝒃|𝒃=𝟎
∫ 𝒙𝒏+𝒃𝒅𝒙𝟏
𝟎
∞
𝒏=𝟎
+
+𝟏
𝟐∑(−𝟏)𝒎
𝝏
𝝏𝒄|𝒄=𝟎
∫ 𝒙𝟐𝒎+𝒄+𝟏𝟏
𝟎
𝒅𝒙
∞
𝒎=𝟎
=
= −𝟏
𝟐∑(−𝟏)𝒌 [
𝟏
𝟐𝒌 + 𝒂 + 𝟏]𝒂=𝟎
′∞
𝒌=𝟎
+𝟏
𝟐∑[
𝟏
𝒏 + 𝒃 + 𝟏]𝒃=𝟎
′∞
𝒏=𝟎
+𝟏
𝟐∑(−𝟏)𝒎 [
𝟏
𝟐𝒎+ 𝒄 + 𝟐]𝒄=𝟎
′∞
𝒎=𝟎
=𝟏
𝟐∑
(−𝟏)𝒌
(𝟐𝒌 + 𝟏)𝟐
∞
𝒏=𝟎
−𝟏
𝟐∑
𝟏
(𝒏 + 𝟏)𝟐
∞
𝒏=𝟎
−𝟏
𝟖∑
(−𝟏)𝒎
(𝒎 + 𝟏)𝟐
∞
𝒎=𝟎
=
=𝟏
𝟐𝑮 −
𝟏
𝟐∑
𝟏
𝒏𝟐
∞
𝒏=𝟏
−𝟏
𝟖∑
(−𝟏)𝒎−𝟏
𝒎𝟐
∞
𝒎=𝟏
=𝟏
𝟐𝑮 −
𝟏
𝟐𝜻(𝟐) −
𝟏
𝟖𝜼(𝟐) =
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24 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝟐𝑮 −
𝟏
𝟐𝜻(𝟐) −
𝟏
𝟏𝟔𝜻(𝟐) =
𝟏
𝟐𝑮 −
𝟗
𝟏𝟔⋅𝝅𝟐
𝟔=𝟏
𝟐𝑮 −
𝟑𝝅𝟐
𝟑𝟐
Therefore,
𝛀 =𝟏
𝟐𝑮 −
𝟑𝝅𝟐
𝟑𝟐
Solution 2 by Rana Ranino-Setif-Algerie
𝛀 = ∫𝒙 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑
𝟏
𝟎
𝒅𝒙 = ∫𝒙(𝟏 + 𝒙) 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟒
𝟏
𝟎
𝒅𝒙 = ∑∫ (𝒙𝟒𝒏+𝟏 + 𝒙𝟒𝒏+𝟐) 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
∞
𝒏=𝟎
=
= −𝟏
𝟏𝟔∑[
𝟏
(𝒏 +𝟏𝟐)𝟐 +
𝟏
(𝒏 +𝟑𝟒)𝟐]
∞
𝒏=𝟎
= −𝟏
𝟏𝟔{𝝍(𝟏) (
𝟏
𝟐) + 𝝍(𝟏) (
𝟑
𝟒)} = −
𝟏
𝟏𝟔(𝟑𝝅𝟐
𝟐− 𝟖𝑮)
Therefore,
𝛀 =𝟏
𝟐𝑮 −
𝟑𝝅𝟐
𝟑𝟐
Solution 3 by Ajetunmobi Abdulquyyum-Nigeria
𝛀 = ∫𝒙 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙 + 𝒙𝟐 − 𝒙𝟑
𝟏
𝟎
𝒅𝒙 = ∫𝒙 𝐥𝐨𝐠𝒙
(𝟏 + 𝒙𝟐)(𝟏 − 𝒙)𝒅𝒙
𝟏
𝟎
=
= ∫ (𝒙 − 𝟏
𝟐(𝒙𝟐 + 𝟏)−
𝟏
𝟐(𝒙 − 𝟏)) 𝐥𝐨𝐠 𝒙𝒅𝒙
𝟏
𝟎
=𝟏
𝟐{∫
𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟐 + 𝟏𝒅𝒙
𝟏
𝟎
−∫𝐥𝐨𝐠𝒙
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
+∫𝐥𝐨𝐠𝒙
𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙}
=𝟏
𝟐{𝑨 − 𝑩 + 𝑪}
𝑨 = ∫𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟐 + 𝟏𝒅𝒙
𝟏
𝟎
=; |𝐥𝐨𝐠 𝒙 = −𝒕 ⇒ 𝒅𝒙 = 𝒆−𝒕(−𝒅𝒕)
𝒙𝟐 = 𝒆−𝟐𝒕| ; = −∫
𝒆−𝟐𝒕
𝟏—𝒆−𝟐𝒕
∞
𝟎
𝒅𝒕 =
= −∑(−𝟏)𝒏∫ 𝒕𝒆−(𝟐𝒏+𝟐)𝒕𝒅𝒕∞
𝟎𝒏≥𝟎
= −∑(−𝟏)𝒏 ⋅𝟏
𝟒(𝒏 + 𝟏)𝟐𝒏≥𝟎
= −𝟏
𝟒∑
(−𝟏)𝒏
(𝒏 + 𝟏)𝟐𝒏≥𝟎
=
= −𝟏
𝟒∑(−𝟏)𝒏−𝟏
𝒏𝟐𝒏≥𝟏
= −𝟏
𝟒𝜼(𝟐) = −
𝝅𝟐
𝟒𝟖
Also,
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25 RMM-CALCULUS MARATHON 1501-1600
𝑩 = ∫𝐥𝐨𝐠𝒙
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
=𝒙=𝐭𝐚𝐧 𝒙
∫ 𝐥𝐨𝐠(𝐭𝐚𝐧𝒙)
𝝅𝟒
𝟎
𝒅𝒙 = −𝑮
𝑪 = ∫𝐥𝐨𝐠𝒙
𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 =∑∫ 𝒙𝒏 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎𝒏≥𝟎
=∑𝟏
(𝒏 + 𝟏)𝟐𝒏≥𝟎
= −𝝅𝟐
𝟔
Hence,
𝛀 =𝟏
𝟐(𝑨 − 𝑩 + 𝑪) =
𝟏
𝟐(−𝝅𝟐
𝟒𝟖+ 𝑮 −
𝝅𝟐
𝟔)
Therefore,
𝛀 =𝟏
𝟐𝑮 −
𝟑𝝅𝟐
𝟑𝟐
1516. Find a closed form:
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠 𝒙
𝒆
𝟎
𝒅𝒙,𝒏 > 𝟎
Proposed by Abdul Mukhtar-Nigeria
Solution 1 by Ajentunmobi Abdulqoyuum-Nigeria
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠𝒙
𝒆
𝟎
𝒅𝒙 =; |𝒕 = √𝟏 − 𝐥𝐨𝐠 𝒙 ;𝒅𝒕 =
𝟏
𝟐𝒙√𝟏 − 𝐥𝐨𝐠 𝒙𝒅𝒙
𝒅𝒙 = −𝟐𝒕𝒆𝟏−𝒕𝟐, 𝒙 = 𝒆𝟏−𝒕
𝟐| ;
= ∫𝟐𝒕𝒆𝟏−𝒕
𝟐𝒆𝟏−𝒕
𝟐
𝒕𝒅𝒕
∞
𝟎
= 𝟐∫ 𝒆(𝒏+𝟏)−(𝒏+𝟏)𝒕𝟐𝒅𝒕
∞
𝟎
=
= 𝟐𝒆𝒏+𝟏∫ 𝒆−(𝒏+𝟏)𝒕𝟐
∞
𝟎
𝒅𝒕;|
|
(𝒏 + 𝟏)𝒕𝟐 = 𝒛; 𝒅𝒕 =𝒅𝒛
𝟐(𝒏 + 𝟏)√𝒛
𝒏 + 𝟏
𝒕 = √𝒛
𝒏 + 𝟏
|
|
𝛀 = 𝟐𝒆𝒏+𝟏∫ 𝒆−𝒛 ⋅𝒅𝒛
𝟐(𝒏 + 𝟏)√𝒛
𝒏+ 𝟏
∞
𝟎
=𝒆𝒏+𝟏
√𝒏 + 𝟏∫ 𝒛
𝟏𝟐−𝟏𝒆−𝒛
∞
𝟎
𝒅𝒛 =√𝝅𝒆𝒏+𝟏
√𝒏 + 𝟏
𝑵𝒐𝒕𝒆: ∫ 𝒛𝟏𝟐−𝟏𝒆−𝒛𝒅𝒛
∞
𝟎
= 𝚪 (𝟏
𝟐) = √𝝅
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26 RMM-CALCULUS MARATHON 1501-1600
Solution 2 by Akerele Olofin-Nigeria
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠𝒙
𝒆
𝟎
𝒅𝒙 =
𝒕=𝟏−𝐥𝐨𝐠 𝒙;
𝒅𝒕=−𝒆𝟏−𝒕𝒅𝒕𝒆𝒏+𝟏∫
𝒆−𝒕(𝒏+𝟏)
√𝒕𝒅𝒕
∞
𝟎
𝓛{𝒕𝒏} = ∫ 𝒆−𝒔𝒕𝒕𝒏𝒅𝒕∞
𝟎
=𝚪(𝒏 + 𝟏)
𝒔𝒏+𝟏, 𝐰𝐡𝐞𝐧 𝒏 = −
𝟏
𝟐; 𝒔 = 𝒏 + 𝟏 ⇒
∫𝒆−𝒕(𝒏+𝟏)
√𝒕
∞
𝟎
𝒅𝒕 =𝚪(𝟏𝟐)
𝒔𝟏𝟐
=√𝝅
√𝒏 + 𝟏⇒ 𝛀(𝒏) =
√𝝅𝒆𝒏+𝟏
√𝒏 + 𝟏
Therefore,
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠 𝒙
𝒆
𝟎
𝒅𝒙 =√𝝅𝒆𝒏+𝟏
√𝒏 + 𝟏
Solution 3 by Muhammad Afzal-Pakistan
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠𝒙
𝒆
𝟎
𝒅𝒙 =𝒖=𝟏−𝐥𝐨𝐠 𝒙
∫𝒆(𝟏−𝒖)(𝒏+𝟏)
√𝒖
∞
𝟎
𝒅𝒖 = 𝒆𝒏+𝟏∫𝒆−𝒖(𝒏+𝟏)
√𝒖𝒅𝒖
∞
𝟎
=
=𝒆𝒏+𝟏
𝒏 + 𝟏∫ 𝒖−
𝟏𝟐𝒆−𝒖(𝒏+𝟏)𝒅𝒖
∞
𝟎
=𝒚=𝒖(𝒏+𝟏) 𝒆𝒏+𝟏
𝒏 + 𝟏∫
𝒚−𝟏𝟐
(𝒏 + 𝟏)−𝟏𝟐
⋅ 𝒆−𝒚∞
𝟎
𝒅𝒚 =
=𝒆𝒏+𝟏
√𝒏 + 𝟏∫ 𝒚
𝟏𝟐−𝟏𝒆−𝒚𝒅𝒚
∞
𝟎
=𝒆𝒏+𝟏
√𝒏 + 𝟏𝚪(𝟏
𝟐)
Therefore,
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠 𝒙
𝒆
𝟎
𝒅𝒙 =√𝝅𝒆𝒏+𝟏
√𝒏 + 𝟏
Solution 4 by Adrian Popa-Romania
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠 𝒙
𝒆
𝟎
𝒅𝒙 =√𝟏−𝐥𝐨𝐠 𝒙=𝒕
𝟐∫ 𝒆𝒏+𝟏 ⋅ 𝒆−𝒕𝟐(𝒏+𝟏)𝒅𝒕
∞
𝟎
= 𝟐𝒆𝒏+𝟏∫ 𝒆−𝒕𝟐(𝒏+𝟏)𝒅𝒕
∞
𝟎
=
=(𝒏+𝟏)𝒕𝟐=𝒖
𝟐𝒆𝒏+𝟏∫ 𝒆−𝒖 ⋅𝟏
𝟐√𝒏 + 𝟏𝒖−
𝟏𝟐
∞
𝟎
𝒅𝒖 =𝒆𝒏+𝟏
√𝒏 + 𝟏∫ 𝒖−
𝟏𝟐𝒆−𝒖
∞
𝟎
𝒅𝒖 =
=𝒆𝒏+𝟏
√𝒏 + 𝟏𝚪(𝟏
𝟐) =
𝒆𝒏+𝟏√𝝅
√𝒏 + 𝟏
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27 RMM-CALCULUS MARATHON 1501-1600
1517. Find a closed form:
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠 𝒙
𝒆
𝟎
𝒅𝒙,𝒏 > 𝟎
Proposed by Abdul Mukhtar-Nigeria
Solution 1 by Ajentunmobi Abdulqoyuum-Nigeria
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠𝒙
𝒆
𝟎
𝒅𝒙 =; |𝒕 = √𝟏 − 𝐥𝐨𝐠 𝒙 ;𝒅𝒕 =
𝟏
𝟐𝒙√𝟏 − 𝐥𝐨𝐠 𝒙𝒅𝒙
𝒅𝒙 = −𝟐𝒕𝒆𝟏−𝒕𝟐, 𝒙 = 𝒆𝟏−𝒕
𝟐| ;
= ∫𝟐𝒕𝒆𝟏−𝒕
𝟐𝒆𝟏−𝒕
𝟐
𝒕𝒅𝒕
∞
𝟎
= 𝟐∫ 𝒆(𝒏+𝟏)−(𝒏+𝟏)𝒕𝟐𝒅𝒕
∞
𝟎
=
= 𝟐𝒆𝒏+𝟏∫ 𝒆−(𝒏+𝟏)𝒕𝟐
∞
𝟎
𝒅𝒕;|
|
(𝒏 + 𝟏)𝒕𝟐 = 𝒛; 𝒅𝒕 =𝒅𝒛
𝟐(𝒏 + 𝟏)√𝒛
𝒏 + 𝟏
𝒕 = √𝒛
𝒏 + 𝟏
|
|
𝛀 = 𝟐𝒆𝒏+𝟏∫ 𝒆−𝒛 ⋅𝒅𝒛
𝟐(𝒏 + 𝟏)√𝒛
𝒏+ 𝟏
∞
𝟎
=𝒆𝒏+𝟏
√𝒏 + 𝟏∫ 𝒛
𝟏𝟐−𝟏𝒆−𝒛
∞
𝟎
𝒅𝒛 =√𝝅𝒆𝒏+𝟏
√𝒏 + 𝟏
𝑵𝒐𝒕𝒆: ∫ 𝒛𝟏𝟐−𝟏𝒆−𝒛𝒅𝒛
∞
𝟎
= 𝚪 (𝟏
𝟐) = √𝝅
Solution 2 by Akerele Olofin-Nigeria
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠𝒙
𝒆
𝟎
𝒅𝒙 =
𝒕=𝟏−𝐥𝐨𝐠 𝒙;
𝒅𝒕=−𝒆𝟏−𝒕𝒅𝒕𝒆𝒏+𝟏∫
𝒆−𝒕(𝒏+𝟏)
√𝒕𝒅𝒕
∞
𝟎
𝓛{𝒕𝒏} = ∫ 𝒆−𝒔𝒕𝒕𝒏𝒅𝒕∞
𝟎
=𝚪(𝒏 + 𝟏)
𝒔𝒏+𝟏, 𝐰𝐡𝐞𝐧 𝒏 = −
𝟏
𝟐; 𝒔 = 𝒏 + 𝟏 ⇒
∫𝒆−𝒕(𝒏+𝟏)
√𝒕
∞
𝟎
𝒅𝒕 =𝚪(𝟏𝟐)
𝒔𝟏𝟐
=√𝝅
√𝒏 + 𝟏⇒ 𝛀(𝒏) =
√𝝅𝒆𝒏+𝟏
√𝒏 + 𝟏
Therefore,
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28 RMM-CALCULUS MARATHON 1501-1600
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠 𝒙
𝒆
𝟎
𝒅𝒙 =√𝝅𝒆𝒏+𝟏
√𝒏 + 𝟏
Solution 3 by Muhammad Afzal-Pakistan
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠𝒙
𝒆
𝟎
𝒅𝒙 =𝒖=𝟏−𝐥𝐨𝐠 𝒙
∫𝒆(𝟏−𝒖)(𝒏+𝟏)
√𝒖
∞
𝟎
𝒅𝒖 = 𝒆𝒏+𝟏∫𝒆−𝒖(𝒏+𝟏)
√𝒖𝒅𝒖
∞
𝟎
=
=𝒆𝒏+𝟏
𝒏 + 𝟏∫ 𝒖−
𝟏𝟐𝒆−𝒖(𝒏+𝟏)𝒅𝒖
∞
𝟎
=𝒚=𝒖(𝒏+𝟏) 𝒆𝒏+𝟏
𝒏 + 𝟏∫
𝒚−𝟏𝟐
(𝒏 + 𝟏)−𝟏𝟐
⋅ 𝒆−𝒚∞
𝟎
𝒅𝒚 =
=𝒆𝒏+𝟏
√𝒏 + 𝟏∫ 𝒚
𝟏𝟐−𝟏𝒆−𝒚𝒅𝒚
∞
𝟎
=𝒆𝒏+𝟏
√𝒏 + 𝟏𝚪(𝟏
𝟐)
Therefore,
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠 𝒙
𝒆
𝟎
𝒅𝒙 =√𝝅𝒆𝒏+𝟏
√𝒏 + 𝟏
Solution 4 by Adrian Popa-Romania
𝛀(𝒏) = ∫𝒙𝒏
√𝟏 − 𝐥𝐨𝐠 𝒙
𝒆
𝟎
𝒅𝒙 =√𝟏−𝐥𝐨𝐠 𝒙=𝒕
𝟐∫ 𝒆𝒏+𝟏 ⋅ 𝒆−𝒕𝟐(𝒏+𝟏)𝒅𝒕
∞
𝟎
= 𝟐𝒆𝒏+𝟏∫ 𝒆−𝒕𝟐(𝒏+𝟏)𝒅𝒕
∞
𝟎
=
=(𝒏+𝟏)𝒕𝟐=𝒖
𝟐𝒆𝒏+𝟏∫ 𝒆−𝒖 ⋅𝟏
𝟐√𝒏 + 𝟏𝒖−
𝟏𝟐
∞
𝟎
𝒅𝒖 =𝒆𝒏+𝟏
√𝒏 + 𝟏∫ 𝒖−
𝟏𝟐𝒆−𝒖
∞
𝟎
𝒅𝒖 =
=𝒆𝒏+𝟏
√𝒏 + 𝟏𝚪(𝟏
𝟐) =
𝒆𝒏+𝟏√𝝅
√𝒏 + 𝟏
1518. Prove that:
∫𝒅𝒙
(𝟒 𝐥𝐨𝐠𝟐 𝒙 + 𝝅𝟐)𝟐(𝒙𝟐 + 𝟏)
∞
𝟎
=𝐥𝐨𝐠𝟐
𝟒𝝅𝟑+
𝟏
𝟗𝟔𝝅
Proposed by Ty Halpen-Florida-SUA
Solution by Rana Ranino-Setif-Algerie
𝛀 = ∫𝒅𝒙
(𝟒 𝐥𝐨𝐠𝟐 𝒙 + 𝝅𝟐)𝟐(𝒙𝟐 + 𝟏)
∞
𝟎
=𝒙=𝒆
−(𝝅𝒕𝟐) 𝟏
𝟐𝝅𝟑∫
𝒆−𝝅𝒕𝟐
(𝒕𝟐 + 𝟐)𝟐(𝟏 + 𝒆−𝝅𝒕)𝒅𝒕
∞
∞
=
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29 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝟒𝝅𝟑∫
𝐬𝐞𝐜𝐡 (𝝅𝒕𝟐 )
(𝒕𝟐 + 𝟏)𝟐𝒅𝒕
∞
∞
=𝟏
𝟐𝝅𝟑∫
𝐬𝐞𝐜𝐡 (𝝅𝒕𝟐 )
(𝒕𝟐 + 𝟏)𝟐
∞
𝟎
𝒅𝒕
𝐔𝐬𝐢𝐧𝐠: 𝐬𝐞𝐜𝐡 (𝝅𝒕
𝟐) =
𝟒
𝝅∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)
𝒕𝟐 + (𝟐𝒌 + 𝟏)𝟐
∞
𝒌=𝟎
𝛀 =𝟐
𝝅𝟒∑(−𝟏)𝒌(𝟐𝒌 + 𝟏)
∞
𝒌=𝟎
∫𝒅𝒕
(𝒕𝟐 + 𝟏)𝟐(𝒕𝟐 + (𝟐𝒌 + 𝟏)𝟐)
∞
𝟎
𝑰(𝒂) = ∫𝒅𝒕
(𝒕𝟐 + 𝟏)𝟐(𝒕𝟐 + 𝒂𝟐)
∞
𝟎
=
=𝟏
𝒂𝟐 − 𝟏∫
𝒅𝒕
(𝟏 + 𝒕𝟐)𝟐
∞
𝟎
−𝟏
(𝒂𝟐 − 𝟏)𝟐∫ (
𝟏
𝟏 + 𝒕𝟐−
𝟏
𝒕𝟐 + 𝒂𝟐)
∞
𝟎
𝒅𝒕 =
=𝝅
𝟒(𝒂𝟐 − 𝟏)−
𝝅(𝒂− 𝟏)
𝟐𝒂(𝒂𝟐 − 𝟏)𝟐
𝑰(𝒂) =𝝅(𝒂𝟑 − 𝟑𝒂 + 𝟐)
𝟒𝒂(𝒂𝟐 − 𝟏)𝟐=𝝅(𝒂− 𝟏)𝟐(𝒂 + 𝟐)
𝟒𝒂(𝒂 − 𝟏)𝟐(𝒂 + 𝟏)𝟐=𝝅(𝒂 + 𝟐)
𝟒𝒂(𝒂 + 𝟏)𝟐
𝛀 =𝟏
𝟐𝝅𝟑∑(−𝟏)𝒌
𝟐𝒌 + 𝟑
(𝟐𝒌 + 𝟐)𝟐
∞
𝒌=𝟎
=𝟏
𝟐𝝅𝟑∑[
(−𝟏)𝒌
𝟐𝒌 + 𝟐+
(−𝟏)𝒌
(𝟐𝒌 + 𝟐)𝟐]
∞
𝒌=𝟎
=
=𝟏
𝟒𝝅𝟑∑(−𝟏)𝒌−𝟏
𝒌
∞
𝒌=𝟏
+𝟏
𝟖𝝅𝟑∑(−𝟏)𝒌−𝟏
𝒌𝟐
∞
𝒌=𝟏
=𝐥𝐨𝐠 𝟐
𝟒𝝅𝟑+
𝟏
𝟗𝟔𝝅
Therefore,
∫𝒅𝒙
(𝟒 𝐥𝐨𝐠𝟐 𝒙 + 𝝅𝟐)𝟐(𝒙𝟐 + 𝟏)
∞
𝟎
=𝐥𝐨𝐠 𝟐
𝟒𝝅𝟑+
𝟏
𝟗𝟔𝝅
1519. Prove that:
𝟏
𝒆< |∫ 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙) 𝑱𝟎(𝟐√𝒙)
∞
𝟎
𝒅𝒙| <𝝅𝟐
𝟔
where 𝑱𝒏(𝒙) is the Bessel function of order 𝒏.
Proposed by Angad Singh-India
Solution 1 by proposer
We know from the definition of Bessel function, that
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30 RMM-CALCULUS MARATHON 1501-1600
𝑱𝟎(𝒙) = ∑(−𝟏)𝒌
𝒌!𝟐
∞
𝒌=𝟎
(𝒙
𝟐)𝟐𝒌 𝒏>𝟎⇒ ∫ 𝒆−𝒏𝒙𝑱𝟎(𝟐√𝒙)
∞
𝟎
𝒅𝒙 =𝒆−𝟏𝒏
𝒏⇒
∫ ∑𝒆−𝒏𝒙
𝒏𝑱𝟎(𝟐√𝒙)
∞
𝒏=𝟎
𝒅𝒙∞
𝟎
= ∑𝒆−𝟏𝒏
𝒏𝟐
∞
𝒏=𝟏
Observe that:
∑𝒆−𝟏𝒏
𝒏𝟐
∞
𝒏=𝟏
>𝟏
𝒆; ∑
𝒆−𝟏𝒏
𝒏𝟐
∞
𝒏=𝟏
<∑𝟏
𝒏𝟐
∞
𝒏=𝟏
=𝝅𝟐
𝟔 𝐚𝐧𝐝 ∑
𝒆−𝒙
𝒏
∞
𝒏=𝟏
= − 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙)
Solution 2 by Akerele Olofin-Nigeria
𝑱𝒏(𝒙) = (𝒙
𝟐)𝒏
∑(−𝟏)𝒌𝒙𝟐𝒌
𝟐𝒌𝒌! 𝚪(𝒏 + 𝒌 + 𝟏)
∞
𝒌=𝟎
⇒ 𝑱𝟎(𝟐√𝒙) =∑(−𝟏)𝒌(𝟐√𝒙)
𝟐𝒌
𝟐𝒌𝒌! 𝚪(𝒌 + 𝟏)
∞
𝒌=𝟎
=∑(−𝟏)𝒌
𝒌!𝟐𝒙𝒌
∞
𝒌=𝟎
⇒ |∫ 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙) 𝑱𝟎(𝟐√𝒙)∞
𝟎
𝒅𝒙| = |∑(−𝟏)𝒌
𝒌!𝟐
∞
𝒌=𝟎
∫ 𝒙𝒌 ⋅ 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙)∞
𝟎
𝒅𝒙| =
= |∑(−𝟏)𝒌+𝟏
𝒌!𝟐
∞
𝒌=𝟎
𝑳𝒊𝒌+𝟐(𝟏)(𝒌!)| = |∑(−𝟏)𝒌𝑳𝒊𝒌+𝟐(𝟏)
𝒌!
∞
𝒌=𝟎
| = |∑(−𝟏)𝒌+𝟏
𝒌!𝜻(𝒌 + 𝟐)
∞
𝒌=𝟎
| =
= |∑𝟏
𝒏𝟐
∞
𝒌=𝟏
∑(−𝟏)𝒌+𝟏
𝒏𝒌(𝒌!)
∞
𝒏=𝟎
| = |−∑𝟏
𝒏𝟐𝒆−𝟏𝒏
∞
𝒌=𝟎
| = ∑ |−𝒆−
𝟏𝒏
𝒏𝟐|
𝒏
𝒌=𝟏
= ∑𝒆−𝟏𝒏
𝒏𝟐
∞
𝒏=𝟏
≅ 𝟎.𝟖𝟒𝟔𝟒𝟐
𝟏
𝒆≅ 𝟎. 𝟑𝟔𝟕𝟖𝟕;
𝝅𝟐
𝟔≅ 𝟏. 𝟔𝟒𝟒𝟗 ⇒ ∑
𝒆−𝟏𝒏
𝒏𝟐
∞
𝒏=𝟏
>𝟏
𝒆; ∑
𝒆−𝟏𝒏
𝒏𝟐
∞
𝒏=𝟏
<𝝅𝟐
𝟔
Therefore,
𝟏
𝒆< |∫ 𝐥𝐨𝐠(𝟏 − 𝒆−𝒙) 𝑱𝟎(𝟐√𝒙)
∞
𝟎
𝒅𝒙| <𝝅𝟐
𝟔
1520. Let 𝒃 > 𝒂 > 𝟏 and 𝒏 be a positive integer. Prove that:
∫√𝒆𝒏𝒙
𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏
𝐥𝐨𝐠𝒃
𝐥𝐨𝐠𝒂
𝒅𝒙 ≤ 𝐥𝐨𝐠( √𝒃
𝒂
𝒏+𝟏
) , 𝒙 ∈ ℝ
Proposed by George Apostolopoulos-Messolonghi-Greece
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31 RMM-CALCULUS MARATHON 1501-1600
Solution 1 by Adrian Popa-Romania
𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏 ≥ (𝒏 + 𝟏) √𝒆𝒏𝒙+(𝒏−𝟏)𝒙+⋯+𝒙+𝟎𝒏+𝟏
=
= (𝒏 + 𝟏) √𝒆𝒏𝒙(𝒏+𝟏)
𝟐
𝒏+𝟏
= (𝒏 + 𝟏)𝒆𝒏𝒙𝟐 = (𝒏 + 𝟏)√𝒆𝒏𝒙 ⇒
∫√𝒆𝒏𝒙
𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 ≤𝟏
𝒏 + 𝟏∫
√𝒆𝒏𝒙
√𝒆𝒏𝒙
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 =
=𝟏
𝒏 + 𝟏𝒙|𝐥𝐨𝐠 𝒂
𝐥𝐨𝐠 𝒃
=𝟏
𝒏 + 𝟏(𝐥𝐨𝐠 𝒃 − 𝐥𝐨𝐠𝒂) =
𝟏
𝒏 + 𝟏𝐥𝐨𝐠 (
𝒃
𝒂) = 𝐥𝐨𝐠( √
𝒃
𝒂
𝒏+𝟏
)
Therefore,
∫√𝒆𝒏𝒙
𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 ≤ 𝐥𝐨𝐠( √𝒃
𝒂
𝒏+𝟏
) , 𝒙 ∈ ℝ
Solution 2 by Ruxandra Daniela Tonilă-Romania
∫√𝒆𝒏𝒙
𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 = ∫√𝒆𝒏𝒙
𝒆(𝒏+𝟏)𝒙 − 𝟏𝒆𝒙 − 𝟏
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 =
= ∫𝒆𝒏𝒙(𝒆𝒙 − 𝟏)
𝒆(𝒏+𝟏)𝒙 − 𝟏⋅𝟏
√𝒆𝒏𝒙
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 = ∫𝒆(𝒏+𝟏)𝒙 − 𝒆𝒏𝒙
𝒆(𝒏+𝟏)𝒙 − 𝟏⋅𝟏
√𝒆𝒏𝒙
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 =
= ∫𝒆(𝒏+𝟏)𝒙 (𝟏 − (
𝟏𝒆)𝒙
)
𝒆(𝒏+𝟏)𝒙(𝟏 − [(𝟏𝒆)𝒙
]𝒏+𝟏 ⋅
𝟏
√𝒆𝒏𝒙
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 =
= ∫𝟏 − (
𝟏𝒆)𝒙
(𝟏 − (𝟏𝒆)𝒙
) (𝟏 + (𝟏𝒆)𝒙
+ (𝟏𝒆)𝟐𝒙
+⋯+ (𝟏𝒆)𝒏𝒙
)
⋅𝟏
√𝒆𝒏𝒙
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 =
=𝟏
𝒏 + 𝟏∫
𝒏 + 𝟏
𝟏 + (𝟏𝒆)𝒙
+ (𝟏𝒆)𝟐𝒙
+⋯+ (𝟏𝒆)𝒏𝒙 ⋅
𝟏
√𝒆𝒏𝒙𝒅𝒙
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
≤𝑯𝑮𝑴
≤𝟏
𝒏 + 𝟏∫
𝟏
√𝒆𝒏𝒙√𝟏 ⋅ 𝒆𝒙 ⋅ 𝒆𝟐𝒙 ⋅ … ⋅ 𝒆𝒏𝒙
𝒏+𝟏𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 =𝟏
𝒏 + 𝟏∫
𝟏
√𝒆𝒏𝒙⋅ √𝒆
𝒏(𝒏+𝟏)𝒙𝟐
𝒏+𝟏
𝒅𝒙𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
=
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32 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝒏 + 𝟏(𝐥𝐨𝐠𝒃 − 𝐥𝐨𝐠 𝒂) = 𝐥𝐨𝐠( √
𝒃
𝒂
𝒏+𝟏
)
Therefore,
∫√𝒆𝒏𝒙
𝒆𝒏𝒙 + 𝒆(𝒏−𝟏)𝒙 +⋯+ 𝒆𝟐𝒙 + 𝒆𝒙 + 𝟏
𝐥𝐨𝐠 𝒃
𝐥𝐨𝐠 𝒂
𝒅𝒙 ≤ 𝐥𝐨𝐠( √𝒃
𝒂
𝒏+𝟏
) , 𝒙 ∈ ℝ
1521. If for 𝒏 ≥ 𝒌 and 𝒏, 𝒌, 𝒂 > 𝟎; 𝒏, 𝒌 ∈ ℕ;
𝝃𝒂(𝒌, 𝒏) =∑√𝒂𝒓
𝒏
𝒓=𝒌
= √𝒂𝒌 + √𝒂
𝒌+𝟏+⋯+ √𝒂
𝒏 𝐚𝐧𝐝 𝜻𝒏(𝒔) = ∑𝟏
𝒌𝒔
𝒏
𝒌=𝟏
=𝟏
𝟏𝒔+𝟏
𝟐𝒔+⋯+
𝟏
𝒏𝒔
then prove:
𝟐(√𝒏 + 𝟏 − 𝟏) + 𝝃𝒆(𝟐, 𝒏 + 𝟏) < 𝜻𝒏(𝟎) + 𝜻𝒏 (𝟏
𝟐) + 𝜻𝒏(𝟏) < 𝝃𝒆(𝟏, 𝒏) + 𝟐√𝒏
𝐰𝐡𝐞𝐫𝐞, 𝒆 = ∑𝟏
𝒏!
∞
𝒏=𝟎
Proposed by Amrit Awasthi-India
Solution by proposer
We can write 𝐥𝐨𝐠 𝒙 as 𝐥𝐨𝐠 𝒙 = ∫𝟏
𝒕𝒅𝒕
𝒙
𝟏 and 𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌) = ∫
𝟏
𝒕𝒅𝒕
𝟏+𝟏
𝒌𝟏
.
Now, as 𝒕 ∈ [𝟏, 𝟏 +𝟏
𝒌] : 𝟏 ≤ 𝒕 ≤ 𝟏 +
𝟏
𝒌⇒
𝟏
𝟏+𝟏
𝒌
≤𝟏
𝒕≤ 𝟏
Integrating under the same interval we have:
∫𝒌
𝒌 + 𝟏
𝟏+𝟏𝒌
𝟏
𝒅𝒕 ≤ ∫𝟏
𝒕
𝟏+𝟏𝒌
𝟏
𝒅𝒕 ≤ ∫ 𝒅𝒕𝟏+𝟏𝒌
𝟏
⇔
𝒌
𝒌 + 𝟏(𝟏 +
𝟏
𝒌− 𝟏) ≤ 𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌) ≤ (𝟏 +
𝟏
𝒌− 𝟏) ⇔
𝟏
𝒌 + 𝟏≤ 𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌) ≤
𝟏
𝒌
Now, summing up from 𝒌 = 𝟏 to 𝒌 = 𝒏 we have:
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33 RMM-CALCULUS MARATHON 1501-1600
∑𝒆𝟏𝒌+𝟏
𝒏
𝒌=𝟏
≤∑(𝟏 +𝟏
𝒌)
𝒏
𝒌=𝟏
≤∑𝒆𝟏𝒌
𝒏
𝒌=𝟏
⇔∑ √𝒆𝒌
𝒏+𝟏
𝒌=𝟐
≤∑𝟏
𝒌𝟎
𝒏
𝒌=𝟏
+∑𝟏
𝒌
𝒏
𝒌=𝟏
≤∑ √𝒆𝒌
𝒏
𝒌=𝟏
𝝃𝒆(𝟐, 𝒏 + 𝟏) ≤ 𝜻𝒏(𝟎) + 𝜻𝒏(𝟏) + 𝝃𝒆(𝟏, 𝒏); (∗)
Now, consider the following
𝟐(√𝒎+ 𝟏 − √𝒎) = 𝟐(√𝒎 + 𝟏 − √𝒎)(√𝒎+ 𝟏 + √𝒎)
√𝒎+ 𝟏 + √𝒎=
𝟐
√𝒎+ 𝟏 + √𝒎<
<𝟐
√𝒎+ √𝒎=𝟏
√𝒎
Thus, 𝟐(√𝒎+ 𝟏 − √𝒎) <𝟏
√𝒎
Proceeding in a similar manner we can prove that:
𝟐(√𝒎 − √𝒎− 𝟏) >𝟏
√𝒎
Combining both results, we have
𝟐(√𝒎+ 𝟏 − √𝒎) <𝟏
√𝒎< 𝟐(√𝒎− √𝒎 − 𝟏)
Summing from 𝒎 = 𝟏 to 𝒎 = 𝒏 we have
∑ 𝟐(√𝒎+ 𝟏 − √𝒎)
𝒏
𝒎=𝟏
< ∑𝟏
√𝒎
𝒏
𝒎=𝟏
< ∑ 𝟐(√𝒎− √𝒎− 𝟏)
𝒏
𝒎=𝟏
Now, both the upper and lower bounds are telescopic sums, hence after canceling the terms we are left with
𝟐(√𝒏 + 𝟏 − 𝟏) < 𝜻𝒏 (𝟏
𝟐) < 𝟐√𝒏; (∗∗)
Adding (∗), (∗∗) the final inequality becomes strict as the second inequality is strict. Hence,
𝟐(√𝒏 + 𝟏 − 𝟏) + 𝝃𝒆(𝟐, 𝒏 + 𝟏) < 𝜻𝒏(𝟎) + 𝜻𝒏 (𝟏
𝟐) + 𝜻𝒏(𝟏) < 𝝃𝒆(𝟏, 𝒏) + 𝟐√𝒏
1522. For 𝒏 ≥ 𝟎 prove or disprove:
𝐥𝐢𝐦𝒌→∞
𝐥𝐢𝐦𝒏→∞
∏(𝟏+∫ (𝟏 − 𝐭𝐚𝐧 𝒙
𝟏 + 𝐭𝐚𝐧 𝒙)𝒏𝝅
𝟒
𝟎
𝒅𝒙)
𝒏𝒎𝒌
𝒎=𝟏
𝟏
√𝒌= 𝒆
𝜸𝟐
where 𝜸 is Euler-Mascheroni constant.
Proposed by Naren Bhandari-Bajura-Nepal
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34 RMM-CALCULUS MARATHON 1501-1600
Solution by Artan Ajredini-Presheva-Serbie
Let 𝒇𝒏(𝒙) = (𝟏−𝐭𝐚𝐧 𝒙
𝟏+𝐭𝐚𝐧 𝒙)𝒏
. For 𝒙 ∈ [𝟎,𝝅
𝟒] we have |𝒇𝒏(𝒙)| ≤ 𝟏 = 𝒈(𝒙), and
∫ 𝒈(𝒙)
𝝅𝟒
𝟎
𝒅𝒙 =𝝅
𝟒. 𝐍𝐨𝐰,
𝐥𝐢𝐦𝒏→∞
𝒇𝒏(𝒙) = {𝟎, 𝒙 ∈ (𝟎,
𝝅
𝟒]
𝟏, 𝒙 = 𝟎
By Lebesgue Dominated Convergence Theorem, we have that:
𝐥𝐢𝐦𝒏→∞
∫ 𝒇𝒏(𝒙)
𝝅𝟒
𝟎
𝒅𝒙 = 𝟎 ⋅ 𝝌(𝟎,𝝅𝟒]+ 𝟏 ⋅ 𝝌{𝟏} = 𝟎; (𝟏)
On the other hand, substituting 𝒖 =𝟏−𝐭𝐚𝐧 𝒙
𝟏+𝐭𝐚𝐧 𝒙 in the integral 𝑰 = ∫ 𝒏 (
𝟏−𝐭𝐚𝐧 𝒙
𝟏+𝐭𝐚𝐧 𝒙)𝒏𝝅
𝟒𝟎
𝒅𝒙 and
integrating by parts, we get
𝑰 =𝟏
𝟐−∫
𝒖𝒏(𝟏 − 𝒖𝟐)
𝟏 + 𝒖𝟐
𝟏
𝟎
𝒅𝒖
Again, let 𝒈𝒏(𝒖) =𝒖𝒏(𝟏−𝒖𝟐)
𝟏+𝒖𝟐. For 𝒖 ∈ [𝟎, 𝟏] we have that 𝒈𝒏(𝒖) ≤
𝟏−𝒖𝟐
𝟏+𝒖𝟐= 𝒉(𝒖), and
∫ 𝒉(𝒖)𝟏
𝟎
𝒅𝒖 =𝝅 − 𝟐
𝟐.𝐍𝐨𝐰, 𝐥𝐢𝐦
𝒏→∞𝒈𝒏(𝒖) = 𝟎, ∀𝒖 ∈ [𝟎, 𝟏], 𝐚𝐧𝐝
using Lebesgue Dominated Convergence Theorem, we obtain
𝐥𝐢𝐦𝒏→∞
∫ 𝒈𝒏(𝒖)𝟏
𝟎
𝒅𝒖 = 𝟎. 𝐓𝐡𝐞𝐫𝐞𝐟𝐨𝐫𝐞,
𝐥𝐢𝐦𝒏→∞
∫ 𝒏(𝟏 − 𝐭𝐚𝐧 𝒙
𝟏 + 𝐭𝐚𝐧 𝒙)𝒏
𝝅𝟒
𝟎
𝒅𝒙 =𝟏
𝟐; (𝟐)
By using (𝟏), (𝟐), we get
𝐥𝐢𝐦𝒌→∞
𝐥𝐢𝐦𝒏→∞
∏(𝟏+∫ (𝟏 − 𝐭𝐚𝐧 𝒙
𝟏 + 𝐭𝐚𝐧 𝒙)𝒏
𝝅𝟒
𝟎
𝒅𝒙)
𝒏𝒎𝒌
𝒎=𝟏
𝟏
√𝒌=
= 𝐥𝐢𝐦𝒌→∞
∏(𝐥𝐢𝐦𝒏→∞
(𝟏 + ∫ (𝟏− 𝐭𝐚𝐧 𝒙
𝟏 + 𝐭𝐚𝐧 𝒙)𝒏
𝒅𝒙
𝝅𝟒
𝟎
)
𝒏𝒎𝒌
𝒎=𝟏
𝟏
√𝒌=
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35 RMM-CALCULUS MARATHON 1501-1600
= 𝐥𝐢𝐦𝒌→∞
∏(𝐞𝐱𝐩 {𝐥𝐢𝐦𝒏→∞
𝟏
𝒎∫ 𝒏(
𝟏− 𝐭𝐚𝐧 𝒙
𝟏 + 𝐭𝐚𝐧 𝒙)𝒏
𝒅𝒙
𝝅𝟒
𝟎
})
𝒌
𝒎=𝟏
𝟏
√𝒌= 𝐥𝐢𝐦𝒌→∞
∏𝟏
√𝒌𝒆𝟏𝟐𝒎
𝒌
𝒎=𝟏
=
= 𝐥𝐢𝐦𝒌→∞
∏𝒆𝟏𝟐(𝟏𝒎−𝐥𝐨𝐠 𝒌)
𝒌
𝒎=𝟏
= 𝒆𝜸𝟐
1523.
𝝓𝒏 =𝟒
𝝅∫
𝐜𝐨𝐭𝐡(𝒏𝒙−𝟏) − 𝒙𝒏−𝟏
𝒏(𝟏 + 𝒙𝟐)𝟐𝒅𝒙
∞
𝟎
, 𝚽𝒏 =𝐜𝐨𝐬(𝒏𝝅)
𝒏; ∀𝒏 ∈ ℕ
Prove that:
𝐥𝐢𝐦𝒎→∞
𝐥𝐢𝐦𝒏→∞
∏∏(𝟏 +𝝓𝒏)𝒏𝒌−𝒓𝚽𝒓
𝒎
𝒓=𝟐
∞
𝒌=𝟏
= 𝒆𝜸
where 𝜸 is Euler-Mascheroni constant.
Proposed by Surjeet Singhania-India
Solution by Naren Bhandari-Bajura-Nepal
Since we have two convergent integrals so we have validity for the linearity of integrals for
𝝓𝒏 that is
𝝅
𝟒𝝓𝒏 = ∫
𝐜𝐨𝐭𝐡 (𝒏𝒙)
𝒏(𝟏 + 𝒙𝟐)𝟐
∞
𝟎
𝒅𝒙⏟
𝑰𝟏
− ∫𝒙𝒅𝒙
𝒏𝟐(𝟏 + 𝒙𝟐)𝟐
∞
𝟎⏟ 𝑰𝟐
Now, let 𝝀𝒏(𝒙) =𝐜𝐨𝐭𝐡(
𝒏
𝒙)
𝒏(𝟏+𝒙𝟐)𝟐 ≤ |
𝟏
𝒏(𝟏+𝒙𝟐)𝟐| ≤ |
𝟏
𝟐𝒏(𝟏+𝒙𝟐)| = 𝑽𝒏(𝒙) and hence
∫ 𝑽𝒏(𝒙)∞
𝟎
𝒅𝒙 =𝝅
𝟒𝒏 𝐚𝐧𝐝 𝐥𝐢𝐦
𝒏→∞𝑽𝒏(𝒙) = 𝟎.
By Lebesgue Dominating convergence theorem, we have:
𝐥𝐢𝐦𝒏→∞
∫ 𝝀𝒏(𝒙)∞
𝟎
𝒅𝒙 = 𝐥𝐢𝐦𝒏→∞
∫ 𝑽𝒏(𝒙)∞
𝟎
𝒅𝒙 = 𝟎; (𝟏)
The integral 𝑰𝟐 has primitive which is
∫𝒙𝒅𝒙
𝒏𝟐(𝟏 + 𝒙𝟐)
∞
𝟎
= ∫𝟐𝒙𝒅𝒙
𝟐𝒏𝟐(𝟏 + 𝒙𝟐)𝟐
∞
𝟎
= −𝟏
𝟐𝒏𝟐(𝟏 + 𝒙𝟐)𝟐|𝟎
∞
=𝟏
𝟐𝒏𝟐
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36 RMM-CALCULUS MARATHON 1501-1600
Giving us 𝐥𝐢𝐦𝒏→∞
𝑰𝟐 = 𝟎 (which we can even perform to show that
𝑰𝟐 = 𝟎 as 𝒏 → ∞ even by LTCD) ;(2)
So, from (1),(2)we have:
𝐥𝐢𝐦𝒏→∞
(𝟏 + 𝝓𝒏)𝒏 ≤ 𝐥𝐢𝐦
𝒏→∞(𝟏 + 𝑰𝟏 − 𝑰𝟐)
𝒏 = 𝐥𝐢𝐦𝒏→∞
(𝟏 +𝟏
𝒏−𝟐
𝒏𝝅𝟐)𝒏
= 𝒆
Now, we have just to evaluate
𝐞𝐱𝐩 ( 𝐥𝐢𝐦𝒎→∞
∑∑𝐜𝐨𝐬(𝝅𝒓)
𝒌𝒓𝒓
𝒎
𝒓=𝟐
∞
𝒌=𝟏
) = 𝐞𝐱𝐩 (𝟏 −∑𝜻(𝒌) − 𝟏
𝒌
∞
𝒌=𝟐
)
= 𝐞𝐱𝐩 (𝟏 −∑𝒙𝒌
𝒌!⋅
𝒅𝒙
𝒙𝒆𝒙(𝒆𝒙 − 𝟏)
∞
𝒌=𝟐
)
Interchanging sum and integral signs we have
∫𝒆𝒙 − 𝒙 − 𝟏
𝒙𝒆𝒙(𝒆𝒙 − 𝟏)
∞
𝟎
𝒅𝒙 = ∫ 𝒆−𝒙𝒅𝒙∞
𝟎
−∫ (𝟏
𝒆𝒙 − 𝟏−𝟏
𝒙𝒆𝒙)
∞
𝟎
𝒅𝒙 = 𝟏 − 𝜸
Since it is well known that ∫ (𝟏
𝒆𝒙−𝟏−
𝟏
𝒙𝒆𝒙)
∞
𝟎𝒅𝒙 = 𝜸 and hence we have desired result 𝒆𝜸.
In other way we calculate
∑𝜻(𝒌) − 𝟏
𝒌
∞
𝒌=𝟐
= ∑∑𝟏
𝒌𝒎𝒌
𝒌≥𝟐𝒎≥𝟐
= 𝐥𝐢𝐦𝑵→∞
∑ (𝟏
𝒎− 𝐥𝐨𝐠𝒎 + 𝐥𝐨𝐠(𝒎+ 𝟏))
𝑵
𝒎=𝟐
=
= − 𝐥𝐢𝐦𝑵→∞
(𝑯𝑵 − 𝟏 − 𝐥𝐨𝐠𝑵) =𝟏 − 𝜸.
1524. Find:
𝛀 = 𝐥𝐢𝐦𝛆→𝟎𝛆>0
∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝜺
𝒅𝒙
Proposed by Vasile Mircea Popa-Romania
Solution 1 by Soumitra Mandal-India
𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝟎
𝒅𝒙 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
𝟏
𝟎
𝒅𝒙 +∫𝒙𝟐 𝐥𝐨𝐠𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝟏
𝒅𝒙 =𝒙=𝟏𝒚
= ∫𝒙𝟐(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔
𝟏
𝟎
𝒅𝒙 −∫𝐥𝐨𝐠𝒚
𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚
𝟏
𝟎
=
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37 RMM-CALCULUS MARATHON 1501-1600
= ∫𝒙𝟐(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔
𝟏
𝟎
𝒅𝒙 − ∫(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔
𝟏
𝟎
=
= 𝟐∫𝒙𝟐 𝐥𝐨𝐠𝒙
𝟏 − 𝒙𝟔
𝟏
𝟎
𝒅𝒙 −∫𝒙𝟒 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔
𝟏
𝟎
𝒅𝒙 − ∫𝐥𝐨𝐠𝒙
𝟏 − 𝒙𝟔
𝟏
𝟎
𝒅𝒙 =
𝟐∑∫ 𝒙𝟔𝒏+𝟐 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
∞
𝒏=𝟎
−∑∫ 𝒙𝟔𝒏+𝟒 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
∞
𝒏=𝟎
−∑∫ 𝒙𝟔𝒏 𝐥𝐨𝐠 𝒙𝟏
𝟎
𝒅𝒙
∞
𝒏=𝟎
=
= −𝟐∑𝟏
(𝟔𝒏 + 𝟑)𝟐
∞
𝒏=𝟎
+∑𝟏
(𝟔𝒏 + 𝟓)𝟐
∞
𝒏=𝟎
+∑𝟏
(𝟔𝒏 + 𝟏)𝟐
∞
𝒏=𝟎
=
= −𝝍(𝟏) (
𝟏𝟐)
𝟏𝟖+𝝍(𝟏) (
𝟓𝟔)
𝟑𝟔+𝝍(𝟏) (
𝟏𝟔)
𝟑𝟔= −
𝝅𝟐
𝟑𝟔+𝝍(𝟏) (𝟏 −
𝟏𝟔) + 𝝍
(𝟏) (𝟏𝟔)
𝟑𝟔=𝝅𝟐
𝟗−𝝅𝟐
𝟑𝟔=𝝅𝟐
𝟏𝟐
Solution 2 by Ajetunmobi Abdulquyyum-Nigeria
𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝟎
𝒅𝒙 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
𝟏
𝟎
𝒅𝒙 + ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝟏
𝒅𝒙⏟
𝑨
𝑨 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝟏
𝒅𝒙 =𝒙=𝟏𝒕−∫
𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐 + 𝒙𝟒
𝟏
𝟎
𝒅𝒙 ⇒
𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
𝟏
𝟎
𝒅𝒙 − ∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐 + 𝒙𝟒
𝟏
𝟎
𝒅𝒙 =
= ∫𝒙𝟐(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
−∫(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐 + 𝒙𝟒𝒅𝒙
𝟏
𝟎
=
= ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔
𝟏
𝟎
𝒅𝒙 − ∫𝒙𝟒 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
−∫𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
+ ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
=
= 𝟐∫𝒙𝟐 𝐥𝐨𝐠𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
−∫𝒙𝟒 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
− ∫𝐥𝐨𝐠𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
= 𝑩− 𝑪 −𝑫
𝑩 = 𝟐∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
=𝒙𝟔=𝒛 𝟏
𝟏𝟖∫𝒛𝟏𝟐−𝟏 𝐥𝐨𝐠 𝒛
𝟏 − 𝒛
𝟏
𝟎
𝒅𝒛 = −𝟏
𝟏𝟖𝝍(𝟏) (
𝟏
𝟐)
𝑪 = ∫𝒙𝟒 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
= ∫𝒛𝟐𝟑 𝐥𝐨𝐠 (𝒛
𝟏𝟔)
𝟏 − 𝒛⋅𝟏
𝟔𝒛−𝟓𝟔
𝟏
𝟎
𝒅𝒛 =𝟏
𝟑𝟔∫𝒛𝟓𝟔−𝟏 𝐥𝐨𝐠 𝒛
𝟏 − 𝒛𝒅𝒛
𝟏
𝟎
= −𝟏
𝟑𝟔𝝍(𝟏) (
𝟓
𝟔)
𝑫 = ∫𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
=𝟏
𝟑𝟔∫𝒛−𝟓𝟔 𝐥𝐨𝐠 𝒛
𝟏 − 𝒛𝒅𝒛
𝟏
𝟎
=𝟏
𝟑𝟔∫𝒛𝟏𝟔−𝟏 𝐥𝐨𝐠 𝒛
𝟏 − 𝒛
𝟏
𝟎
𝒅𝒛 = −𝟏
𝟑𝟔𝝍(𝟏) (
𝟏
𝟔)
Thus,
𝛀 = 𝑩 − 𝑪 −𝑫 = −𝟐
𝟑𝟔𝝍(𝟏) +
𝟏
𝟑𝟔𝝍(𝟏) (
𝟓
𝟔) +
𝟏
𝟑𝟔𝝍(𝟏) (
𝟏
𝟔) =
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38 RMM-CALCULUS MARATHON 1501-1600
𝝅𝟐
𝟗−𝝅𝟐
𝟑𝟔=𝝅𝟐
𝟏𝟐
Therefore,
𝛀 = 𝐥𝐢𝐦𝛆→𝟎𝛆>0
∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝜺
𝒅𝒙 =𝝅𝟐
𝟏𝟐
Solution 3 by Ose Favour-Nigeria
𝛀 = 𝐥𝐢𝐦𝛆→𝟎𝛆>0
∫𝒙𝟐 𝐥𝐨𝐠𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝜺
𝒅𝒙 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
𝟏
𝟎
𝒅𝒙 +∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝟏
𝒅𝒙 = 𝚽 +𝚿
𝚿 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏𝒅𝒙
∞
𝟏
=𝒖=𝟏𝒙− ∫
𝐥𝐨𝐠 𝒖
𝒖𝟒 + 𝒖𝟐 + 𝟏𝒅𝒖
𝟏
𝟎
𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙 − 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏𝒅𝒙
𝟏
𝟎
= ∫(𝟏 − 𝒙𝟐)(𝒙𝟐 − 𝟏) 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
= −∫(𝒙𝟐 − 𝟏)𝟐 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
=𝒕=𝒙𝟔
−𝟏
𝟑𝟔∫(𝒕𝟒𝟔−𝟓𝟔 − 𝟐𝒕
𝟐𝟔−𝟓𝟔 + 𝒕−
𝟓𝟔) 𝐥𝐨𝐠 𝒕
𝟏 − 𝒕𝒅𝒕
𝟏
𝟎
𝐋𝐞𝐭: 𝛀(𝒏) = ∫𝒕−𝟏𝟔+𝒏 − 𝟐𝒕−
𝟏𝟐+𝒏 + 𝒕−
𝟓𝟔+𝒏
𝟏 − 𝒕𝒅𝒕
𝟏
𝟎
𝛀 = −𝟏
𝟑𝟔𝛀′(𝟎)
𝛀(𝒏) = ∫𝒕−𝟏𝟔+𝒏 − 𝟐𝒕−
𝟏𝟐+𝒏 + 𝒕−
𝟓𝟔+𝒏
𝟏 − 𝒕𝒅𝒕
𝟏
𝟎
= −𝝍(𝟎) (𝒏 +𝟏
𝟔) −𝝍(𝟎) (𝒏 +
𝟓
𝟔) + 𝟐𝝍(𝟎) (𝒏 +
𝟏
𝟐)
𝛀′(𝒏) = −𝝍(𝟎)′(𝒏 +
𝟏
𝟔) −𝝍(𝟎)
′(𝒏 +
𝟓
𝟔) + 𝟐𝝍(𝟎)
′(𝒏 +
𝟏
𝟐)
𝛀′(𝟎) = −𝝍(𝟎)′(𝟏
𝟔) − 𝝍(𝟎)
′(𝟓
𝟔) + 𝟐𝝍(𝟎)
′(𝟏
𝟐) = −(𝝅𝟐 𝐜𝐬𝐜𝟐 (
𝝅
𝟔)) + 𝟐
𝝅𝟐
𝟐= −𝟑𝝅𝟐
𝛀 = −𝟏
𝟑𝟔𝛀′(𝟎) =
𝟏
𝟑𝟔⋅ 𝟑𝝅𝟐 =
𝝅𝟐
𝟏𝟐
Solution 4 by Mohammad Rostami-Afghanistan
𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝟎
𝒅𝒙 =𝒙=𝟏𝒚∫
−𝐥𝐨𝐠 𝒚
𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚
∞
𝟎
= ∫− 𝐥𝐨𝐠 𝒚
𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚
𝟏
𝟎
+∫− 𝐥𝐨𝐠 𝒚
𝒚𝟒 + 𝒚𝟐 + 𝟏
∞
𝟏
= 𝑰𝟏 + 𝑰𝟐
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39 RMM-CALCULUS MARATHON 1501-1600
𝑰𝟏 = ∫− 𝐥𝐨𝐠𝒚
𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚
𝟏
𝟎
= ∫−(𝟏 − 𝒚𝟐) 𝐥𝐨𝐠𝒚
𝟏 − 𝒚𝟔
𝟏
𝟎
𝒅𝒚 = −∫𝐥𝐨𝐠 𝒚
𝟏 − 𝒚𝟔𝒅𝒚
𝟏
𝟎
+∫𝒚𝟐 𝐥𝐨𝐠 𝒚
𝟏 − 𝒚𝟔𝒅𝒚
𝟏
𝟎
=
= −∫ ∑𝒚𝟔𝒏 𝝏
𝝏𝒂|𝒂=𝟎𝒚𝒂𝒅𝒚
∞
𝒏=𝟎
𝟏
𝟎
+∫ 𝒚𝟐∑𝒚𝟔𝒌𝝏
𝝏𝒃|𝒃=𝟎𝒚𝒃
∞
𝒌=𝟎
𝒅𝒚𝟏
𝟎
= −∑𝝏
𝝏𝒂|𝒂=𝟎
∞
𝒏=𝟎
∫ 𝒚𝟔𝒏+𝒂𝒅𝒚𝟏
𝟎
+
+∑𝝏
𝝏𝒃|𝒃=𝟎
∞
𝒌=𝟎
∫ 𝒚𝟔𝒌+𝒃+𝟐𝒅𝒚𝟏
𝟎
= −∑ [𝟏
𝟔𝒏 + 𝒂 + 𝟏]𝒂=𝟎
′∞
𝒏=𝟎
+∑[𝟏
𝟔𝒌 + 𝒃 + 𝟑]𝒃=𝟎
′∞
𝒌=𝟎
=
= ∑𝟏
(𝟔𝒏 + 𝟏)𝟐
∞
𝒏=𝟎
−∑𝟏
(𝟔𝒌 + 𝟑)𝟐
∞
𝒌=𝟎
=𝟏
𝟑𝟔[∑
𝟏
(𝒏 +𝟏𝟔)𝟐
∞
𝒏=𝟎
−∑𝟏
(𝒌 +𝟏𝟐)𝟐
∞
𝒌=𝟎
] =
=𝟏
𝟑𝟔[𝝍(𝟏) (
𝟏
𝟔) − 𝝍(𝟏) (
𝟏
𝟐)]
𝑰𝟐 = −∫𝐥𝐨𝐠 𝒚
𝒚𝟒 + 𝒚𝟐 + 𝟏𝒅𝒚
∞
𝟏
=𝒚=𝟏𝒙∫
𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏𝒅𝒙
𝟏
𝟎
= ∫𝒙𝟐(𝟏 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
=
= ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
−∫𝒙𝟒 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟔𝒅𝒙
𝟏
𝟎
=
= ∫ 𝒙𝟐∑𝒙𝟔𝒏𝝏
𝝏𝒂|𝒂=𝟎𝒙𝒂𝒅𝒙
∞
𝒏=𝟎
𝟏
𝟎
−∫ 𝒙𝟒∑𝒙𝟔𝒌𝝏
𝝏𝒃|𝒃=𝟎𝒙𝒃
∞
𝒌=𝟎
𝒅𝒙𝟏
𝟎
=
= ∑𝝏
𝝏𝒂|𝒂=𝟎
∫ 𝒙𝟔𝒏+𝒂+𝟐𝒅𝒙𝟏
𝟎
∞
𝒏=𝟎
−∑𝝏
𝝏𝒃|𝒃=𝟎
∞
𝒌=𝟎
∫ 𝒙𝟔𝒌+𝒃+𝟒𝟏
𝟎
𝒅𝒙 =
= ∑[𝟏
𝟔𝒏 + 𝒂 + 𝟑]𝒂=𝟎
′∞
𝒏=𝟎
−∑[𝟏
𝟔𝒌 + 𝒃 + 𝟓]𝒃=𝟎
′∞
𝒌=𝟎
= −∑𝟏
(𝟔𝒏 + 𝟑)𝟐
∞
𝒏=𝟎
+∑𝟏
(𝟔𝒏 + 𝟓)𝟐
∞
𝒌=𝟎
=
=𝟏
𝟑𝟔[∑
𝟏
(𝒌 +𝟓𝟔)𝟐
∞
𝒌=𝟎
−∑𝟏
(𝒏 +𝟏𝟐)𝟐
∞
𝒏=𝟎
] =𝟏
𝟑𝟔[𝝍(𝟏) (
𝟓
𝟔) − 𝝍(𝟏) (
𝟏
𝟐)]
Therefore,
𝛀 = 𝑰𝟏 + 𝑰𝟐 =𝟏
𝟑𝟔[𝝍(𝟏) (
𝟓
𝟔) ∓ 𝝍(𝟏) (
𝟓
𝟔) − 𝟐𝝍(𝟏) (
𝟏
𝟐)]
Solution 5 by Kartick Chandra Betal-India
𝛀 = ∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟒 + 𝒙𝟐 + 𝟏
∞
𝟎
𝒅𝒙 = −∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐 + 𝒙𝟒𝒅𝒙
∞
𝟎
𝟐𝛀 = 𝟐∫(𝒙𝟐 − 𝟏) 𝐥𝐨𝐠 𝒙
𝟏 + 𝒙𝟐 + 𝒙𝟒𝒅𝒙
𝟏
𝟎
= 𝟐∫(𝟏 −
𝟏𝒙𝟐) 𝐥𝐨𝐠 𝒙
(𝒙 +𝟏𝒙)𝟏
− 𝟏
𝟏
𝟎
𝒅𝒙
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40 RMM-CALCULUS MARATHON 1501-1600
𝛀 =𝟏
𝟐𝐥𝐨𝐠(
𝒙𝟐 − 𝒙 + 𝟏
𝒙𝟐 + 𝒙 + 𝟏) 𝐥𝐨𝐠 𝒙|
𝟎
𝟏
−𝟏
𝟐∫ 𝐥𝐨𝐠(
𝒙𝟐 − 𝒙 + 𝟏
𝒙𝟐 + 𝒙 + 𝟏)𝒅𝒙
𝒙
𝟏
𝟎
=
= −𝟏
𝟐∫ {𝐥𝐨𝐠(
𝟏 + 𝒙𝟑
𝟏 − 𝒙𝟑) + 𝐥𝐨𝐠 (
𝟏 − 𝒙
𝟏+ 𝒙)}𝒅𝒙
𝟏
𝟎
=
=𝟏
𝟔∫ 𝐥𝐨𝐠 (
𝟏 − 𝒙
𝟏+ 𝒙)𝒅𝒙
𝒙
𝟏
𝟎
−𝟏
𝟐∫𝐥𝐨𝐠 (
𝟏 − 𝒙𝟏 + 𝒙)
𝒙𝒅𝒙
𝟏
𝟎
=
=𝟏
𝟑∫𝐥𝐨𝐠(𝟏 + 𝒙)
𝒙𝒅𝒙
𝟏
𝟎
−𝟏
𝟑∫𝐥𝐨𝐠(𝟏 − 𝒙)
𝒙𝒅𝒙
𝟏
𝟎
=
=𝟏
𝟑𝜼(𝟐) +
𝟏
𝟑𝜻(𝟐) =
𝟏
𝟐𝜻(𝟐) =
𝝅𝟐
𝟏𝟐
1525. Find a closed form:
𝛀 = ∑ 𝐭𝐚𝐧−𝟏 ((𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟓) ⋅ 𝒏!
𝟏 + (𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ (𝒏!)𝟐)
∞
𝒏=𝟎
Proposed by Daniel Sitaru-Romania
Solution 1 by Asmat Qatea-Afghanistan
∵ 𝐭𝐚𝐧−𝟏 𝒙 − 𝐭𝐚𝐧−𝟏 𝒚 = 𝐭𝐚𝐧−𝟏 (𝒙 − 𝒚
𝟏+ 𝒙𝒚),
∵ 𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔 = (𝒏 + 𝟏)(𝒏 + 𝟐)(𝒏 + 𝟑)
𝛀 = ∑𝐭𝐚𝐧−𝟏 ((𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟓) ⋅ 𝒏!
𝟏 + (𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ (𝒏!)𝟐)
∞
𝒏=𝟎
=
= ∑𝐭𝐚𝐧−𝟏 ((𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ 𝒏! − 𝒏!
𝟏 + (𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ (𝒏!)𝟐)
∞
𝒏=𝟎
=
= ∑𝐭𝐚𝐧−𝟏 ((𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔) ⋅ 𝒏!)
∞
𝒏=𝟎
−∑𝐭𝐚𝐧−𝟏(𝒏!)
∞
𝒏=𝟎
=
= ∑𝐭𝐚𝐧−𝟏((𝒏 + 𝟑)!) −
∞
𝒏=𝟎
∑𝐭𝐚𝐧−𝟏(𝒏!)
∞
𝒏=𝟎
=
= ∑ 𝐭𝐚𝐧−𝟏((𝒏 + 𝟑)!)
∞
𝒏=𝟎
− 𝟐 𝐭𝐚𝐧−𝟏(𝟏) − 𝐭𝐚𝐧−𝟏(𝟐) −∑𝐭𝐚𝐧−𝟏(𝒏!)
∞
𝒏=𝟑
= 𝝅− 𝐭𝐚𝐧−𝟏(𝟐)
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41 RMM-CALCULUS MARATHON 1501-1600
Solution 2 by Naren Bhandari-Bajura-Nepal
Note the 𝑮(𝒏) = 𝒏𝟑 + 𝟔𝒏𝟐 + 𝟏𝟏𝒏 + 𝟔 = (𝒏 + 𝟏)(𝒏+ 𝟐)(𝒏 + 𝟑) and it is easy to see that
∑𝒕𝒂𝒏−𝟏 ((𝑮(𝒏) − 𝟏)𝒏!
𝟏 + 𝑮(𝒏)(𝒏!)𝟐)
∞
𝒏=𝟎
= ∑(𝒕𝒂𝒏−𝟏((𝒏 + 𝟑)!) − 𝒕𝒂𝒏−𝟏(𝒏!))
∞
𝒏=𝟎
We have telescoping series giving us
−𝒕𝒂𝒏−𝟏(𝟎!) − 𝒕𝒂𝒏−𝟏(𝟏!) − 𝒕𝒂𝒏−𝟏(𝟐!) + 𝐥𝐢𝐦𝒏→∞
(𝒕𝒂𝒏−𝟏((𝒏 + 𝟑)!) + 𝒕𝒂𝒏−𝟏((𝒏 + 𝟐)!) +
𝒕𝒂𝒏−𝟏((𝒏 + 𝟏)!)) .For all 𝒏 > 𝟐 and on the further solving we have:
−𝝅
𝟐− 𝒕𝒂𝒏−𝟏(𝟐) +
𝟑𝝅
𝟐= 𝝅− 𝒕𝒂𝒏−𝟏(𝟐)
1526. 𝒙𝟎 = 𝟏, 𝒙𝟏 = 𝟎, 𝒙𝒏 = (𝒏 − 𝟏)(𝒙𝒏−𝟏 + 𝒙𝒏−𝟐), 𝒏 ≥ 𝟐, 𝒏 ∈ ℕ. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒙𝒏𝒏!
Proposed by Daniel Sitaru-Romania
Solution 1 by Ravi Prakash-New Delhi-India
𝒙𝟎 = 𝟏, 𝒙𝟏 = 𝟎, 𝒙𝒏 = (𝒏 − 𝟏)(𝒙𝒏−𝟏 + 𝒙𝒏−𝟐), 𝒏 ≥ 𝟐,𝒏 ∈ ℕ; (𝟏) ⇒
𝒙𝒏 − 𝒏𝒙𝒏−𝟏 = −[𝒙𝒏−𝟏 − (𝒏 −)𝒙𝒏−𝟐]. Put: 𝒂𝒏 = 𝒙𝒏 − 𝒏𝒙𝒏−𝟏, ∀𝒏 ≥ 𝟏, 𝒂𝟏 = −𝟏
Also, (1) gives 𝒂𝒏 = −𝒂𝒏−𝟏, ∀𝒏 ≥ 𝟐 ⇒ (𝒂𝒏)𝒏≥𝟐 −geometric progression with ratio 𝒒 = −𝟏.
Thus, 𝒂𝒏 = (−𝟏)𝒏−𝟏𝒂𝟏, ∀𝒏 ≥ 𝟏 ⇒ 𝒙𝒏 − 𝒏𝒙𝒏−𝟏 = (−𝟏)
𝒏, ∀𝒏 ≥ 𝟏
⇒𝒙𝒏𝒏!−
𝒙𝒏−𝟏(𝒏 − 𝟏)!
=(−𝟏)𝒏
𝒏!⇒∑(
𝒙𝒓𝒓!−
𝒙𝒓−𝟏(𝒓 − 𝟏)!
)
𝒏
𝒓=𝟏
=∑(−𝟏)𝒓
𝒓!
𝒏
𝒓=𝟏
, ∀𝒏 ≥ 𝟏
⇒𝒙𝒏𝒏!=∑
(−𝟏)𝒓
𝒓!
𝒏
𝒓=𝟏
+𝒙𝟎𝟎!
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒙𝒏𝒏!= 𝐥𝐢𝐦𝒏→∞
∑(−𝟏)𝒓
𝒓!
𝒏
𝒓=𝟎
=𝟏
𝒆
Solution 2 by Naren Bhandari-Bajura-Nepal
Since 𝒙𝟎 = 𝟏 and 𝒙𝟏 = 𝟎 and given recurrence relation 𝒙𝒏 = (𝒏 − 𝟏)(𝒙𝒏−𝟐 + 𝒙𝒏−𝟏) and
on expanding the recurrence relation we can observe that
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42 RMM-CALCULUS MARATHON 1501-1600
𝑷 = {𝒙𝒏|𝒏 ∈ ℤ≥𝟎} = {𝟏, 𝟎, 𝟏, 𝟐, 𝟗, 𝟒𝟒, 𝟐𝟔𝟓, 𝟏𝟖𝟓𝟒,… }
Since the sequence we have well know from Derangement so in other word we can
rewrite the recurrence in terms subfactorial, that is
! 𝒏 = (𝒏 − 𝟏)(! (𝒏 − 𝟐) + ! (𝒏 − 𝟏)), ∀𝒏 ≥ 𝟐 and hence required limit to be evaluated is
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒙𝒏𝒏!= 𝐥𝐢𝐦𝐧→∞
! 𝒏
𝒏!= 𝐥𝐢𝐦𝒏→∞
∑(−𝟏)𝒌
𝒌!
𝒏
𝒌=𝟎
=𝟏
𝒆
𝐚𝐬 ! 𝒏 = 𝒏! ∑(−𝟏)𝒌
𝒌!
𝒏
𝒌=𝟎
1527. 𝑮(𝒏) −Barnes 𝑮-function, 𝑲(𝒏) − 𝑲 function. Find:
𝛀 =∑ √𝒏!
𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐)
𝒏∞
𝒏=𝟐
Proposed by Daniel Sitaru-Romania
Solution 1 by Asmat Qatea-Afghanistan
𝑮(𝒏) =(𝚪(𝒏))
𝒏−𝟏
𝑲(𝒏); 𝑮(𝒏 + 𝟐) =
((𝒏 + 𝟏)!)𝒏+𝟏
𝑲(𝒏 + 𝟐)
𝑲(𝒏 + 𝟏) = 𝟏𝟏 ⋅ 𝟐𝟐 ⋅ 𝟑𝟑 ⋅ … ⋅ 𝒏𝒏 ⇒𝑲(𝒏 + 𝟏)
𝑲(𝒏 + 𝟐)=
𝟏
(𝒏 + 𝟏)𝒏+𝟏
√𝒏!
𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐)
𝒏
=√
𝒏!
𝑲(𝒏 + 𝟏) ⋅((𝒏 + 𝟏)!)
𝒏+𝟏
𝑲(𝒏 + 𝟐)
𝒏 =√
𝒏!
((𝒏 + 𝟏)!)𝒏+𝟏
(𝒏 + 𝟏)𝒏+𝟏
𝒏 =
=√
𝒏!
(𝒏 + 𝟏)𝒏+𝟏 ⋅ (𝒏!)𝒏+𝟏
(𝒏 + 𝟏)𝒏+𝟏
𝒏 =𝟏
𝒏!
Therefore,
𝛀 = ∑ √𝒏!
𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐)
𝒏∞
𝒏=𝟐
= ∑𝟏
𝒏!
∞
𝒏=𝟐
= 𝒆 − 𝟐
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43 RMM-CALCULUS MARATHON 1501-1600
Solution 2 by Amrit Awasthi-India
We know: 𝑲(𝒏 + 𝟏) = 𝟏𝟏 ⋅ 𝟐𝟐 ⋅ 𝟑𝟑 ⋅ … ⋅ 𝒏𝒏 and 𝑮(𝒏+ 𝟐) = 𝟏! ⋅ 𝟐! ⋅ 𝟑! ⋅ … ⋅ 𝒏!
⇒ 𝑮(𝒏 + 𝟐) = 𝟏𝒏 ⋅ 𝟐𝒏−𝟏 ⋅ 𝟑𝒏−𝟐 ⋅ … ⋅ (𝒏 − 𝟏)𝟐 ⋅ 𝒏𝟏
𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐) = (𝟏𝟏 ⋅ 𝟐𝟐 ⋅ 𝟑𝟑 ⋅ … ⋅ 𝒏𝒏) ⋅ (𝟏𝒏 ⋅ 𝟐𝒏−𝟏 ⋅ 𝟑𝒏−𝟐 ⋅ … ⋅ (𝒏 − 𝟏)𝟐 ⋅ 𝒏𝟏) =
= 𝟏𝒏+𝟏 ⋅ 𝟐𝒏+𝟏 ⋅ … ⋅ (𝒏 − 𝟏)𝒏+𝟏 ⋅ 𝒏𝒏+𝟏 = (𝒏!)𝒏+𝟏
Therefore,
𝛀 = ∑ √𝒏!
𝑲(𝒏 + 𝟏) ⋅ 𝑮(𝒏 + 𝟐)
𝒏∞
𝒏=𝟐
= ∑ √𝒏!
(𝒏!)𝒏+𝟏𝒏
∞
𝒏=𝟐
= ∑ √𝟏
(𝒏!)𝒏𝒏
∞
𝒏=𝟐
=
= ∑𝟏
𝒏!
∞
𝒏=𝟐
= 𝒆 − 𝟐
1528. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
√(𝟐𝒏)! ⋅ (𝟐∑𝟏
(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!
𝒏
𝒌=𝟎
−𝟒𝒏
(𝟐𝒏)!)
𝒏
Proposed by Daniel Sitaru-Romania
Solution 1 by Adrian Popa-Romania
(𝟐𝒏
𝒏 − 𝒌) =
(𝟐𝒏)!
(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!⇒
𝟏
(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!=( 𝟐𝒏𝒏−𝒌)
(𝟐𝒏)!
∑(𝟐𝒏
𝒏− 𝒌)
𝒏
𝒌=𝟎
= (𝟐𝒏
𝒏) + (
𝟐𝒏
𝒏 − 𝟏) + (
𝟐𝒏
𝒏 − 𝟐) +⋯+ (
𝟐𝒏
𝟎)
= (𝟐𝒏
𝒏 + 𝟏) + (
𝟐𝒏
𝒏 + 𝟐) +⋯+ (
𝟐𝒏
𝟐𝒏) = 𝑺
∵ (𝟐𝒏
𝟎) + (
𝟐𝒏
𝟏) + (
𝟐𝒏
𝟐) +⋯+ (
𝟐𝒏
𝟐𝒏) = 𝟐𝟐𝒏 ⇒ 𝟐𝑺 + (
𝟐𝒏
𝒏) = 𝟐𝟐𝒏 ⇒ 𝟐𝑺 = 𝟒𝒏 − (
𝟐𝒏
𝒏)
⇒ 𝑺 =𝟒𝒏 − (𝟐𝒏
𝒏)
𝟐⇒∑(
𝟐𝒏
𝒏 − 𝒌)
𝒏
𝒌=𝟎
= 𝑺 + (𝟐𝒏
𝒏) =
(𝟒𝒏 + (𝟐𝒏𝒏))
𝟐
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44 RMM-CALCULUS MARATHON 1501-1600
𝛀 = 𝐥𝐢𝐦𝒏→∞
√(𝟐𝒏)! ⋅ (𝟐∑𝟏
(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!
𝒏
𝒌=𝟎
−𝟒𝒏
(𝟐𝒏)!)
𝒏
=
= 𝐥𝐢𝐦𝒏→∞
√(𝟐𝒏)! ⋅ (𝟒𝒏 + (𝟐𝒏
𝒏)
𝟐(𝟐𝒏)!⋅ 𝟐 −
𝟒𝒏
(𝟐𝒏)!)
𝒏
= 𝐥𝐢𝐦𝒏→∞
√(𝟐𝒏
𝒏)
𝒏
= 𝐥𝐢𝐦𝒏→∞
√(𝟐𝒏)!
(𝒏!)𝟐𝒏
=𝑪−𝑫
= 𝐥𝐢𝐦𝒏→∞
(𝟐𝒏 + 𝟐)!
((𝒏 + 𝟏)!)𝟐 ⋅(𝒏!)𝟐
(𝟐𝒏)!= 𝐥𝐢𝐦𝒏→∞
(𝟐𝒏)! (𝟐𝒏 + 𝟏)(𝟐𝒏+ 𝟐)(𝒏!)𝟐
(𝒏!)𝟐(𝒏 + 𝟏)𝟐(𝟐𝒏)!= 𝟒
Solution 2 by Ahmed Yackoube Chach-Mauritania
𝑰 = ∑𝟏
(𝒏 − 𝒌)! (𝒏 + 𝒌)!
𝒏
𝒌=𝟎
=𝟏
𝒏! ⋅ 𝒏!+
𝟏
(𝒏 − 𝟏)! ⋅ (𝒏 + 𝟏)!+ ⋯+
𝟏
𝟏! ⋅ (𝟐𝒏 − 𝟏)!+
𝟏
𝟎! ⋅ (𝟐𝒏)!
=𝟏
𝟎! ⋅ (𝟐𝒏)!+
𝟏
𝟏! ⋅ (𝟐𝒏 − 𝟏)!+ ⋯+
𝟏
𝒏! ⋅ 𝒏!= ∑
𝟏
(𝟐𝒏 − 𝒌) ⋅ 𝒌!
𝒏
𝒌=𝟎
∑𝟏
(𝟐𝒏− 𝒌)! ⋅ 𝒌!
𝟐𝒏
𝒌=𝒏+𝟏
=𝟏
(𝒏 − 𝟏)! ⋅ (𝒏 + 𝟏)!+⋯+
𝟏
𝟏! ⋅ (𝟐𝒏 − 𝟏)!+
𝟏
𝟎! ⋅ (𝟐𝒏)!+
𝟏
(𝒏!)𝟐−
𝟏
(𝒏!)𝟐
= 𝑰 −𝟏
(𝒏!)𝟐
𝟐𝑰 = ∑𝟏
(𝟐𝒏 − 𝒌)! ⋅ 𝒌!
𝟐𝒏
𝒌=𝟎
+𝟏
(𝒏!)𝟐=
𝟏
(𝒏!)𝟐+𝟒𝒏
(𝟐𝒏)!
𝛀𝐧 = √(𝟐𝒏)! ⋅ (𝟐∑𝟏
(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!
𝒏
𝒌=𝟎
−𝟒𝒏
(𝟐𝒏)!)
𝒏
= √(𝟐𝒏)! (𝟐𝑰 −𝟒𝒏
(𝟐𝒏)!)
𝒏
=
= √(𝟐𝒏)!
(𝒏!)𝟐𝒏
= √𝟐𝟐𝒏𝚪(𝒏 +
𝟏𝟐)
√𝝅𝚪(𝒏 + 𝟏)
𝒏
= 𝟒(𝟏
√𝝅⋅𝚪 (𝒏 +
𝟏𝟐)
𝚪(𝒏 + 𝟏))
𝟏𝒏
= 𝟒 ⋅ 𝒆
𝟏𝒏⋅𝐥𝐨𝐠(
𝟏
√𝝅⋅𝚪(𝒏+
𝟏𝟐)
𝚪(𝒏+𝟏))𝒏→∞→ 𝟒 ⋅ 𝒆𝟎
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
√(𝟐𝒏)! ⋅ (𝟐∑𝟏
(𝒏 − 𝒌)! ⋅ (𝒏 + 𝒌)!
𝒏
𝒌=𝟎
−𝟒𝒏
(𝟐𝒏)!)
𝒏
= 𝟒
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45 RMM-CALCULUS MARATHON 1501-1600
1529. For 𝒂, 𝒃, 𝒑, 𝒒 ∈ ℕ such that 𝒑(𝒒 − 𝒃) = 𝒂 + 𝟏. Find:
𝛀 = 𝐥𝐢𝐦𝒎→∞
𝟏
𝒎⋅ 𝐥𝐢𝐦𝒏→∞
∑∑𝒊𝒂𝒔𝒊𝒏𝒑 (𝒊𝒃
𝒏𝒒)
𝒏
𝒊=𝟏
𝟐𝒎
𝒑=𝟏
Proposed by Florică Anastase-Romania
Solution by Ruxandra Daniela Tonilă-Romania
𝒂𝒏 =∑𝒊𝒂𝒔𝒊𝒏𝒑 (𝒊𝒃
𝒏𝒒)
𝒏
𝒊=𝟏
=∑(𝒔𝒊𝒏(
𝒊𝒃
𝒏𝒒)
𝒊𝒃
𝒏𝒒
)
𝒑
𝒊𝒂+𝒃𝒑
𝒏𝒑𝒒
𝒏
𝒊=𝟏
=∑(𝒔𝒊𝒏(
𝒊𝒃
𝒏𝒒)
𝒊𝒃
𝒏𝒒
)
𝒑
𝒊𝒂+𝒃𝒑
𝒏𝒂+𝒃𝒑+𝟏
𝒏
𝒊=𝟏
⇔
∀𝒏 ∈ ℕ, ∃𝜻𝒏 > 0 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡: 1 − 𝜻𝒏 ≤ (𝒔𝒊𝒏(
𝒊𝒃
𝒏𝒒)
𝒊𝒃
𝒏𝒒
)
𝒑
≤ 𝟏 + 𝜻𝒏
( 𝟏 − 𝜻𝒏)∑𝒊𝒂+𝒃𝒑
𝒏𝒂+𝒃𝒑+𝟏≤ 𝒂𝒏 ≤
𝒏
𝒊=𝟏
( 𝟏 + 𝜻𝒏)∑𝒊𝒂+𝒃𝒑
𝒏𝒂+𝒃𝒑+𝟏
𝒏
𝒊=𝟏
𝒍𝒊𝒎𝒏→∞
∑𝒊𝒂+𝒃𝒑
𝒏𝒂+𝒃𝒑+𝟏= 𝒍𝒊𝒎𝒏→∞
𝟏
𝒏∑(
𝒊
𝒏)𝒂+𝒃𝒑
= ∫𝒙𝒂+𝒃𝒑𝒅𝒙 =𝟏
𝒂 + 𝒃𝒑 + 𝟏
𝟏
𝟎
𝒏
𝒊=𝟏
𝒏
𝒊=𝟏
Hence,
𝒍𝒊𝒎𝒏→∞
∑𝒊𝒂𝒔𝒊𝒏𝒑 (𝒊𝒃
𝒏𝒒) =
𝟏
𝒂 + 𝒃𝒑 + 𝟏
𝒏
𝒊=𝟏
∑∑𝒊𝒂𝒔𝒊𝒏𝒑 (𝒊𝒃
𝒏𝒒)
𝒏
𝒊=𝟏
𝟐𝒎
𝒑=𝟏
= ∑𝟏
𝒂+ 𝒃𝒑 + 𝟏
𝟐𝒎
𝒑=𝟏
𝟎 ≤𝟏
𝒎⋅∑
𝟏
𝒂 + 𝒃𝒑 + 𝟏
𝟐𝒎
𝒑=𝟏
≤⏞𝑨𝑴−𝑮𝑴 𝟏
𝒎∑
𝟏
𝟑√𝒂𝒃𝒑𝟑
𝟐𝒎
𝒑=𝟏
=𝟏
𝟑√𝒂𝒃𝟑 ∑
𝟏
𝒎√𝒑𝟑
𝟐𝒎
𝒑=𝟏
=𝟏
𝟑√𝒂𝒃𝟑 ⋅
∑𝟏
√𝒑𝟑
𝟐𝒎
𝒑=𝟏
𝒎
𝐥𝐢𝐦𝒎→∞
𝟏
𝒎⋅∑
𝟏
𝒂 + 𝒃𝒑 + 𝟏
𝟐𝒎
𝒑=𝟏
= 𝐥𝐢𝐦𝒎→∞
𝟏
𝟑√𝒂𝒃𝟑 ⋅
∑𝟏
√𝒑𝟑
𝟐𝒎
𝒑=𝟏
𝒎=⏞
𝑳.𝑪−𝑺 𝟏
𝟑√𝒂𝒃𝟑 ⋅ 𝐥𝐢𝐦
𝒏→∞
𝟏
√𝟐𝒎𝟑
(𝒎 + 𝟏) −𝒎= 𝟎
Therefore,
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46 RMM-CALCULUS MARATHON 1501-1600
𝛀 = 𝐥𝐢𝐦𝒎→∞
𝟏
𝒎⋅ 𝐥𝐢𝐦𝒏→∞
∑∑𝒊𝒂𝒔𝒊𝒏𝒑 (𝒊𝒃
𝒏𝒒)
𝒏
𝒊=𝟏
𝟐𝒎
𝒑=𝟏
= 𝟎
1530. If (𝒂𝒏)𝒏≥𝟎, (𝒃𝒏)𝒏≥𝟎 are given by 𝒂𝟎 = 𝒃𝟎 = 𝟏,𝒂𝒏+𝟏 = 𝒂𝒏 + 𝒃𝒏,
𝒃𝒏+𝟏 = (𝒏𝟐 + 𝒏 + 𝟏)𝒂𝒏 + 𝒃𝒏, 𝒏 ≥ 𝟏. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒏 ⋅ √∏(𝒂𝒌𝒃𝒌)
𝒏
𝒌=𝟏
𝒏
Proposed by Neculai Stanciu-Romania
Solution by George Florin Șerban-Romania
𝒂𝒏+𝟏 = 𝒂𝒏 + 𝒃𝒏 ⇒ 𝒃𝒏 = 𝒂𝒏+𝟏 − 𝒂𝒏 and 𝒃𝒏+𝟏 = (𝒏𝟐 + 𝒏+ 𝟏)𝒃𝒏, then
𝒃𝒏+𝟏 − 𝒃𝒏 = (𝒏𝟐 + 𝒏+ 𝟏)𝒂𝒏 ⇒ 𝒂𝒏+𝟐 − 𝒂𝒏+𝟏 − 𝒂𝒏+𝟏 + 𝒂𝒏 = (𝒏
𝟐 + 𝒏)𝒂𝒏 + 𝒂𝒏
𝒂𝒏+𝟐 − 𝟐𝒂𝒏+𝟏 = (𝒏𝟐 + 𝒏)𝒂𝒏 ⇒
𝒂𝒏+𝟐𝒂𝒏
− 𝟐𝒂𝒏+𝟏𝒂𝒏
= 𝒏𝟐 + 𝒏. 𝐋𝐞𝐭 𝒙𝒏 =𝒂𝒏+𝟏𝒂𝒏
⇒
𝒙𝒏+𝟏𝒙𝒏 − 𝟐𝒙𝒏 = 𝒏𝟐 + 𝒏. Applying mathematical induction to 𝑷(𝒏): 𝒙𝒏 = 𝒏+ 𝟏 from 𝒏 ≥
𝟎, we have: 𝑷(𝟎): 𝒙𝟎 =𝒂𝟏
𝒂𝟎= 𝟏 true.
Suppose that: 𝑷(𝒌): 𝒙𝒌 = 𝒌 + 𝟏 ⇒ 𝒙𝒌+𝟏 ⋅ (𝒌 + 𝟏) − 𝟐(𝒌 + 𝟏) = 𝒌𝟐 + 𝒌 ⇒
𝒙𝒌+𝟏 ⋅ (𝒌 + 𝟏) = 𝒌(𝒌 + 𝟏) + 𝟐(𝒌 + 𝟏) ⇒ 𝒙𝒌+𝟏 = 𝒌+ 𝟐. Hence,
𝒂𝒏+𝟏𝒂𝒏
= 𝒏+ 𝟏 ⇒∏𝒂𝒌+𝟏𝒂𝒌
𝒏
𝒌=𝟎
= (𝒏 + 𝟏)! ⇒ 𝒂𝒏+𝟏 = (𝒏 + 𝟏)! ⇒ 𝒂𝒏 = 𝒏!
∏𝒃𝒌𝒂𝒌
𝒏
𝒌=𝟏
=∏𝒌 ⋅ 𝒌!
𝒌!
𝒏
𝒌=𝟏
=∏𝒌
𝒏
𝒌=𝟏
= 𝒏!
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒏 ⋅ √∏(𝒂𝒌𝒃𝒌)
𝒏
𝒌=𝟏
𝒏
= 𝐥𝐢𝐦𝒏→∞
𝒏
√𝒏!𝒏 = 𝐥𝐢𝐦
𝒏→∞√𝒏𝒏
𝒏!
𝒏
=𝑪−𝑫′𝑨
𝐥𝐢𝐦𝒏→∞
(𝒏 + 𝟏)𝒏+𝟏
(𝒏 + 𝟏)!⋅𝒏!
𝒏𝒏=
= 𝐥𝐢𝐦𝒏→∞
(𝟏 +𝟏
𝒏)𝒏
= 𝒆.
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47 RMM-CALCULUS MARATHON 1501-1600
1531. Find:
𝛀(𝒏) = 𝐥𝐢𝐦𝒙→𝟎
𝟐𝐭𝐚𝐧𝟐𝒙 ⋅ 𝟒𝐭𝐚𝐧𝟒𝒙 ⋅ … ⋅ (𝟐𝒏)𝐭𝐚𝐧(𝟐𝒏𝒙) − 𝟏
𝟑𝐭𝐚𝐧𝟑𝒙 ⋅ 𝟓𝐭𝐚𝐧𝟓𝒙 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝐭𝐚𝐧((𝟐𝒏+𝟏)𝒙) − 𝟏;𝒏 ∈ ℕ, 𝒏 ≥ 𝟏
Proposed by Mohammad Hamed Nasery-Afghanistan
Solution 1 by Amrit Awasthi-India
Rewriting we have:
𝛀 = 𝐥𝐢𝐦𝒙→𝟎
𝒆𝐭𝐚𝐧 𝟐𝒙 𝐥𝐨𝐠 𝟐+𝐭𝐚𝐧 𝟒𝒙 𝐥𝐨𝐠 𝟒+⋯+𝐭𝐚𝐧(𝟐𝒏𝒙) 𝐥𝐨𝐠(𝟐𝒏) − 𝟏
𝒆𝐭𝐚𝐧 𝟑𝒙 𝐥𝐨𝐠 𝟑+𝐭𝐚𝐧 𝟓𝒙 𝐥𝐨𝐠 𝟓+⋯+𝐭𝐚𝐧(𝟐𝒏+𝟏)𝒙 𝐥𝐨𝐠(𝟐𝒏+𝟏) − 𝟏
𝒇(𝒙) = 𝒆𝐭𝐚𝐧 𝟐𝒙 𝐥𝐨𝐠 𝟐+𝐭𝐚𝐧 𝟒𝒙 𝐥𝐨𝐠 𝟒+⋯+𝐭𝐚𝐧(𝟐𝒏𝒙) 𝐥𝐨𝐠(𝟐𝒏),
𝒈(𝒙) = 𝒆𝐭𝐚𝐧 𝟑𝒙 𝐥𝐨𝐠 𝟑+𝐭𝐚𝐧 𝟓𝒙 𝐥𝐨𝐠 𝟓+⋯+𝐭𝐚𝐧(𝟐𝒏+𝟏)𝒙 𝐥𝐨𝐠(𝟐𝒏+𝟏)
⇒ 𝒇′(𝒙) = 𝒆𝐭𝐚𝐧 𝟐𝒙 𝐥𝐨𝐠 𝟐+𝐭𝐚𝐧 𝟒𝒙 𝐥𝐨𝐠 𝟒+⋯+𝐭𝐚𝐧(𝟐𝒏𝒙) 𝐥𝐨𝐠(𝟐𝒏) ⋅
⋅ (𝐥𝐨𝐠 𝟐 𝐬𝐞𝐜𝟐(𝟐𝒙) ⋅ 𝟐 + ⋯+ 𝐥𝐨𝐠(𝟐𝒏) 𝐬𝐞𝐜𝟐((𝟐𝒏 + 𝟏)𝒙) ⋅ 𝟐𝒏)
𝒈′(𝒙) = 𝒆𝐭𝐚𝐧 𝟑𝒙 𝐥𝐨𝐠 𝟑+𝐭𝐚𝐧 𝟓𝒙 𝐥𝐨𝐠 𝟓+⋯+𝐭𝐚𝐧(𝟐𝒏+𝟏)𝒙 𝐥𝐨𝐠(𝟐𝒏+𝟏) ⋅
⋅ (𝐥𝐨𝐠 𝟑 𝐬𝐞𝐜𝟐(𝟑𝒙) ⋅ 𝟑 +⋯+ 𝐥𝐨𝐠(𝟐𝒏 + 𝟏) 𝐬𝐞𝐜𝟐(𝟐𝒏 + 𝟏) 𝐬𝐞𝐜𝟐((𝟐𝒏 + 𝟏)𝒙) ⋅ (𝟐𝒏 + 𝟏))
That implies,
𝛀 = 𝐥𝐢𝐦𝒙→𝟎
𝒇′(𝒙)
𝒈′(𝒙)=
𝟐 𝐥𝐨𝐠 𝟐 + 𝟒 𝐥𝐨𝐠 𝟒 + ⋯+ 𝟐𝒏 𝐥𝐨𝐠𝟐𝒏
𝟑 𝐥𝐨𝐠 𝟑 + 𝟓 𝐥𝐨𝐠𝟓 + ⋯+ (𝟐𝒏 + 𝟏) 𝐥𝐨𝐠(𝟐𝒏 + 𝟏)=
=𝐥𝐨𝐠 𝟐𝟐 + 𝐥𝐨𝐠 𝟒𝟒 +⋯+ 𝐥𝐨𝐠(𝟐𝒏)𝟐𝒏
𝐥𝐨𝐠𝟑𝟑 + 𝐥𝐨𝐠 𝟓𝟓 +⋯+ 𝐥𝐨𝐠(𝟐𝒏 + 𝟏)𝟐𝒏+𝟏=
𝐥𝐨𝐠(𝟐𝟐 ⋅ 𝟒𝟒 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏)
𝐥𝐨𝐠(𝟑𝟑 ⋅ 𝟓𝟓 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝟐𝒏+𝟏)
Solution 2 by Serlea Kabay-Liberia
Recall, 𝐭𝐚𝐧(𝜶𝒙)~𝜶𝒙 and 𝒆𝜶𝒙 − 𝟏~𝜶𝒙;∀𝜶 ∈ ℝ.
𝛀(𝒏)~ 𝐥𝐢𝐦𝒙→𝟎
𝟐𝟐𝒙 ⋅ 𝟒𝟒𝒙 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏𝒙 − 𝟏
𝟑𝟑𝒙 ⋅ 𝟓𝟓𝒙 ⋅ … ⋅ (𝟐𝒏 + 𝟏)(𝟐𝒏+𝟏)𝒙 − 𝟏=
= 𝐥𝐢𝐦𝒙→𝟎
𝒆𝐥𝐨𝐠(𝟐𝟐𝒙⋅𝟒𝟒𝒙⋅…⋅(𝟐𝒏)𝟐𝒏𝒙) − 𝟏
𝒆𝐥𝐨𝐠(𝟑𝟑𝒙⋅𝟓𝟓𝒙⋅…⋅(𝟐𝒏+𝟏)(𝟐𝒏+𝟏)𝒙) − 𝟏
=
= 𝐥𝐢𝐦𝒙→𝟎
𝒆𝒙 𝐥𝐨𝐠 𝟐𝟐+𝒙 𝐥𝐨𝐠 𝟒𝟒+⋯+𝒙 𝐥𝐨𝐠(𝟐𝒏)𝟐𝒏 − 𝟏
𝒆𝒙 𝐥𝐨𝐠 𝟑𝟑+𝒙 𝐥𝐨𝐠 𝟓𝟓+⋯+𝒙 𝐥𝐨𝐠(𝟐𝒏+𝟏)𝟐𝒏+𝟏 − 𝟏
=
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48 RMM-CALCULUS MARATHON 1501-1600
= 𝐥𝐢𝐦𝒙→𝟎
𝒆𝒙 𝐥𝐨𝐠(∏ (𝟐𝒌)𝟐𝒌𝒏𝒌=𝟏 ) − 𝟏
𝒆𝒙 𝐥𝐨𝐠(∏ (𝟐𝒌+𝟏)𝟐𝒌+𝟏𝒏𝒌=𝟏 ) − 𝟏
= 𝐥𝐢𝐦𝒙→𝟎
𝒙 𝐥𝐨𝐠(∏ (𝟐𝒌)𝟐𝒌𝒏𝒌=𝟏 )
𝒙 𝐥𝐨𝐠(∏ (𝟐𝒌 + 𝟏)𝟐𝒌+𝟏𝒏𝒌=𝟏 )
=
= 𝐥𝐢𝐦𝒙→𝟎
𝐥𝐨𝐠(∏ (𝟐𝒌)𝟐𝒌𝒏𝒌=𝟏 )
𝐥𝐨𝐠(∏ (𝟐𝒌 + 𝟏)𝟐𝒌+𝟏𝒏𝒌=𝟏 )
=𝐥𝐨𝐠(𝟐𝟐 ⋅ 𝟒𝟒 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏)
𝐥𝐨𝐠(𝟑𝟑 ⋅ 𝟓𝟓 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝟐𝒏+𝟏)
Solution 3 by Obaidullah Jaihon-Afghanistan
𝟐𝐭𝐚𝐧 𝟐𝒙 ⋅ 𝟒𝐭𝐚𝐧 𝟒𝒙 ⋅ … ⋅ (𝟐𝒏)𝐭𝐚𝐧(𝟐𝒏𝒙) = 𝒕 ⇒
𝐥𝐨𝐠 𝒕 = 𝐥𝐨𝐠𝟐 𝐭𝐚𝐧 𝟐𝒙 + 𝐥𝐨𝐠 𝟒 𝐭𝐚𝐧 𝟒𝒙 +⋯+ 𝟐𝒏 𝐥𝐨𝐠(𝟐𝒏) 𝐭𝐚𝐧(𝟐𝒏𝒙)
(𝐥𝐨𝐠 𝒕)′ = 𝟐 𝐥𝐨𝐠𝟐 𝐬𝐞𝐜𝟐 𝟐𝒙 + 𝟒 𝐥𝐨𝐠𝟒 𝐬𝐞𝐜𝟐 𝟒𝒙 +⋯+ 𝟐𝒏 𝐥𝐨𝐠(𝟐𝒏) 𝐬𝐞𝐜𝟐(𝟐𝒏)
𝐥𝐢𝐦𝒙→𝟎(𝐥𝐨𝐠 𝒕)′ = 𝟐 𝐥𝐨𝐠𝟐 + 𝟒 𝐥𝐨𝐠𝟒 + ⋯+ 𝟐𝒏 𝐥𝐨𝐠(𝟐𝒏) = 𝐥𝐨𝐠(𝟐𝟐 ⋅ 𝟒𝟒 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏)
𝒑 = 𝟑𝐭𝐚𝐧 𝟑𝒙 ⋅ 𝟓𝐭𝐚𝐧 𝟓𝒙 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝐭𝐚𝐧(𝟐𝒏+𝟏)𝒙 − 𝟏
𝐥𝐨𝐠𝒑 = 𝐥𝐨𝐠𝟑 𝐭𝐚𝐧 𝟑𝒙 + 𝐥𝐨𝐠𝟓 𝐭𝐚𝐧 𝟓𝒙 +⋯+ 𝐥𝐨𝐠(𝟐𝒏+ 𝟏) 𝐭𝐚𝐧(𝟐𝒏+ 𝟏)
𝐥𝐢𝐦𝒙→𝟎(𝐥𝐨𝐠𝒑)′ = 𝐥𝐢𝐦
𝒙→𝟎(𝟑 𝐥𝐨𝐠𝟑 𝐬𝐞𝐜𝟐 𝟑𝒙 + 𝟓 𝐥𝐨𝐠𝟓 𝐬𝐞𝐜𝟐 𝟓𝒙 +⋯
+ (𝟐𝒏+ 𝟏) 𝐥𝐨𝐠(𝟐𝒏 + 𝟏) 𝐬𝐞𝐜𝟐(𝟐𝒏 + 𝟏)𝒙) =
= 𝟑 𝐥𝐨𝐠𝟑 + 𝟓 𝐥𝐨𝐠𝟓 + ⋯+ (𝟐𝒏+ 𝟏) 𝐥𝐨𝐠(𝟐𝒏 + 𝟏) = 𝐥𝐨𝐠(𝟑𝟑 ⋅ 𝟓𝟓 ⋅ … ⋅ (𝟐𝒏 + 𝟏)𝟐𝒏+𝟏)
𝛀 =𝐥𝐨𝐠 (𝟐𝟐 ⋅ 𝟒𝟒 ⋅ … ⋅ (𝟐𝒏)𝟐𝒏)
𝐥𝐨𝐠 (𝟑𝟑 ⋅ 𝟓𝟓⋅ … ⋅ (𝟐𝒏+ 𝟏)𝟐𝒏+𝟏)
1532. Solve for integers:
𝟐𝒙𝟐 + 𝒙
𝒙𝟐 + 𝒙 + 𝟏+𝟏𝟖𝒙𝟐 + 𝟏𝟔𝒙 + 𝟑𝟎
𝒙𝟐 + 𝒙+ 𝟐+𝟖𝟒𝒙𝟐 + 𝟖𝟏𝒙 + 𝟐𝟒𝟎
𝒙𝟐 + 𝒙 + 𝟑+⋯+
𝒂𝒏 ⋅ 𝒙𝟐 + 𝒃𝒏 ⋅ 𝒙 + 𝒄𝒏𝒙𝟐 + 𝒙 + 𝒏
=𝟔𝒏𝟓 + 𝟏𝟓𝒏𝟒 + 𝟏𝟎𝒏𝟑 − 𝒏
𝟑𝟎;𝒏 ∈ ℕ∗ 𝐚𝐧𝐝 𝐟𝐢𝐧𝐝:
𝛀 = 𝐥𝐢𝐦𝐧→∞
(𝒂𝒏𝒃𝒏)
𝒄𝒏𝒏𝟐
Proposed by Costel Florea-Romania
Solution by George Florin Șerban-Romania
𝟔𝒏𝟓 + 𝟏𝟓𝒏𝟒 + 𝟏𝟎𝒏𝟑 − 𝒏 = 𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)(𝟑𝒏𝟐 + 𝟑𝒏 − 𝟏)
∑𝒌𝟒𝒏
𝒌=𝟏
=𝒏(𝒏 + 𝟏)(𝟐𝒏 + 𝟏)(𝟑𝒏𝟐 + 𝟑𝒏 − 𝟏)
𝟑𝟎
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49 RMM-CALCULUS MARATHON 1501-1600
𝒂𝟏 = 𝟐, 𝒂𝟐 = 𝟏𝟖, 𝒂𝟑 = 𝟖𝟒, … , 𝒂𝒏 = 𝒏𝟒 + 𝒏
𝒃𝟏 = 𝟏, 𝒃𝟐 = 𝟏𝟔, 𝒃𝟑 = 𝟖𝟏,… , 𝒃𝒏 = 𝒏𝟒
𝒄𝟏 = 𝟎, 𝒄𝟐 = 𝟑𝟎, 𝒄𝟑 = 𝟐𝟒𝟎, … , 𝒄𝒏 = 𝒏𝟓 − 𝒏
(𝟐𝒙𝟐 + 𝒙
𝒙𝟐 + 𝒙 + 𝟏− 𝟏𝟒) + (
𝟏𝟖𝒙𝟐 + 𝟏𝟔𝒙 + 𝟑𝟎
𝒙𝟐 + 𝒙 + 𝟐− 𝟐𝟒) + (
𝟖𝟒𝒙𝟐 + 𝟖𝟏𝒙 + 𝟐𝟒𝟎
𝒙𝟐 + 𝒙 + 𝟑− 𝟑𝟒) +⋯
+ ((𝒏𝟒 + 𝒏)𝒙𝟐 + 𝒏𝟒𝒙 + 𝒏𝟓 − 𝒏)
𝒙𝟐 + 𝒙 + 𝒏− 𝒏𝟒) = 𝟎 ⇒
𝒙𝟐 − 𝟏
𝒙𝟐 + 𝒙 + 𝟏+𝟐𝒙𝟐 − 𝟐
𝒙𝟐 + 𝒙 + 𝟐+𝟑𝒙𝟐 − 𝟑
𝒙𝟐 + 𝒙 + 𝟑+⋯+
𝒏𝒙𝟐 − 𝒏
𝒙𝟐 + 𝒙 + 𝒏= 𝟎 ⟺
(𝒙𝟐 − 𝟏) (𝟏
𝒙𝟐 + 𝒙 + 𝟏+
𝟐
𝒙𝟐 + 𝒙 + 𝟐+⋯+
𝒏
𝒙𝟐 + 𝒙 + 𝒏) = 𝟎
Because 𝟏
𝒙𝟐+𝒙+𝟏+
𝟐
𝒙𝟐+𝒙+𝟐+⋯+
𝒏
𝒙𝟐+𝒙+𝒏≠ 𝟎 from 𝒙𝟐 + 𝒙 + 𝒏 > 0, ∀𝑛 ∈ ℕ
⇒ 𝒙𝟐 − 𝟏 = 𝟎 ⟺ 𝒙 ∈ {−𝟏, 𝟏}
𝛀 = 𝐥𝐢𝐦𝐧→∞
(𝒂𝒏𝒃𝒏)
𝒄𝒏𝒏𝟐= 𝐥𝐢𝐦𝒏→∞
(𝒏𝟒 + 𝒏
𝒏𝟒)
𝒏𝟓−𝒏
𝒏𝟐
= 𝐥𝐢𝐦𝒏→∞
[(𝟏 +𝟏
𝒏𝟑)𝒏𝟑
]
𝒏𝟒−𝟏
𝒏𝟒
= 𝒆𝐥𝐢𝐦𝒏→∞
𝒏𝟒−𝟏
𝒏𝟒 = 𝒆
1533. Let (𝒂𝒏)𝒏≥𝟏 −be a sequence of real numbers with 𝒂𝟎 = 𝟏 and
[(𝒂𝒏 − 𝒂𝒏−𝟏)(𝒏 + 𝟏)! 𝒏 − 𝒂𝒏𝒂𝒏−𝟏](𝒏 + 𝟏) = 𝒏𝟐𝒂𝒏𝒂𝒏−𝟏; 𝒏 ≥ 𝟎. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
( √𝒂𝒏+𝟏𝒏 + 𝟐
𝒏+𝟏)
𝒂
− (√𝒂𝒏𝒏 + 𝟏
𝒏)
𝒂
(√𝒂𝒏𝒏 + 𝟏
𝒏)
𝒂−𝟏
Proposed by Florică Anastase-Romania
Solution by Mikael Bernardo-Mozambique
𝒂𝟎 = 𝟏; [(𝒂𝒏 − 𝒂𝒏−𝟏)(𝒏 + 𝟏)!𝒏 − 𝒂𝒏𝒂𝒏−𝟏](𝒏 + 𝟏) = 𝒏𝟐𝒂𝒏𝒂𝒏−𝟏; 𝒏 ≥ 𝟏 ⟺
(𝒂𝒏 − 𝒂𝒏−𝟏𝒏(𝒏 + 𝟏) ⋅ (𝒏 + 𝟏)! = 𝒂𝒏𝒂𝒏−𝟏(𝒏𝟐 + 𝒏 + 𝟏) ⟺
𝒂𝒏 − 𝒂𝒏−𝟏𝒂𝒏𝒂𝒏−𝟏
=(𝒏 + 𝟏)𝟐 − 𝒏
𝒏(𝒏 + 𝟏)(𝒏 + 𝟏)!⟺
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50 RMM-CALCULUS MARATHON 1501-1600
𝟏
𝒂𝒏−𝟏−𝟏
𝒂𝒏=
𝒏 + 𝟏
𝒏(𝒏 + 𝟏)!−
𝟏
(𝒏 + 𝟏)(𝒏 + 𝟏)!
𝟏
𝒂𝒏−
𝟏
𝒂𝒏−𝟏=
𝟏
(𝒏 + 𝟏) ⋅ (𝒏 + 𝟏)!−
𝟏
𝒏 ⋅ 𝒏!
𝟏
𝒂𝟏−𝟏
𝒂𝟎=
𝟏
𝟐 ⋅ 𝟐!−
𝟏
𝟏 ⋅ 𝟏!⇒ 𝒂𝒏 = (𝒏 + 𝟏) ⋅ (𝒏 + 𝟏)!
Hence,
𝛀 = 𝐥𝐢𝐦𝒏→∞
( √𝒂𝒏+𝟏𝒏 + 𝟐
𝒏+𝟏)
𝒂
− (√𝒂𝒏𝒏 + 𝟏
𝒏)
𝒂
(√𝒂𝒏𝒏 + 𝟏
𝒏)
𝒂−𝟏 = 𝐥𝐢𝐦𝒏→∞
( √(𝒏 + 𝟐)!𝒏+𝟏
)𝒂
− (√(𝒏 + 𝟏)!𝒏
)𝒂
(√(𝒏 + 𝟏)!𝒏
)𝒂−𝟏 =
= 𝐥𝐢𝐦𝒏→∞
(√(𝒏 + 𝟏)!𝒏
)𝒂+𝟏−𝒂
⋅ ((√(𝒏 + 𝟐)!
𝒏+𝟏
√(𝒏 + 𝟏)!𝒏
)
𝒂
− 𝟏) =
= 𝐥𝐢𝐦𝒏→∞
(√(𝒏 + 𝟏)!𝒏
𝒏) ⋅ 𝒏 ⋅
(√(𝒏 + 𝟐)!
𝒏+𝟏
√(𝒏 + 𝟏)!𝒏
)
𝒂
− 𝟏
𝐥𝐨𝐠 ((√(𝒏 + 𝟐)!
𝒏+𝟏
√(𝒏 + 𝟏)!𝒏
)
𝒂
)
⋅ 𝐥𝐨𝐠((√(𝒏 + 𝟐)!
𝒏+𝟏
√(𝒏 + 𝟏)!𝒏
)
𝒂
)
𝐥𝐢𝐦𝒏→∞
(√(𝒏 + 𝟏)!𝒏
𝒏) =𝑪−𝑫
𝐥𝐢𝐦𝒏→∞
((𝒏 + 𝟐)!
(𝒏 + 𝟏)!⋅
𝒏𝒏
(𝒏 + 𝟏)𝒏+𝟏=𝟏
𝒆; (𝟏)
𝐥𝐢𝐦𝒏→∞
√(𝒏 + 𝟐)!𝒏+𝟏
√(𝒏 + 𝟏)!𝒏
= 𝟏 ⇒ 𝐥𝐢𝐦𝒏→∞
(√(𝒏 + 𝟐)!
𝒏+𝟏
√(𝒏 + 𝟏)!𝒏
)
𝒂
− 𝟏
𝐥𝐨𝐠((√(𝒏 + 𝟐)!
𝒏+𝟏
√(𝒏 + 𝟏)!𝒏
)
𝒂
)
= 𝟏; (𝟐)
𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠 ((√(𝒏 + 𝟐)!
𝒏+𝟏
√(𝒏 + 𝟏)!𝒏
)
𝒏𝒂
) = 𝜶 ⋅ 𝐥𝐨𝐠 (𝐥𝐢𝐦𝒏→∞
(𝒏 + 𝟐)!
(𝒏 + 𝟏)!⋅
𝒏 + 𝟏
√(𝒏 + 𝟐)!𝒏+𝟏
⋅𝟏
𝒏 + 𝟏) = 𝒂 ⋅ 𝐥𝐨𝐠 𝒆
= 𝒂; (𝟑)
From (1),(2),(3) it follows that:
𝛀 = 𝐥𝐢𝐦𝒏→∞
( √𝒂𝒏+𝟏𝒏 + 𝟐
𝒏+𝟏)
𝒂
− (√𝒂𝒏𝒏 + 𝟏
𝒏)
𝒂
(√𝒂𝒏𝒏 + 𝟏
𝒏)
𝒂−𝟏 =𝟏
𝒆⋅ 𝟏 ⋅ 𝒂 =
𝒂
𝒆
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51 RMM-CALCULUS MARATHON 1501-1600
1534.
(𝒂𝒏)𝒏≥𝟏, (𝒃𝒏)𝒏≥𝟏; 𝒂𝒏 = ∫ [𝒏𝟐
𝒙]
𝒏
𝟏
𝒅𝒙, 𝒃𝟏 > 1,
𝒃𝒏+𝟏 = 𝟏 + 𝐥𝐨𝐠(𝒃𝒏) , [∗] − 𝑮𝑰𝑭 . Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒂𝒏 ⋅ 𝐥𝐨𝐠 √𝒃𝒏𝒏
𝐥𝐨𝐠𝒏
Proposed by Florică Anastase-Romania
Solution 1 by Ruxandra Daniela Tonilă-Romania
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒂𝒏 ⋅ 𝐥𝐨𝐠 √𝒃𝒏𝒏
𝐥𝐨𝐠𝒏= 𝐥𝐢𝐦𝒏→∞
𝒂𝒏 ⋅ 𝐥𝐨𝐠 𝒃𝒏𝒏 ⋅ 𝐥𝐨𝐠 𝒏
= 𝐥𝐢𝐦𝒏→∞
𝒂𝒏𝒏𝟐 ⋅ 𝐥𝐨𝐠 𝒏
⋅ 𝒏 𝐥𝐨𝐠 𝒃𝒏
= 𝛀𝟏 ⋅ 𝛀𝟐; (𝟏)
𝛀𝟏 = 𝐥𝐢𝐦𝒏→∞
𝒂𝒏𝒏𝟐 ⋅ 𝐥𝐨𝐠 𝒏
𝐚𝐧𝐝 𝛀𝟐 = 𝐥𝐢𝐦𝒏→∞
𝒏 ⋅ 𝐥𝐨𝐠 𝒃𝒏
We have: 𝒏𝟐
𝒙− 𝟏 < [
𝒏𝟐
𝒙] ≤
𝒏𝟐
𝒙; ∀𝒙 ∈ ℝ, 𝒏 ∈ ℕ ⇒
∫ (𝒏𝟐
𝒙− 𝟏)
𝒏
𝟏
𝒅𝒙 < 𝒂𝒏 ≤ ∫𝒏𝟐
𝒙
𝒏
𝟏
𝒅𝒙 ⟺ 𝒏𝟐 𝐥𝐨𝐠𝒏 − (𝒏 − 𝟏) < 𝒂𝒏 ≤ 𝒏𝟐 𝐥𝐨𝐠 𝒏 ⟺
𝟏 −𝒏 − 𝟏
𝒏𝟐 𝐥𝐨𝐠 𝒏<
𝒂𝒏𝒏𝟐 𝐥𝐨𝐠 𝒏
≤ 𝟏
𝐥𝐢𝐦𝒏→∞
(𝟏 −𝒏 − 𝟏
𝒏𝟐 𝐥𝐨𝐠 𝒏) < 𝐥𝐢𝐦
𝒏→∞
𝒂𝒏𝒏𝟐 ⋅ 𝐥𝐨𝐠𝒏
≤ 𝟏 ⇒ 𝐥𝐢𝐦𝒏→∞
𝒂𝒏𝒏𝟐 ⋅ 𝐥𝐨𝐠 𝒏
= 𝟏; (𝟐)
Now, 𝒃𝟏 > 1. Suppose that 𝒃𝒌 > 1;∀𝑘 ∈ ℕ and from 𝒃𝒌+𝟏 = 𝟏 + 𝐥𝐨𝐠 𝒃𝒌 we get
𝒃𝒌+𝟏 > 1;∀𝑘 ∈ ℕ. Thus, 𝒃𝒏 > 1;∀𝑛 ∈ ℕ.
𝒃𝒏+𝟏 = 𝟏 + 𝐥𝐨𝐠 𝒃𝒏 ⇒ 𝒃𝒏+𝟏 − 𝒃𝒏 = 𝟏 − 𝒃𝒏 + 𝐥𝐨𝐠 𝒃𝒏 ; (𝟑)
Let be the function 𝒇(𝒙) = 𝐥𝐨𝐠𝒙 − 𝒙 − 𝟏; (𝒙 > 1) with 𝒇′(𝒙) =𝟏
𝒙− 𝟏 < 0;∀𝑥 > 1
⇒ 𝒇−decreasing on (𝟏,∞) ⇒ 𝒇(𝒙) < 𝑓(𝟏) = 𝟎; ∀𝒙 > 1 ⇒
𝒇(𝒃𝒏) < 0 ⇒ 𝒃𝒏+𝟏 < 𝒃𝒏.
Since (𝒃𝒏)𝒏≥𝟏 −is decreasing and bounded, then (𝒃𝒏)𝒏≥𝟏 converges.
So, ∃𝒍 ∈ ℝ such that 𝐥𝐢𝐦𝒏→∞
𝒃𝒏 = 𝒍 ⇒ 𝒍 = 𝟏 + 𝐥𝐨𝐠 𝒍 ⇒ 𝒍 = 𝟏.
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52 RMM-CALCULUS MARATHON 1501-1600
𝛀𝟐 = 𝐥𝐢𝐦𝒏→∞
𝒏 ⋅ 𝐥𝐨𝐠 𝒃𝒏 = 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠 𝒃𝒏𝒏 = 𝐥𝐨𝐠 (𝐥𝐢𝐦
𝒏→∞𝒃𝒏𝒏) =
= 𝐥𝐨𝐠 (𝐥𝐢𝐦𝒏→∞
[(𝟏 + 𝒃𝒏 − 𝟏)𝟏
𝒃𝒏−𝟏]
𝒏(𝒃𝒏−𝟏)
) = 𝐥𝐨𝐠 (𝒆𝐥𝐢𝐦𝒏→∞
𝒏(𝒃𝒏−𝟏)) = 𝐥𝐢𝐦𝒏→∞
𝒏(𝒃𝒏 − 𝟏) =
= 𝐥𝐢𝐦𝒏→∞
𝒏
𝟏𝒃𝒏 − 𝟏
=𝑪−𝑺
𝐥𝐢𝐦𝒏→∞
𝒏 + 𝟏 − 𝒏
𝟏𝒃𝒏+𝟏 − 𝟏
−𝟏
𝒃𝒏 − 𝟏
= 𝐥𝐢𝐦𝒏→∞
(𝒃𝒏+𝟏 − 𝟏)(𝒃𝒏 − 𝟏)
𝒃𝒏 − 𝒃𝒏+𝟏=
= 𝐥𝐢𝐦𝒏→∞
(𝒃𝒏 − 𝟏)𝐥𝐨𝐠𝒃𝒏𝒃𝒏 − 𝐥𝐨𝐠 𝒃𝒏 − 𝟏
= 𝐥𝐢𝐦𝒏→∞
(𝒃𝒏 − 𝟏)𝟐
𝒃𝒏 − 𝐥𝐨𝐠 𝒃𝒏 − 𝟏=
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒃𝒏 − 𝐥𝐨𝐠𝒃𝒏 − 𝟏(𝒃𝒏 − 𝟏)𝟐
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝟏𝒃𝒏 − 𝟏
−𝐥𝐨𝐠 𝒃𝒏(𝒃𝒏 − 𝟏)𝟐
=
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝟏𝒃𝒏 − 𝟏
−𝐥𝐨𝐠(𝟏 + 𝒃𝒏 − 𝟏)
𝒃𝒏 − 𝟏⋅
𝟏𝒃𝒏 − 𝟏
= +∞; (𝟒)
From (1),(2),(3),(4) it follows that:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒂𝒏 ⋅ 𝐥𝐨𝐠 √𝒃𝒏𝒏
𝐥𝐨𝐠𝒏= ∞
Solution 2 by proposer
𝒙 ∈ [𝟏, 𝒏] ⇒ 𝒏 ≤𝒏𝟐
𝒙≥ 𝒏𝟐
Let [𝒙𝟐
𝒏] = 𝒕, 𝒕 ∈ {𝒏, 𝒏 + 𝟏,… , 𝒏𝟐} ⇒
𝒏𝟐
𝒙− 𝟏 < 𝑡 ≤
𝒏𝟐
𝒙⇔
𝒏𝟐
𝒕+𝟏< 𝑥 ≤
𝒏𝟐
𝒕
𝒂𝒏 = ∑ ∫ 𝒕
𝒏𝟐
𝒕
𝒏𝟐
𝒕+𝟏
𝒅𝒙
𝒏𝟐−𝟏
𝒕=𝒏
= ∑ 𝒕(𝒏𝟐
𝒕−𝒏𝟐
𝒕 + 𝟏)
𝒏𝟐−𝟏
𝒕=𝒏
= 𝒏𝟐 (𝟏
𝒏 + 𝟏+
𝟏
𝒏 + 𝟐+⋯+
𝟏
𝒏𝟐)
Now, 𝒃𝟏 > 1,𝒃𝒏+𝟏 = 𝟏 + 𝐥𝐨𝐠(𝒃𝒏) ⇒ 𝒃𝒏 > 1,∀𝑛 ∈ ℕ (induction from 𝒏 ∈ ℕ) and
from 𝒃𝒏+𝟏 − 𝒃𝒏 = 𝟏 + 𝐥𝐨𝐠 𝒃𝒏 − 𝒃𝒏 ≤ 𝟎, because 𝐥𝐨𝐠(𝟏 + 𝒙) ≤ 𝒙, ∀𝒙 > −1 ⇒
(𝒃𝒏)𝒏≥𝟏−decreasing.
So, (𝒃𝒏)𝒏≥𝟏 −convergent sequence, then
∃ 𝒍 ∈ ℝ, 𝒍 > 0 such that 𝒍 = 𝐥𝐢𝐦𝒏→∞
𝒃𝒏 ⇒ 𝒍 = 𝟏 + 𝐥𝐨𝐠 𝒍 ⇒ 𝒍 = 𝟏.
Let 𝒄𝒏 = 𝟏 +𝟏
𝟐+𝟏
𝟑+⋯+
𝟏
𝒏− 𝐥𝐨𝐠 𝒏.
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53 RMM-CALCULUS MARATHON 1501-1600
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝒂𝒏 ⋅ 𝐥𝐨𝐠 √𝒃𝒏𝒏
𝐥𝐨𝐠 𝒏= 𝐥𝐢𝐦𝒏→∞
𝒂𝒏 ⋅ 𝐥𝐨𝐠 𝒃𝒏𝒏 ⋅ 𝐥𝐨𝐠𝒏
= 𝐥𝐢𝐦𝒏→∞
(𝒂𝒏
𝒏𝟐 ⋅ 𝐥𝐨𝐠 𝒏⋅ 𝒏 𝐥𝐨𝐠 𝒃𝒏) =
= 𝐥𝐢𝐦𝒏→∞
[(𝒄𝒏𝟐 − 𝒄𝒏𝐥𝐨𝐠𝒏
+ 𝟏) ⋅ 𝒏 𝐥𝐨𝐠 𝒃𝒏] = 𝐥𝐢𝐦𝒏→∞
(𝒏 𝐥𝐨𝐠 𝒃𝒏) =
= 𝐥𝐢𝐦𝒏→∞
𝒏(𝒃𝒏 − 𝟏) 𝐥𝐨𝐠(𝟏 + 𝒃𝒏 − 𝟏)𝟏
𝒃𝒏−𝟏 = 𝐥𝐢𝐦𝒏→∞
𝒏(𝒃𝒏 − 𝟏) = 𝐥𝐢𝐦𝒏→∞
𝒏 𝐥𝐨𝐠 𝒃𝒏−𝟏 =
= ⋯ = 𝐥𝐢𝐦𝒏→∞
𝒏 ⋅ 𝒃𝟏 = +∞
1535. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
(𝟏
𝒏(∑∑
𝒊𝟑 + 𝒋𝟑
𝒊𝟒 + 𝒋𝟒
𝒏
𝒋=𝟏
𝒏
𝒊=𝟏
−∑∑𝒌𝟑 − 𝒍𝟑
𝒌𝟒 − 𝒍𝟒
𝒏
𝒍=𝟏
𝒏
𝒌=𝟏
))
Proposed by Mikael Bernardo-Mozambique
Solution by Ty Halpen-Florida-USA
𝛀 = 𝐥𝐢𝐦𝒏→∞
(𝟏
𝒏(∑∑
𝒊𝟑 + 𝒋𝟑
𝒊𝟒 + 𝒋𝟒
𝒏
𝒋=𝟏
𝒏
𝒊=𝟏
−∑∑𝒌𝟑 − 𝒍𝟑
𝒌𝟒 − 𝒍𝟒
𝒏
𝒍=𝟏
𝒏
𝒌=𝟏
))
= 𝐥𝐢𝐦𝒏→∞
(∑𝟏
𝒏∑𝟏
𝒏⋅(𝒊𝒏)
𝟑
+ (𝒋𝒏)
𝟑
(𝒊𝒏)𝟒
+ (𝒋𝒏)𝟒
𝒏
𝒋=𝟏
𝒏
𝒊=𝟏
−∑𝟏
𝒏∑𝟏
𝒏⋅(𝒌𝒏)
𝟑
− (𝒍𝒏)
𝟑
(𝒌𝒏)𝟒
− (𝒍𝒏)𝟒
𝒏
𝒍=𝟏
𝒏
𝒌=𝟏
) =
= ∫ ∫ (𝒙𝟑 + 𝒚𝟑
𝒙𝟒 + 𝒚𝟒+𝒙𝟑 − 𝒚𝟑
𝒙𝟒 − 𝒚𝟒)
𝟏
𝟎
𝒅𝒙𝟏
𝟎
𝒅𝒚 =
= ∫ ∫ (𝒙𝟑
𝒙𝟒 + 𝒚𝟒−
𝒙
𝟐(𝒙𝟐 + 𝒚𝟐)−
𝟏
𝟐(𝒙 + 𝒚))
𝟏
𝟎
𝒅𝒙𝟏
𝟎
𝒅𝒚 +∫ ∫ (𝒚𝟑
𝒙𝟒 + 𝒚𝟒−
𝒚
𝟐(𝒙𝟐 + 𝒚𝟐))𝒅𝒚
𝟏
𝟎
𝒅𝒙𝟏
𝟎
=
=𝟏
𝟐∫ (− 𝐥𝐨𝐠(𝟏 + 𝒕) + 𝐥𝐨𝐠 𝒕 − 𝐥𝐨𝐠(𝟏 + 𝒕𝟒) − 𝟒 𝐥𝐨𝐠 𝒕 − 𝐥𝐨𝐠(𝟏 + 𝒕𝟐) + 𝟐 𝐥𝐨𝐠 𝒕)𝒅𝒕𝟏
𝟎
=
=𝟏
𝟐∫ (− 𝐥𝐨𝐠 𝒕 − 𝐥𝐨𝐠(𝟏 + 𝒕) − 𝐥𝐨𝐠(𝟏 + 𝒕𝟐) + 𝐥𝐨𝐠(𝟏 + 𝒕𝟒))𝒅𝒕𝟏
𝟎
=
=𝑰𝑩𝑷 𝟏
𝟐(−𝟐 𝐥𝐨𝐠𝟐 − 𝟐∫ (
𝟏
𝟏 + 𝒕𝟐−
𝟐
𝟏 + 𝒕𝟒)
𝟏
𝟎
𝒅𝒕 =
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54 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝟐(−𝟐 𝐥𝐨𝐠 𝟐 −
𝝅
𝟐+𝟏
√𝟐∫ (
𝒕 + √𝟐
𝒕𝟐 + √𝟐𝒕 + 𝟏+
𝒕 − √𝟐
−𝒕𝟐 + √𝟐𝒕 − 𝟏)𝒅𝒕
𝟏
𝟎
=
=𝟏
𝟐(−𝟐 𝐥𝐨𝐠𝟐 −
𝝅
𝟐+√𝟐
𝟐(𝝅 + 𝟐 𝐥𝐨𝐠(𝟐 + √𝟐) − 𝐥𝐨𝐠 𝟐) =
=𝝅
𝟒(√𝟐 − 𝟏) −
𝟒 + √𝟐
𝟒⋅ 𝐥𝐨𝐠 𝟐 +
√𝟐
𝟐𝐥𝐨𝐠(𝟐 + √𝟐) ≅ 𝟎. 𝟐𝟓𝟓𝟒
1536. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏√∑(−𝟏)𝒌 (
𝒏
𝒌) (𝒏 − 𝒌)𝒏
𝒏−𝟏
𝒌=𝟎
𝒏
Proposed by Daniel Sitaru-Romania
Solution 1 by Ravi Prakash-New Delhi-India
∑(−𝟏)𝒌 (𝒏
𝒌) (𝒏 − 𝒌)𝒏
𝒏−𝟏
𝒌=𝟎
= ∑(−𝟏)𝒏(−𝟏)𝒌 (𝒏
𝒏 − 𝒌)𝒌𝒏
𝒏
𝒌=𝟏
=
= (−𝟏)𝒏∑(−𝟏)𝒌 (𝒏
𝒌)𝒌𝒏; (𝟏)
𝒏
𝒌=𝟏
We have:
∑(𝒏
𝒌) (−𝟏)𝒌𝒙𝒌
𝒏
𝒌=𝟎
= (𝟏 − 𝒙)𝒏
Differentiating w.r.t. 𝒙, we get:
∑(𝒏
𝒌) (−𝟏)𝒌𝒌𝒙𝒌−𝟏
𝒏
𝒌=𝟏
= (−𝟏)(𝟏 − 𝒙)𝒏−𝟏(𝒏) ⇒
∑(𝒏
𝒌) (−𝟏)𝒌𝒌𝒙𝒌
𝒏
𝒌=𝟏
= (−𝟏)(𝟏 − 𝒙)𝒏−𝟏(𝒏𝒙)
Differentiating w.r.t. 𝒙, we get:
∑(𝒏
𝒌)(−𝟏)𝒌𝒌𝟐𝒙𝒌−𝟏
𝒏
𝒌=𝟏
= (−𝟏)(𝟏 − 𝒙)𝒏−𝟏(𝒏) + 𝒏(𝒏 − 𝟏)(−𝟏)𝟐(𝟏 − 𝒙)𝒏−𝟐(𝒙)
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55 RMM-CALCULUS MARATHON 1501-1600
Multiply again by 𝒙, we get:
∑(𝒏
𝒌) (−𝟏)𝒌𝒌𝟐𝒙𝒌
𝒏
𝒌=𝟏
= (−𝟏)(𝟏 − 𝒙)𝒏−𝟏(𝒏𝒙) + 𝒏(𝒏 − 𝟏)(−𝟏)𝟐(𝟏 − 𝒙)𝒏−𝟐𝒙𝟐
Differentiating again, we get:
∑(−𝟏)𝒌 (𝒏
𝒌)𝒌𝟑𝒙𝒌−𝟏
𝒏
𝒌=𝟏
= (−𝟏)(𝟏 − 𝒙)∗𝒏−𝟏 + 𝒏(𝒏 − 𝟏)(−𝟏)𝟐(𝟐𝒙)(𝟏 − 𝒙)𝒏−𝟐 +
+𝒏(𝒏 − 𝟏)(𝒏− 𝟐)(−𝟏)𝟑(𝟏 − 𝒙)𝒏−𝟑𝒙𝟐
Continuing this way, we get:
∑(−𝟏)𝒌 (𝒏
𝒌)𝒌𝒏𝒙𝒌−𝟏
𝒏
𝒌=𝟏
= (𝟏 − 𝒙)𝒈(𝒙) + 𝒏! (−𝟏)𝒏𝒙𝒏
Putting 𝒙 = 𝟏, we get
∑(−𝟏)𝒌 (𝒏
𝒌)𝒌𝒏
𝒏
𝒌=𝟏
= 𝒏! (−𝟏)𝒏; (𝟐)
From (1),(2) it follows
∑(−𝟏)𝒏−𝒌 (𝒏
𝒌) (𝒏 − 𝒌)𝒏
𝒏
𝒌=𝟏
− (−𝟏)𝒏𝒏! (−𝟏)𝒏 = 𝒏!
Now,
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏√∑(−𝟏)𝒌 (
𝒏
𝒌) (𝒏 − 𝒌)𝒏
𝒏−𝟏
𝒌=𝟎
𝒏
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏√𝒏!𝒏
=𝑪−𝑫
𝐥𝐢𝐦𝒏→∞
(𝒏 + 𝟏)!
(𝒏 + 𝟏)𝒏+𝟏⋅𝒏𝒏
𝒏!=
= 𝐥𝐢𝐦𝒏→∞
𝟏
(𝟏 +𝟏𝒏)
𝒏 =𝟏
𝒆
Solution 2 by Felix Marin-Romania
∑(−𝟏)𝒌 (𝒏
𝒌) (𝒏 − 𝒌)𝒏
𝒏−𝟏
𝒌=𝟎
= ∑(−𝟏)𝒏−𝒌 (𝒏
𝒏 − 𝒌) [𝒏 − (𝒏 − 𝒌)]𝒏
𝒏
𝒌=𝟎
= (−𝟏)𝒏∑(−𝟏)𝒌 (𝒏
𝒌)𝒌𝒏
𝒏
𝒌=𝟎
= (−𝟏)𝒏∑(−𝟏)𝒌 (𝒏
𝒌) {𝒏! [𝒛𝒏]𝒆𝒌𝒛}
𝒏
𝒌=𝟎
= (−𝟏)𝒏𝒏! [𝒛𝒏]∑(𝒏
𝒌) (−𝒆𝒛)𝒌
𝒏
𝒌=𝟎
=
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56 RMM-CALCULUS MARATHON 1501-1600
= (−𝟏)𝒏𝒏! [𝒛𝒏](𝟏 − 𝒆𝒛)𝒏 = 𝒏! [𝒛𝒏](𝒆𝒛 − 𝟏)𝒏 = 𝒏! [𝒛𝒏] [𝒏!∑{𝒋
𝒏}𝒛𝒋
𝒋!
∞
𝒋=𝒏
]
{𝒋𝒏} −is the Stirling Number of the Second Kind and {𝒏
𝒏} = 𝟏.
∑(−𝟏)𝒌 (𝒏
𝒌) (𝒏 − 𝒌)𝒏
𝒏−𝟏
𝒌=𝟎
= 𝒏!
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏√∑(−𝟏)𝒌 (
𝒏
𝒌) (𝒏 − 𝒌)𝒏
𝒏−𝟏
𝒌=𝟎
𝒏
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏√𝒏!𝒏
= 𝐥𝐢𝐦𝒏→∞
√√𝟐𝝅𝒏𝒏+𝟏𝟐𝒆−𝒏
𝒏
𝒏=
= 𝐥𝐢𝐦𝒏→∞
(𝟐𝝅)𝟏𝟐𝒏𝒏 ⋅ 𝒏
𝟏𝟐𝒏𝒆−𝟏
𝒏=𝟏
𝒆≅ 𝟎. 𝟑𝟔𝟕𝟗
1537. 𝒙𝟎 = 𝟏, 𝒙𝟏 = √𝟐, 𝒙𝒏+𝟏 + 𝒙𝒏−𝟏 = √𝟐𝒙𝒏, 𝒏 ∈ ℕ∗. Find:
𝛀(𝒏) = ∑∑(𝒙𝟐𝒌+𝒊 + 𝒙𝟑𝒌+𝒊 + 𝒙𝟓𝒌+𝒊)
𝟖
𝒊=𝟏
𝒏
𝒌=𝟏
Proposed by Daniel Sitaru-Romania
Solution 1 by Kamel Gandouli Rezgui-Tunisia
𝒙𝒏+𝟐 = √𝟐𝒙𝒏+𝟏 − 𝒙𝒏; 𝒂 = √𝟐, 𝒃 = −𝟏
𝑬: 𝒙𝟐 = √𝟐𝒙− 𝟏 characteristic equation⇒ 𝒙𝟐 − √𝟐𝒙+ 𝟏 = 𝟎,𝚫 = −𝟏, 𝒛𝟏,𝟐 =√𝟐+√𝟐𝒊
𝟐⇒
𝒙𝒏 = 𝝀𝐜𝐨𝐬𝒏𝝅
𝟒+ 𝝁𝐬𝐢𝐧
𝒏𝝅
𝟒, 𝒙𝟎 = 𝟏 ⇒ 𝝀 = 𝟏 and 𝒙𝟏 = √𝟐 ⇒ 𝝁 = 𝟏 ⇒
𝒙𝒏 = 𝐜𝐨𝐬𝒏𝝅
𝟒+ 𝐬𝐢𝐧
𝒏𝝅
𝟒= √𝟐 𝐜𝐨𝐬 (
(𝒏 − 𝟏)𝝅
𝟒)
𝒙𝟐𝒌+𝟏 = √𝟐𝐜𝐨𝐬 ((𝟐𝒌 + 𝒊 − 𝟏)𝝅
𝟒) = √𝟐𝐜𝐨𝐬 (
𝒌𝝅
𝟐+(𝒊 − 𝟏)𝝅
𝟒) ; (𝒊 = 𝟏, 𝟖̅̅ ̅̅ ̅) ⇒
∑𝒙𝟐𝒌+𝒊
𝟖
𝒊=𝟏
= √𝟐𝐜𝐨𝐬 (𝒌𝝅
𝟐) + √𝟐𝐜𝐨𝐬 (
𝒌𝝅
𝟐+𝝅
𝟒) + √𝟐𝐜𝐨𝐬 (
𝒌𝝅
𝟐+𝝅
𝟐) + 𝟒√𝟐𝐜𝐨𝐬 (
𝒌𝝅
𝟐+𝟑𝝅
𝟒)
+
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57 RMM-CALCULUS MARATHON 1501-1600
+√𝟐𝐜𝐨𝐬 (𝒌𝝅
𝟐+ 𝝅) + √𝟐𝐜𝐨𝐬 (
𝒌𝝅
𝟐+𝟓𝝅
𝟒) + √𝟐𝐜𝐨𝐬 (
𝒌𝝅
𝟐+𝟑𝝅
𝟐) + √𝟐 𝐜𝐨𝐬 (
𝒌𝝅
𝟐+𝟕𝝅
𝟒) = 𝟎
𝒙𝟑𝒌+𝒊 = √𝟐𝐜𝐨𝐬 ((𝟑𝒌 + 𝒊 − 𝟏)𝝅
𝟒) , 𝒙𝟓𝒌+𝒊 = √𝟐𝐜𝐨𝐬 (
(𝟓𝒌 + 𝒊 − 𝟏)𝝅
𝟒) ⇒
𝒙𝟑𝒌+𝒊 + 𝒙𝟓𝒌+𝒊 = 𝟐√𝟐𝐜𝐨𝐬 ((𝟖𝒌 + 𝟐𝒊 − 𝟐)𝝅
𝟖) 𝐜𝐨𝐬 (
𝒌𝝅
𝟒)
= 𝟐√𝟐𝐜𝐨𝐬 (𝒌𝝅 +(𝒊 − 𝟏)𝝅
𝟒) 𝐜𝐨𝐬 (
𝒌𝝅
𝟒)
∑(𝒙𝟓𝒌+𝒊 + 𝒙𝟑𝒌+𝒊)
𝟖
𝒊=𝟏
= 𝟐√𝟐∑𝐜𝐨𝐬(𝒌𝝅 +(𝒊 − 𝟏)𝝅
𝟒)𝐜𝐨𝐬 (
𝒌𝝅
𝟒)
𝟖
𝒊=𝟏
= 𝟎
Therefore,
𝛀(𝒏) =∑∑(𝒙𝟐𝒌+𝒊 + 𝒙𝟑𝒌+𝒊 + 𝒙𝟓𝒌+𝒊)
𝟖
𝒊=𝟏
𝒏
𝒌=𝟏
= 𝟎
Solution 2 by Ravi Prakash-New Delhi-India
Let generating function of (𝒙𝒏)𝒏≥𝟏 be
𝑨(𝒕) = 𝒙𝟎 + 𝒙𝟏𝒕 + 𝒙𝟐𝒕𝟐 + 𝒙𝟑𝒕
𝟑 +⋯
−√𝟐𝒕𝑨(𝒕) = −√𝟐𝒙𝟎𝒕 − √𝟐𝒙𝟏𝒕𝟐 − √𝟐𝒙𝟐𝒕
𝟑 −⋯
𝒕𝟐𝑨(𝒕) = 𝒙𝟎𝒕𝟐 + 𝒙𝟏𝒕
𝟑 +⋯
(𝟏 − √𝟐𝒕 + 𝒕𝟐)𝑨(𝒕) = 𝟏 ⇒ 𝑨(𝒕) =𝟏
𝟏 − √𝟐𝒕 + 𝒕𝟐=
𝟏
(𝟏 − 𝜶𝒕)(𝟏 − 𝜷𝒕), 𝐰𝐡𝐞𝐫𝐞
𝜶 =𝟏
𝟐(√𝟐 + √𝟐𝒊) =
𝟏
√𝟐(𝟏 + 𝒊) 𝐚𝐧𝐝 𝜷 =
𝟏
√𝟐(𝟏 − 𝒊).
𝑨(𝒕) =𝟏
(𝜶 − 𝜷)𝒕(
𝟏
𝟏 − 𝜶𝒕−
𝟏
𝟏 − 𝜷𝒕) =
𝟏
(𝜶 − 𝜷)𝒕[(𝟏 − 𝜶𝒕)−𝟏 − (𝟏 − 𝜷𝒕)−𝟏] =
=𝟏
𝜶− 𝜷(𝜶𝒏+𝟏 − 𝜷𝒏+𝟏) =
=𝟏
√𝟐𝒊[𝐜𝐨𝐬
(𝒏 + 𝟏)𝝅
𝟒+ 𝒊 𝐬𝐢𝐧
(𝒏 + 𝟏)𝝅
𝟒− 𝐜𝐨𝐬
(𝒏 + 𝟏)𝝅
𝟒+ 𝒊 𝐬𝐢𝐧
(𝒏 + 𝟏)𝝅
𝟒]
Thus, 𝒙𝒏 = √𝟐𝐬𝐢𝐧(𝒏+𝟏)𝝅
𝟒. Now,
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58 RMM-CALCULUS MARATHON 1501-1600
∑𝒙𝟐𝒌+𝒊
𝟖
𝒊=𝟏
= √𝟐∑𝐬𝐢𝐧 ((𝟐𝒌 + 𝒊 + 𝟏)𝝅
𝟒)
𝟖
𝒊=𝟏
=
= √𝟐 [𝟐 𝐬𝐢𝐧 ((𝟐𝒌 + 𝟏𝟏)𝝅
𝟖)(𝐜𝐨𝐬
𝟕𝝅
𝟖+ 𝐜𝐨𝐬
𝟓𝝅
𝟖+ 𝐜𝐨𝐬
𝟑𝝅
𝟖+ 𝐜𝐨𝐬
𝝅
𝟖)] =
= √𝟐 [𝟐 𝐬𝐢𝐧 ((𝟐𝒌 + 𝟏𝟏)𝝅
𝟖)(𝐜𝐨𝐬 (𝝅 −
𝝅
𝟖) + 𝐜𝐨𝐬 (𝝅 −
𝟑𝝅
𝟖) + 𝐜𝐨𝐬
𝟑𝝅
𝟖+ 𝐜𝐨𝐬
𝝅
𝟖)] = 𝟎
Similarly,
∑𝒙𝟑𝒌+𝒊
𝟖
𝒊=𝟏
= 𝟎 𝐚𝐧𝐝 ∑𝒙𝟓𝒌+𝒊
𝟖
𝒊=𝟏
= 𝟎
Therefore,
𝛀(𝒏) =∑∑(𝒙𝟐𝒌+𝒊 + 𝒙𝟑𝒌+𝒊 + 𝒙𝟓𝒌+𝒊)
𝟖
𝒊=𝟏
𝒏
𝒌=𝟏
= 𝟎
1538. For 𝒂, 𝒃 > 𝟎 prove that:
∫𝒙𝟐 − 𝒂
𝒙𝟐 + 𝒃
∞
−∞
𝐬𝐢𝐧 (𝒙
√𝒃𝐥𝐨𝐠 (
𝒂 + 𝒃
𝒂))𝒅𝒙
𝒙= 𝟎
Proposed by Srinivasa Raghava-AIRMC-India
Solution by Kartick Chandra Betal-India
∫𝒙𝟐 − 𝒂
𝒙𝟐 + 𝒃
∞
−∞
𝐬𝐢𝐧 (𝒙
√𝒃𝐥𝐨𝐠 (
𝒂 + 𝒃
𝒂))𝒅𝒙
𝒙= 𝟐∫
𝒙𝟐 − 𝒂
𝒙𝟐 + 𝒃
∞
𝟎
𝐬𝐢𝐧 (𝒙
√𝒃𝐥𝐨𝐠 (
𝒂 + 𝒃
𝒂))𝒅𝒙
𝒙=
= 𝟐∫𝟏
𝒙⋅ 𝐬𝐢𝐧 (
𝒙
√𝒃𝐥𝐨𝐠 (
𝒂 + 𝒃
𝒂))
∞
𝟎
𝒅𝒙 − 𝟐(𝒂 + 𝒃)∫𝟏
𝒙(𝒙𝟐 + 𝒃)⋅𝒙
√𝒃𝐥𝐨𝐠 (𝟏 +
𝒃
𝒂)
∞
𝟎
𝒅𝒙 =
= 𝟐 ⋅𝝅
𝟐− 𝟐(𝒂 + 𝒃)∫
𝟏
𝒙𝟐 + 𝒃∫ 𝐜𝐨𝐬(𝒙𝒚)
𝟏
√𝒃𝐥𝐨𝐠(𝟏+
𝒃𝒂)
𝟎
𝒅𝒚∞
𝟎
𝒅𝒙 =
= 𝝅 − 𝟐(𝒂 + 𝒃)∫ ∫𝐜𝐨𝐬(√𝒃𝒙𝒚)
√𝒃(𝟏 + 𝒙𝟐)
∞
𝟎
𝒅𝒙
𝟏
√𝒃𝐥𝐨𝐠(𝟏+
𝒃𝒂)
𝟎
𝒅𝒚 =
= 𝝅 − 𝟐(𝒂 + 𝒃)∫𝝅
𝟐√𝒃⋅ 𝒆−√𝒃𝒚
𝟏
√𝒃𝐥𝐨𝐠(𝟏+
𝒃𝒂)
𝟎
𝒅𝒚 =
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59 RMM-CALCULUS MARATHON 1501-1600
= 𝝅−𝝅(𝒂 + 𝒃)
√𝒃⋅𝟏
√𝒃[𝟏 − 𝒆− 𝐥𝐨𝐠(𝟏+
𝒃𝒂)] = 𝝅 −
𝝅(𝒂 + 𝒃)
𝒃(𝟏 −
𝒂
𝒂 + 𝒃) = 𝟎
1539. Prove that:
𝑭𝟏(𝒌, 𝒌; 𝒌 + 𝟏; 𝟏 − 𝒎)𝟐 =𝚪(𝒌 + 𝟏)
𝚪(𝒌)∫
𝒅𝒙
(𝟏 + 𝒙)(𝒎 + 𝒙)𝒌
∞
𝟎
where 𝑭𝟏(. , . ; . ; . ) −𝟐 𝒉𝒚𝒑𝒆𝒓𝒈𝒆𝒐𝒎𝒆𝒕𝒓𝒊𝒄 function, 𝒎 ∈ ℝ+, 𝒌 ∈ ℕ.
Proposed by Simon Peter-Madagascar
Solution by Syed Shahabudeen-Kerala-India
𝑭𝟏(𝒌,𝒌; 𝒌 + 𝟏; 𝟏 −𝒎)𝟐 =
= (𝒎)−𝒌 𝑭𝟏 (𝒌, 𝟏; 𝒌 + 𝟏;𝒎 − 𝟏
𝒎)𝟐 (𝒂𝒑𝒑𝒍𝒚 𝑷𝒇𝒂𝒇𝒇 𝑻𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎𝒂𝒕𝒊𝒐𝒏)
=𝚪(𝒌 + 𝟏)
𝒎𝒌𝚪(𝒌)∫
(𝟏 − 𝒕)𝒌−𝟏
(𝟏 − (𝒎− 𝟏𝒎 ) 𝒕)
𝒌
𝟏
𝟎
𝒅𝒕; (𝒂𝒑𝒑𝒍𝒚 𝑬𝒖𝒍𝒆𝒓 𝑰𝒏𝒕𝒆𝒈𝒓𝒂𝒍)
=𝚪(𝒌 + 𝟏)
𝚪(𝒌)∫
(𝟏 − 𝒕)𝒌−𝟏
(𝒎 − (𝒎− 𝟏)𝒕)𝒌
𝟏
𝟎
𝒅𝒕 =𝚪(𝒌 + 𝟏)
𝚪(𝒌)∫
𝟏
(𝟏 − 𝒕) (𝒎+𝒕
𝟏 − 𝒕)𝒌𝒅𝒕
𝟏
𝟎
=𝒙=
𝒕𝟏−𝒕
=𝚪(𝒌 + 𝟏)
𝚪(𝒌)∫
𝒙 + 𝟏
(𝒎+ 𝒙)𝒌𝒅𝒙
(𝒙 + 𝟏)𝟐
∞
𝟎
=𝚪(𝒌 + 𝟏)
𝚪(𝒌)∫
𝒅𝒙
(𝒙 + 𝟏)(𝒎+ 𝒙)𝒌
∞
𝟎
1540. Prove that:
∏𝒏+ 𝐜𝐨𝐬 (
𝒏𝝅𝟑)
𝒏 + 𝐜𝐨𝐭 (𝝅𝟑) 𝐬𝐢𝐧 (
𝒏𝝅𝟑)
∞
𝒏=𝟏
=𝝅√𝟐 ⋅ 𝚪 (
𝟏𝟐) 𝚪 (
𝟓𝟏𝟐)
𝚪𝟐 (𝟏𝟒) 𝚪 (
𝟏𝟑) 𝚪 (
𝟕𝟔) 𝚪 (
𝟏𝟏𝟏𝟐)
Proposed by Asmat Qatea-Afghanistan
Solution by Amrit Awasthi-India
𝐜𝐨𝐬 (𝒏𝝅
𝟑) and 𝐬𝐢𝐧 (
𝒏𝝅
𝟑) have same sighn, ∀𝒏 = 𝟔𝒌 + 𝟏 or 𝒏 = 𝟔𝒌 + 𝟓.
Therefore, rewriting the product we get:
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60 RMM-CALCULUS MARATHON 1501-1600
𝑷 =∏𝟔𝒌+ 𝟐 −
𝟏𝟐
𝟔𝒌 + 𝟐 +𝟏𝟐
∞
𝒌=𝟎
⋅𝟔𝒌 + 𝟑 − 𝟏
𝟔𝒌 + 𝟑⋅𝟔𝒌 + 𝟔 + 𝟏
𝟔𝒌 + 𝟔=
=𝟏𝟏 ⋅ 𝟕
𝟓 ⋅ 𝟑 ⋅ 𝟗𝐥𝐢𝐦𝒏→∞
∏
𝚪(𝒏+ 𝟏 +𝟏𝟒)
𝚪 (𝟏 +𝟏𝟒)
𝚪 (𝒏 + 𝟏 +𝟏𝟑)
𝚪 (𝟏 +𝟏𝟑)
𝚪(𝒏 + 𝟏 +𝟏𝟏𝟏𝟐)
𝚪(𝟏 +𝟏𝟏𝟏𝟐)
𝚪(𝒏 + 𝟏 +𝟕𝟔)
𝚪(𝟏 +𝟕𝟔)
𝚪 (𝒏 + 𝟏 +𝟓𝟏𝟐)
𝚪 (𝟏 +𝟓𝟏𝟐)
𝚪 (𝒏 + 𝟏 +𝟏𝟐)
𝚪 (𝟏 +𝟏𝟐)
𝚪(𝒏 + 𝟏 +𝟑𝟒)
𝚪(𝟏 +𝟑𝟒)
𝚪(𝒏 + 𝟏)𝚪(𝟐)
∞
𝒌=𝟏
=
=𝟏𝟏 ⋅ 𝟕
𝟓 ⋅ 𝟑 ⋅ 𝟗⋅
𝚪 (𝟏 +𝟓𝟏𝟐)𝚪 (𝟏 +
𝟏𝟐) 𝚪(𝟏 +
𝟑𝟒)
𝚪 (𝟏 +𝟏𝟒) 𝚪(𝟏 +
𝟏𝟑) 𝚪 (𝟏 +
𝟏𝟏𝟐)𝚪 (𝟏 +
𝟕𝟔)=
=𝟏𝟏 ⋅ 𝟕
𝟓 ⋅ 𝟑 ⋅ 𝟗⋅
𝟓𝟏𝟐 ⋅
𝟏𝟐 ⋅𝟑𝟒 ⋅ 𝚪 (
𝟓𝟏𝟐)𝚪 (
𝟏𝟐)𝚪 (
𝟑𝟒)
𝟏𝟒 ⋅𝟏𝟑 ⋅𝟏𝟏𝟏𝟐 ⋅
𝟕𝟔 ⋅ 𝚪 (
𝟏𝟒)𝚪 (
𝟏𝟑)𝚪 (
𝟏𝟏𝟐)𝚪 (
𝟕𝟔)=
=𝚪(𝟓𝟏𝟐)𝚪 (
𝟏𝟐)𝚪 (
𝟑𝟒)
𝚪(𝟏𝟒)𝚪 (
𝟏𝟑)𝚪 (
𝟏𝟏𝟐)𝚪 (
𝟕𝟔)=𝚪 (𝟓𝟏𝟐)𝚪(
𝟏𝟐) 𝚪(𝟏 −
𝟏𝟒)
𝚪(𝟏𝟒) 𝚪(
𝟏𝟑) 𝚪(
𝟏𝟏𝟐)𝚪 (
𝟕𝟔)=
𝝅√𝟐 ⋅ 𝚪 (𝟏𝟐)𝚪 (
𝟓𝟏𝟐)
𝚪𝟐 (𝟏𝟒)𝚪 (
𝟏𝟑)𝚪 (
𝟕𝟔)𝚪 (
𝟏𝟏𝟏𝟐)
Therefore,
∏𝒏+ 𝐜𝐨𝐬 (
𝒏𝝅𝟑 )
𝒏 + 𝐜𝐨𝐭 (𝝅𝟑) 𝐬𝐢𝐧 (
𝒏𝝅𝟑 )
∞
𝒏=𝟏
=𝝅√𝟐 ⋅ 𝚪 (
𝟏𝟐)𝚪 (
𝟓𝟏𝟐)
𝚪𝟐 (𝟏𝟒)𝚪 (
𝟏𝟑)𝚪 (
𝟕𝟔)𝚪 (
𝟏𝟏𝟏𝟐)
1541. If all the derivatives of 𝒇(𝒙) are defined at 𝒙 = 𝟏, then prove that:
𝒇(𝒆−𝒙) = ∑(−𝒙)𝒌
𝒌![𝑩𝒌(𝑫)𝒇(𝒙)]|𝒙=𝟏
∞
𝒌=𝟎
where, 𝑫 ≔𝒅
𝒅𝒙 and 𝑩𝒌(𝒙) is the Bell polynomial.
Proposed by Angad Singh-India
Solution by proposer
𝐋𝐞𝐭 𝒇(𝒙) = ∑𝒂𝒏𝒙𝒏
∞
𝒏=𝟎
, 𝐭𝐡𝐞𝐧
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61 RMM-CALCULUS MARATHON 1501-1600
𝒇(𝒆−𝒙) = ∑𝒂𝒏𝒆−𝒏𝒙
∞
𝒏=𝟎
= ∑𝒂𝒏
∞
𝒏=𝟎
∑(−𝒏𝒙)𝒌
𝒌!
∞
𝒌=𝟎
Hence,
𝒇(𝒆−𝒙) = ∑(−𝒙)𝒌
𝒌!
∞
𝒌=𝟎
∑𝒏𝒌𝒂𝒏
∞
𝒏=𝟎
𝐍𝐨𝐰, 𝐥𝐞𝐭: 𝑭𝒎(𝒙) = ∑𝒏𝒎𝒂𝒏𝒙𝒏
∞
𝒏=𝟎
= ∑𝒄(𝒎, 𝒏)𝒇(𝒏)(𝒙)𝒙𝒏𝒎
𝒏=𝟎
𝒙𝑭𝒎′ (𝒙) = 𝑭𝒎+𝟏(𝒙) ⇒ 𝒄(𝒎,𝒏) = 𝒄(𝒎− 𝟏, 𝒏 − 𝟏) + 𝒏𝒄(𝒎− 𝟏, 𝒏), where
𝟎 ≤ 𝒏 ≤ 𝒎, 𝒄(𝒎, 𝟎) = 𝜹𝒎𝟎 and 𝒄(𝒏, 𝒏) = 𝟏, where 𝜹𝒎𝒏 is the Kronecker delta. It is known from the definition of Bell polynomials that,
𝑩𝒎(𝒙) = ∑𝑺(𝒎,𝒌)𝒙𝒌𝒎
𝒌=𝟎
; 𝑺(𝒎, 𝒌) − (𝑺𝒕𝒊𝒓𝒍𝒊𝒏𝒈 𝒏𝒖𝒎𝒃𝒆𝒓𝒔 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒆𝒄𝒐𝒏𝒅 𝒌𝒊𝒏𝒅)
𝑺(𝒎, 𝒌) = 𝑺(𝒎− 𝟏,𝒌 − 𝟏) + 𝒌𝑺(𝒎− 𝟏, 𝒌), where 𝟎 ≤ 𝒌 ≤ 𝒎, using this property and knowing the fact that 𝑺(𝒏,𝒏) = 𝟏 and 𝑺(𝒏, 𝟎) = 𝜹𝒏𝟎, we can show that:
𝒙𝑩𝒎′ (𝒙𝑫) = 𝑩𝒎+𝟏(𝒙𝑫) since 𝒄(𝒎,𝒏) and 𝑺(𝒎, 𝒌) satisfies the same recurrence relation
with same boundary/initial conditions, we conclude that: 𝑺(𝒎, 𝒌) = 𝒄(𝒎, 𝒌) ⇒ 𝑭𝒎(𝒙) = 𝑩𝒎(𝒙𝑫)𝒇(𝒙) ⇒ 𝑭𝒎(𝟏) = [𝑩𝒌(𝑫)𝒇(𝒙)]|𝒙=𝟏
1542. If 𝚽𝒏 = ∑ ∑ (𝒏𝒊) (𝒏
𝒋)𝒏
𝒋=𝟎 𝐜𝐨𝐬 (𝟐𝝅(𝒋−𝒊)
𝟕)𝒏
𝒊=𝟎 −
𝟐∑ (𝒏𝒊) (𝒏
𝒋)𝟎≤𝒊<𝑗≤𝑛 𝐜𝐨𝐬 (
𝟐𝝅(𝒊−𝒋)
𝟕). Define 𝑴 = {√𝚽𝒏
𝐜𝐨𝐬(𝒏𝝅)𝒏
𝐬𝐢𝐧 (𝒏𝝅
𝟒) |𝒏 ∈ ℕ}.
Find 𝑴′ −derived set.
Proposed by Surjeet Singhania-India
Solution by proposer
𝚽𝒏 =∑∑(𝒏
𝒊)(𝒏
𝒋)
𝒏
𝒋=𝟎
𝐜𝐨𝐬 (𝟐𝝅(𝒋 − 𝒊)
𝟕)
𝒏
𝒊=𝟎
− 𝟐 ∑ (𝒏
𝒊) (𝒏
𝒋)
𝟎≤𝒊<𝑗≤𝑛
𝐜𝐨𝐬 (𝟐𝝅(𝒊 − 𝒋)
𝟕)
Evaluate these finite series one by one
∑∑(𝒏
𝒋) (𝒏
𝒌)𝐜𝐨𝐬 (
𝟐𝝅(𝒌 − 𝒋)
𝟕)
𝒏
𝒌=𝟎
𝒏
𝒋=𝟎
= 𝓡(∑(𝒏
𝒌) 𝒆−
𝟐𝒊𝝅𝒌𝟕
𝐧
𝐤=𝟎
∑(𝒏
𝒋)
𝒏
𝒋=𝟎
𝒆𝟐𝝅𝒊𝒋𝟕 ) =
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62 RMM-CALCULUS MARATHON 1501-1600
= (𝟏 + 𝐞𝐱𝐩 (𝟐𝝅
𝟕))𝒏
(𝟏 + 𝐞𝐱𝐩 (−𝟐𝝅
𝟕))𝒏
= 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏 (𝝅
𝟕)
∑ (𝒏
𝒌)(𝒏
𝒋) 𝐜𝐨𝐬 (
𝟐𝝅(𝒋 − 𝒌)
𝟕)
𝟎≤𝒌<𝑗≤𝑛
= 𝟐𝟐𝒏−𝟏 𝐜𝐨𝐬𝟐𝒏 (𝝅
𝟕) −
𝟏
𝟐(𝟐𝒏
𝒏)
Hence our 𝚽𝒏 = (𝟐𝒏𝒏). Denote 𝑿𝒏 = √𝚽𝒏
𝐜𝐨𝐬(𝒏𝝅)𝒏
𝐬𝐢𝐧 (𝒏𝝅
𝟒).
For finding derived set we need to find possible convergent sequences
𝑿𝟒𝒏 = 𝟎,𝑿𝟒𝒏+𝟏, 𝑿𝟒𝒏+𝟑 → ±√𝟐
𝟖 and 𝑿𝟒𝒏+𝟐 → ±𝟒.
Hence, 𝑴′ = {𝟎, ±𝟒, ±√𝟐
𝟖}
1543. If 𝒂, 𝒃, 𝒏, 𝒌 ∈ ℕ and 𝑺(𝒂, 𝒃, 𝒏) = {𝒌|𝒌 ≡ 𝒂(𝒎𝒐𝒅 𝒃), 𝒌|𝒏} then prove
that:
∑𝒙𝒂𝒌
𝟏 − 𝒙𝒃𝒌
∞
𝒌=𝟏
= ∑|𝑺(𝒂, 𝒃, 𝒏)|𝒙𝒏∞
𝒏=𝟏
Proposed by Angad Singh-India
Solution by proposer
Observe that if |𝒙| < 1, then
𝒙𝒂
𝟏 − 𝒙𝒃= 𝒙𝒂(𝟏 + 𝒙𝒃 + 𝒙𝟐𝒃 + 𝒙𝟑𝒃 + 𝒙𝟒𝒃 +⋯)
𝒙𝟐𝒂
𝟏 − 𝒙𝟐𝒃= 𝒙𝟐𝒂(𝟏 + 𝒙𝟐𝒃 + 𝒙𝟒𝒃 + 𝒙𝟔𝒃 + 𝒙𝟖𝒃 +⋯)
𝒙𝟑𝒂
𝟏 − 𝒙𝟑𝒃= 𝒙𝟑𝒂(𝟏 + 𝒙𝟑𝒃 + 𝒙𝟔𝒃 + 𝒙𝟗𝒃 + 𝒙𝟏𝟐𝒃 +⋯)
Adding them, we have:
∑𝒙𝒂𝒌
𝟏 − 𝒙𝒃𝒌
∞
𝒌=𝟏
= ∑𝒂𝒏𝒙𝒏
∞
𝒏=𝟏
Where 𝒂𝒏 is the number of solutions of 𝒂𝒑 + 𝒃𝒑𝒒 = 𝒏, where 𝒑 ∈ ℕ and 𝒒 ∈ ℕ + {𝟎} for
the some given values of 𝒂 and 𝒃, thus 𝒂𝒏 is the number of divisors of 𝒏 of the form 𝒃𝒒 +
𝒂.
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63 RMM-CALCULUS MARATHON 1501-1600
Using the definition of 𝑺(𝒂, 𝒃, 𝒏) we can show that 𝒂𝒏 = |𝑺(𝒂, 𝒃, 𝒏)| and this completes
the proof.
1544. Find:
𝛀 = ∫√𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟑 + 𝒙√𝒙 + 𝟏𝒅𝒙
∞
𝟏
Proposed by Vasile Mircea Popa-Romania
Solution 1 by Rana Ranino-Setif-Algerie
𝛀 = ∫√𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟑 + 𝒙√𝒙+ 𝟏𝒅𝒙
∞
𝟏
=𝒙=𝒙𝟐
𝟒∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟔 + 𝒙𝟑 + 𝟏
∞
𝟏
𝒅𝒙 =𝒙=𝟏𝒙− 𝟒∫
𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟔 + 𝒙𝟑 + 𝟏𝒅𝒙
𝟏
𝟎
=
= 𝟒∫𝒙𝟐(𝒙𝟑 − 𝟏) 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟗
𝟏
𝟎
𝒅𝒙 = 𝟒∫(𝒙𝟓 − 𝒙𝟐) 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟗
𝟏
𝟎
𝒅𝒙 =
= 𝟒∑∫ (𝒙𝟗𝒏+𝟓 − 𝒙𝟗𝒏+𝟐) 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
=
∞
𝒏=𝟎
𝟒∑(𝟏
(𝟗𝒏 + 𝟑)𝟐−
𝟏
(𝟗𝒏 + 𝟔)𝟐)
∞
𝒏=𝟎
=
=𝟒
𝟖𝟏∑(
𝟏
(𝒏+𝟏𝟑)𝟐 −
𝟏
(𝒏 +𝟐𝟑)𝟐)
∞
𝒏=𝟎
Therefore,
𝛀 = ∫√𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟑 + 𝒙√𝒙+ 𝟏𝒅𝒙
∞
𝟏
=𝟒
𝟖𝟏[𝝍(𝟏) (
𝟏
𝟑) − 𝝍(𝟏) (
𝟐
𝟑)]
Solution 2 by Ajetunmobi Abdulqoyyum-Nigeria
𝛀 = ∫√𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟑 + 𝒙√𝒙 + 𝟏𝒅𝒙
∞
𝟏
=𝒙=𝒕𝟐
𝟒∫𝒕𝟐 𝐥𝐨𝐠 𝒕
𝒕𝟔 + 𝒕𝟑 + 𝟏𝒅𝒕
∞
𝟏
𝛀 = ∫𝒕𝟐 𝐥𝐨𝐠 𝒕
𝒕𝟔 + 𝒕𝟑 + 𝟏𝒅𝒕
∞
𝟏
=𝒕=𝟏𝒕− 𝟒∫
𝒕𝟐 𝐥𝐨𝐠 𝒕
𝒕𝟔 + 𝒕𝟑 + 𝟏𝒅𝒕
𝟏
𝟎
𝛀 = −𝟒∫(𝟏 − 𝒕𝟑)𝒕𝟐 𝐥𝐨𝐠 𝒕
𝟏 − 𝒕𝟗𝒅𝒕
𝟏
𝟎
= −∫𝒕𝟐 𝐥𝐨𝐠 𝒕
𝟏 − 𝒕𝟗𝒅𝒕
𝟏
𝟎
+ 𝟒∫𝒕𝟓 𝐥𝐨𝐠 𝒕
𝟏 − 𝒕𝟗𝒅𝒕
𝟏
𝟎
=𝒕𝟗=𝒙
=𝟒
𝟖𝟏(−∫
𝒕𝟏𝟑−𝟏 𝐥𝐨𝐠 𝒕
𝟏 − 𝒕𝒅𝒕
𝟏
𝟎
+∫𝒕𝟐𝟑−𝟏 𝐥𝐨𝐠 𝒕
𝟏 − 𝒕𝒅𝒕
𝟏
𝟎
) =
Therefore,
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64 RMM-CALCULUS MARATHON 1501-1600
𝛀 = ∫√𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟑 + 𝒙√𝒙+ 𝟏𝒅𝒙
∞
𝟏
=𝟒
𝟖𝟏[𝝍(𝟏) (
𝟏
𝟑) − 𝝍(𝟏) (
𝟐
𝟑)]
Solution 3 by Muhammad Afzal-Pakistan
𝝍(𝒎) = −∫𝒕𝒛−𝟏
𝟏 − 𝒕𝐥𝐨𝐠𝒎 𝒕 𝒅𝒕
𝟏
𝟎
𝛀 = ∫√𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟑 + 𝒙√𝒙+ 𝟏𝒅𝒙
∞
𝟏
=𝒙=√𝒙
𝟒∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟔 + 𝒙𝟑 + 𝟏𝒅𝒙
∞
𝟎
=𝒙=𝟏𝒙
= −𝟒∫𝒙𝟐 𝐥𝐨𝐠 𝒙
𝒙𝟔 + 𝒙𝟑 + 𝟏𝒅𝒙
𝟏
𝟎
= 𝟒{∫𝒙𝟓 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟗𝒅𝒙
𝟏
𝟎
−∫𝒙𝟐 𝐥𝐨𝐠𝒙
𝟏 − 𝒙𝟗𝒅𝒙
𝟏
𝟎
}
𝑨 = ∫𝒙𝟓 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝟗𝒅𝒙
𝟏
𝟎
=𝟏
𝟖𝟏∫𝒙−𝟏𝟑 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝒅𝒙
𝟏
𝟎
= −𝟏
𝟖𝟏𝝍(𝟏) (
𝟐
𝟑)
𝑩 = ∫𝒙𝟐 𝐥𝐨𝐠𝒙
𝟏 − 𝒙𝟗𝒅𝒙
𝟏
𝟎
=𝟏
𝟖𝟏∫𝒙−𝟐𝟑 𝐥𝐨𝐠 𝒙
𝟏 − 𝒙𝒅𝒙
𝟏
𝟎
= −𝟏
𝟖𝟏𝝍(𝟏) (
𝟏
𝟑)
Therefore,
𝛀 = 𝟒(𝑨 − 𝑩) =𝟒
𝟖𝟏[𝝍(𝟏) (
𝟏
𝟑) − 𝝍(𝟏) (
𝟐
𝟑)]
Solution 4 by Probal Chakraborty-Kolkata-India
𝝍(𝒎) = −∫𝒕𝒛−𝟏
𝟏 − 𝒕𝐥𝐨𝐠𝒎 𝒕 𝒅𝒕
𝟏
𝟎
𝛀 = ∫√𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟑 + 𝒙√𝒙 + 𝟏𝒅𝒙
∞
𝟏
= ∫√𝒙 𝐥𝐨𝐠 𝒙
(𝒙𝟑𝟐)𝟐
+ 𝒙𝟑𝟐 + 𝟏
𝒅𝒙∞
𝟎
=𝒙𝟑𝟐=𝒕 𝟐
𝟑∫
𝐥𝐨𝐠 (𝒕𝟐𝟑)
𝒕𝟐 + 𝒕 + 𝟏𝒅𝒕
𝟏
𝟎
=
=𝟒
𝟗∫
𝐥𝐨𝐠 𝒕
𝒕𝟐 + 𝒕 + 𝟏𝒅𝒕
𝟏
𝟎
=𝟒
𝟗∫
𝟏 − 𝒕
𝟏 − 𝒕𝟑𝐥𝐨𝐠 𝒕𝒅𝒕
𝟏
𝟎
=𝟒
𝟗∫
𝐥𝐨𝐠 𝒕
𝟏 − 𝒕𝟑𝒅𝒕
𝟏
𝟎
−𝟒
𝟗∫𝒕 𝐥𝐨𝐠 𝒕
𝟏 − 𝒕𝟑𝒅𝒕
𝟏
𝟎
=𝒕=𝒚
𝟏𝟑
=𝟒
𝟖𝟏∫𝒚−𝟐𝟑 𝐥𝐨𝐠 𝒚
𝟏 − 𝒚𝒅𝒚
𝟏
𝟎
=𝟒
𝟖𝟏∫𝒚−𝟏𝟑 𝐥𝐨𝐠 𝒚
𝟏 − 𝒚𝒅𝒚
𝟏
𝟎
=𝟒
𝟖𝟏[𝝍(𝟏) (
𝟏
𝟑) − 𝝍(𝟏) (
𝟐
𝟑)]
1545. Find:
𝛀(𝟏, 𝟐) = ∫𝐥𝐨𝐠 𝒙
(𝟏 + 𝐥𝐨𝐠 𝒙)𝟐𝒅𝒙 ;𝛀(𝟐, 𝟑) = ∫
𝐥𝐨𝐠𝟐 𝒙
(𝟏 + 𝐥𝐨𝐠 𝒙)𝟑𝒅𝒙
𝛀(𝒎,𝒏) = ∫𝐥𝐨𝐠𝒎 𝒙
(𝟏 + 𝐥𝐨𝐠 𝒙)𝒏𝒅𝒙 ,𝒎, 𝒏 ∈ ℕ
Proposed by Durmuş Ogmen-Turkyie
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65 RMM-CALCULUS MARATHON 1501-1600
Solution by Mikael Bernardo-Mozambique
𝛀(𝟏, 𝟐) = ∫𝐥𝐨𝐠𝒙
(𝟏 + 𝐥𝐨𝐠𝒙)𝟐𝒅𝒙 =
𝐥𝐨𝐠 𝒙=𝒖∫
𝒖𝒆𝒖
(𝟏 + 𝒖)𝟐𝒅𝒖 = ∫
𝒆𝒖
𝟏 + 𝒖𝒅𝒖 −∫
𝒆𝒖
(𝟏 + 𝒖)𝟐𝒅𝒖 =
𝑰𝑩𝑷
= ∫𝒆𝒖
𝟏 + 𝒖𝒅𝒖 + ∫
𝒆𝒖
𝟏 + 𝒖𝒅𝒖 − ∫
𝒆𝒖
𝟏 + 𝒖𝒅𝒖 =
𝒆𝒖
𝟏 + 𝒖+ 𝑪 =
𝒙
𝟏 + 𝐥𝐨𝐠 𝒙+ 𝑪
𝛀(𝟐, 𝟑) = ∫𝐥𝐨𝐠𝟐 𝒙
(𝟏 + 𝐥𝐨𝐠 𝒙)𝟑𝒅𝒙 =
𝐥𝐨𝐠 𝒙=𝒖∫
𝒖𝟐𝒆𝒖
(𝟏 + 𝒖)𝟑𝒅𝒖 = ∫
𝒖𝒆𝒖
(𝟏 + 𝒖)𝟐𝒅𝒖 −∫
𝒖𝒆𝒖
(𝟏 + 𝒖)𝟑𝒅𝒖
𝛀(𝟐, 𝟑) = 𝛀(𝟏, 𝟐) − ∫𝒖𝒆𝒖
(𝟏 + 𝒖)𝟐𝒅𝒖 + ∫
𝒖𝒆𝒖
(𝟏 + 𝒖)𝟑𝒅𝒖 =
𝑰𝑩𝑷
=𝒙
𝟏 + 𝐥𝐨𝐠𝒙− ∫
𝒖𝒆𝒖
(𝟏 + 𝒖)𝟐𝒅𝒖 −
𝒆𝒖
(𝟏 + 𝒖)𝟐+∫
𝒆𝒖
(𝟏 + 𝒖)𝟐𝒅𝒖 =
=𝒙
𝟏 + 𝐥𝐨𝐠𝒙−
𝒆𝒖
(𝟏 + 𝒖)𝟐=
𝒙
𝟏 + 𝐥𝐨𝐠 𝒙−
𝒙
(𝟏 + 𝐥𝐨𝐠 𝒙)𝟐+ 𝑪
𝛀(𝒎,𝒏) = ∫𝐥𝐨𝐠𝒎 𝒙
(𝟏 + 𝐥𝐨𝐠 𝒙)𝒏𝒅𝒙 ,𝒎, 𝒏 ∈ ℕ
∵𝟏
(𝟏 + 𝒖)𝒏=∑(−𝟏)𝒌−(𝒏−𝟏) ⋅
(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!
(𝒌 − (𝒏 − 𝟏))!⋅ 𝒖𝒌−(𝒏−𝟏)
∞
𝒌=𝟎
, ∀𝒏 ≥ 𝟐
𝐏𝐮𝐭 𝒖 = 𝐥𝐨𝐠𝒙 ⇒𝟏
(𝟏 + 𝐥𝐨𝐠 𝒙)𝒏=∑(−𝟏)𝒌−(𝒏−𝟏) ⋅
(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!
(𝒌 − (𝒏 − 𝟏))!⋅ (𝐥𝐨𝐠𝒙)𝒌−(𝒏−𝟏)
∞
𝒌=𝟎
𝛀(𝒎, 𝒏) = ∑(−𝟏)𝒌−(𝒏−𝟏) ⋅(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!
(𝒌 − (𝒏 − 𝟏))!⋅ ∫(𝐥𝐨𝐠 𝒙)𝒎+𝒌−(𝒏−𝟏) 𝒅𝒙
∞
𝒌=𝟎
=
= ∑(−𝟏)𝒌−(𝒏−𝟏) ⋅(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!
(𝒌 − (𝒏 − 𝟏))!⋅𝝏𝒎+𝒌−(𝒏−𝟏)
𝝏𝒂𝒎+𝒌−(𝒏−𝟏)|𝒂=𝟎
∞
𝒌=𝟎
∫𝒙𝒂 𝒅𝒙 =
=∑(−𝟏)𝒌−(𝒏−𝟏) ⋅(𝒌 + 𝟏 − (𝒏 − 𝟏)!)!
(𝒌 − (𝒏 − 𝟏))!⋅𝝏𝒎+𝒌−(𝒏−𝟏)
𝝏𝒂𝒎+𝒌−(𝒏−𝟏)|𝒂=𝟎
⋅𝒙𝒂+𝟏
𝒂 + 𝟏+ 𝑪
∞
𝒌=𝟎
1546. Find:
𝛀 = ∫ 𝐬𝐢𝐧−𝟏 (𝟏
𝟏 + 𝒙𝟐)𝒅𝒙
∞
𝟎
Proposed by Ajetunmobi Abdulqoyyum-Nigeria
www.ssmrmh.ro
66 RMM-CALCULUS MARATHON 1501-1600
Solution 1 by Amrit Awasthi-India
𝛀 = ∫ 𝐬𝐢𝐧−𝟏 (𝟏
𝟏 + 𝒙𝟐)𝒅𝒙
∞
𝟎
=𝑰𝑩𝑷𝒙 ⋅ 𝐬𝐢𝐧−𝟏 (
𝟏
𝟏 + 𝒙𝟐) + 𝟐∫
𝒙
(𝟏 + 𝒙𝟐)√𝒙𝟐 + 𝟐𝒅𝒙
∞
𝟎
=𝒖=√𝒙𝟐+𝟐
= 𝒙 ⋅ 𝐬𝐢𝐧−𝟏 (𝟏
𝟏 + 𝒙𝟐) + 𝟐∫
𝟏
𝒖𝟐 − 𝟏𝒅𝒖
∞
𝟏
=
= 𝒙 ⋅ 𝐬𝐢𝐧−𝟏 (𝟏
𝟏 + 𝒙𝟐) + 𝟐∫ (
𝟏
𝟐(𝒖 − 𝟏)−
𝟏
𝟐(𝒖 + 𝟏))𝒅𝒖
∞
𝟏
=
= [𝒙 ⋅ 𝐬𝐢𝐧−𝟏 (𝟏
𝟏 + 𝒙𝟐) + 𝐥𝐨𝐠 (
√𝒙𝟐 + 𝟐 − 𝟏
√𝒙𝟐 + 𝟐 + 𝟏)]𝟎
∞
= − 𝐥𝐨𝐠 (√𝟐− 𝟏
√𝟐+ 𝟏)
𝛀 = ∫ 𝐬𝐢𝐧−𝟏 (𝟏
𝟏 + 𝒙𝟐)𝒅𝒙
∞
𝟎
= 𝐥𝐨𝐠(√𝟐 + 𝟏) − 𝐥𝐨𝐠(√𝟐 − 𝟏)
Solution 2 by Abdul Mukhtar-Nigeria
𝛀 = ∫ 𝐬𝐢𝐧−𝟏 (𝟏
𝟏 + 𝒙𝟐)𝒅𝒙
∞
𝟎
= ∫ 𝐜𝐬𝐜−𝟏(𝟏 + 𝒙𝟐) 𝒅𝒙∞
𝟎
=𝑰𝑩𝑷
= [𝒙 ⋅ 𝐜𝐬𝐜−𝟏(𝟏 + 𝒙𝟐)]𝟎∞ + 𝟐∫
𝒙
(𝟏 + 𝒙𝟐)√𝒙𝟐 + 𝟐𝒅𝒙
∞
𝟎
𝛀 = 𝟐∫𝒙
(𝟏 + 𝒙𝟐)√𝒙𝟐 + 𝟐𝒅𝒙
∞
𝟎
=𝒚=√𝒙𝟐+𝟐
∫𝟐𝒚
(𝒚𝟐 − 𝟏)𝒚𝒅𝒚
∞
√𝟐
=𝟏
𝟐∫
𝟐(𝒚 + 𝟏) − 𝟐(𝒚 − 𝟏)
𝒚𝟐 − 𝟏𝒅𝒚
∞
√𝟐
= [𝐥𝐨𝐠 |𝒚 − 𝟏
𝒚 + 𝟏|]√𝟐
∞
= 𝐥𝐨𝐠√𝟐 + 𝟏
√𝟐 − 𝟏
1547. Prove that:
∫ 𝒆−𝒙∏(𝟏 − 𝒆−𝟔𝒙𝒌)(𝟏 + 𝒆−𝟔𝒙𝒌+𝒙)(𝟏 + 𝒆−𝟔𝒙𝒌+𝟓𝒙)𝒅𝒙
∞
𝒌=𝟏
∞
𝟎
=𝝅
√𝟐⋅
𝐬𝐢𝐧𝐡 (𝟐𝝅√𝟐𝟑 )
𝐜𝐨𝐬𝐡(𝟐𝝅√𝟐𝟑 ) − 𝐜𝐨𝐬 (
𝟐𝝅𝟑 )
Proposed by Syed Shahabudeen-India
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67 RMM-CALCULUS MARATHON 1501-1600
Solution by Kaushik Mahanta-Assam-India
Recall definition of Jacobi’s triple product identity:
∑ 𝒒𝒌𝟐𝒛𝒌
∞
𝒌=−∞
=∏(𝟏 − 𝒒𝟐𝒌)(𝟏 − 𝒛 − 𝟏 𝒒𝟐𝒌−𝟏)(𝟏 + 𝒛𝒒𝟐𝒌−𝟏)
∞
𝒌=𝟏
Comparing, we get:
𝒒𝟐𝒌 = 𝒆−𝟔𝒙𝒌, 𝒒 = 𝒆−𝟑𝒙; 𝒛𝒒𝟐𝒌−𝟏 = 𝒆𝒙𝒆−𝟔𝒙𝒌 ⇒𝒛 ⋅ 𝒆−𝟔𝒙𝒌
𝒆−𝟑𝒙= 𝒆−𝟔𝒙𝒌 ⋅ 𝒆𝒙 ⇒ 𝒛 = 𝒆−𝟐𝒙
𝑰 = ∫ 𝒆−𝒙 ∑ (𝒆−𝟑𝒙)𝒌𝟐(𝒆−𝟐𝒙)𝒌𝒅𝒙
∞
𝒌=−∞
∞
𝟎
= ∑ ∫ 𝒆−𝒙(𝟏+𝟐𝒌+𝟑𝒌𝟐)𝒅𝒙
∞
𝟎
∞
𝒌=−∞
=∑𝟏
𝟑𝒌𝟐 + 𝟐𝒌+ 𝟏
∞
𝒌=𝟏
∑𝟏
𝒂𝒌𝟐 + 𝒃𝒌 + 𝒄=
∞
𝒌=−∞
𝟐𝝅
√𝚫⋅
𝐬𝐢𝐧 (𝝅√𝚫𝒂 )
𝐜𝐨𝐬 (𝝅√𝚫𝒂 ) − 𝒄𝒐𝒔 (
𝝅𝒃𝒂 )
For 𝒂 = 𝟑, 𝒃 = 𝟐, 𝒄 = 𝟏,𝚫 = −𝟖 ⇒
∑𝟏
𝟑𝒌𝟐 + 𝟐𝒌 + 𝟏
∞
𝒌=𝟏
=𝟐𝝅
𝒊𝟐√𝟐⋅
𝐬𝐢𝐧 (𝒊𝟐√𝟐𝝅𝟑 )
𝐜𝐨𝐬 (𝒊𝟐√𝟐𝝅𝟑 ) − 𝐜𝐨𝐬 (
𝟐𝒊𝟑 )
=𝝅
√𝟐⋅
𝐬𝐢𝐧𝐡 (𝟐𝝅√𝟐𝟑 )
𝐜𝐨𝐬𝐡(𝟐𝝅√𝟐𝟑 ) − 𝐜𝐨𝐬 (
𝟐𝝅𝟑 )
1548. Find without any software:
𝛀 = ∫ ∫𝒙
√𝒙𝟐 + 𝒚𝟐
√𝟐−𝒙𝟐
𝒙
𝒅𝒙𝟏
𝟎
𝒅𝒚
Proposed by Durmuş Ogmen-Turkiye
Solution 1 by Yen Tung Chung-Taichung-Taiwan
Let 𝒙 = 𝒓 𝐬𝐢𝐧𝜽 , 𝒚 = 𝒓 𝐬𝐢𝐧𝜽, then 𝒙𝟐 + 𝒚𝟐 = 𝒓𝟐, 𝒅𝒙𝒅𝒚 = 𝒓 𝒅𝒓𝒅𝜽 and
𝑹 = {(𝒙, 𝒚)|𝒙 ≤ 𝒚 ≤ √𝟐 − 𝒙𝟐, 𝟎 ≤ 𝒙 ≤ 𝟏} = {(𝒓, 𝜽)|𝟎 ≤ 𝒓 ≤ √𝟐,𝝅𝟒 ≤ 𝜽 ≤
𝝅𝟐}
𝛀 = ∫ ∫𝒙
√𝒙𝟐 + 𝒚𝟐
√𝟐−𝒙𝟐
𝒙
𝒅𝒙𝟏
𝟎
𝒅𝒚 = ∫ ∫𝒓𝐜𝐨𝐬𝜽
𝒓⋅ 𝒓 𝒅𝒓𝒅𝜽
√𝟐
𝟎
𝝅𝟐
𝝅𝟒
=
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68 RMM-CALCULUS MARATHON 1501-1600
= (∫ 𝐜𝐨𝐬𝜽𝒅𝜽
𝝅𝟐
𝝅𝟒
)(∫ 𝒓√𝟐
𝟎
𝒅𝒓) = 𝐬𝐢𝐧𝜽|𝝅𝟒
𝝅𝟐 ⋅𝟏
𝟐𝒓𝟐|
𝟎
√𝟐
=𝟐 − √𝟐
𝟐
Solution 2 by Yen Tung Chung-Taichung-Taiwan
𝛀 = ∫ ∫𝒙
√𝒙𝟐 + 𝒚𝟐
√𝟐−𝒙𝟐
𝒙
𝒅𝒙𝟏
𝟎
𝒅𝒚
= ∫ ∫𝒙
√𝒙𝟐 + 𝒚𝟐𝒅𝒙𝒅𝒚
√𝟐
𝟎
𝟏
𝟎
+∫ ∫𝒙
√𝒙𝟐 + 𝒚𝟐
√𝟐−𝒚𝟐
𝟎
𝒅𝒙√𝟐
𝟏
𝒅𝒚
= ∫ (√𝒙𝟐 + 𝒚𝟐)𝟏
𝟎
|𝟎
𝒚
𝒅𝒚 +∫ (√𝒙𝟐 + 𝒚𝟐)√𝟐
𝟏
|
𝟎
√𝟐−𝒚𝟐
𝒅𝒚 =
= ∫ (√𝟐− 𝟏)𝒚𝟏
𝟎
𝒅𝒚 +∫ (√𝟐 − 𝒚)√𝟐
𝟏
𝒅𝒚 = (√𝟐 − 𝟏)𝒚𝟐
𝟐|𝟎
𝟏
+ (√𝟐𝒚 −𝒚𝟐
𝟐)|𝟏
√𝟐
=𝟐 − √𝟐
𝟐
Solution 3 by Katrick Chandra Betal-India
𝛀 = ∫ ∫𝒙
√𝒙𝟐 + 𝒚𝟐
√𝟐−𝒙𝟐
𝒙
𝒅𝒙𝟏
𝟎
𝒅𝒚 = ∫ 𝒙 [𝐥𝐨𝐠 (𝒚 + √𝒙𝟐 + 𝒚𝟐)]𝒙
√𝟐−𝒙𝟐𝟏
𝟎
𝒅𝒙 =
= ∫ 𝒙 𝐥𝐨𝐠(√𝟐 − 𝒙𝟐 + √𝟐
𝒙 + √𝟐𝒙)𝒅𝒙 =
𝟏
𝟎
= ∫ 𝒙 𝐥𝐨𝐠 (√𝟐 + √𝟐 − 𝒙𝟐)𝒅𝒙𝟏
𝟎
− 𝐥𝐨𝐠(𝟏 + √𝟐 )∫ 𝒙𝟏
𝟎
𝒅𝒙 = ∫ 𝒙 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
=
=𝟏
𝟐∫ 𝐥𝐨𝐠(√𝟐 + √𝟐 − 𝒙)𝒅𝒙𝟏
𝟎
−𝟏
𝟐𝐥𝐨𝐠(𝟏 + √𝟐) − [
𝒙𝟐
𝟐𝐥𝐨𝐠𝒙 −
𝒙𝟐
𝟒]𝟎
𝟏
=
=𝟏
𝟐∫ 𝐥𝐨𝐠(√𝟐 + √𝒙 + 𝟏)𝒅𝒙𝟏
𝟎
−𝟏
𝟐𝐥𝐨𝐠(𝟏 + √𝟐) +
𝟏
𝟒=
=𝟏
𝟒−𝟏
𝟐𝐥𝐨𝐠(𝟏 + √𝟐) +
𝟏
𝟐[𝒙 𝐥𝐨𝐠(√𝟐 + √𝟏 + 𝒙)]
𝟎
𝟏−𝟏
𝟐∫
𝒙
√𝟐+ √𝟏 + 𝒙⋅
𝒅𝒙
𝟐√𝟏 + 𝒙
𝟏
𝟎
=
=𝟏
𝟒−𝟏
𝟐𝐥𝐨𝐠(𝟏 + √𝟐) +
𝟏
𝟐𝐥𝐨𝐠(𝟐√𝟐) −
𝟏
𝟒∫(𝒙 − 𝟏)(√𝒙 − √𝟐)
(𝒙 − 𝟐)√𝒙
𝟏
𝟎
𝒅𝒙 =
=𝟏
𝟒−𝟏
𝟐𝐥𝐨𝐠 (
𝟏 + √𝟐
𝟐√𝟐) −
𝟏
𝟐∫ (𝒙 − √𝟐)𝒅𝒙√𝟐
𝟏
−𝟏
𝟐[𝐥𝐨𝐠(𝒙 + √𝟐)]
𝟏
√𝟐=
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69 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝟒−𝟏
𝟐𝐥𝐨𝐠 (
𝟏 + √𝟐
𝟐√𝟐) −
𝟏
𝟐[𝒙𝟐
𝟐− √𝟐𝒙]
𝟏
√𝟐
− 𝐥𝐨𝐠 (𝟐√𝟐
𝟏+ √𝟐) =
=𝟏
𝟒−𝟏
𝟒+𝟐 − √𝟐
𝟐= 𝟏 −
𝟏
√𝟐
1549. Find without any software:
𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐
(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙𝒙 + 𝟑𝒙𝒆𝒙 + 𝒙)𝒅𝒙
Proposed by Daniel Sitaru-Romania
Solution 1 by Hussain Reza Zadah-Afghanistan
𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐
(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙𝒙 + 𝟑𝒙𝒆𝒙 + 𝒙)𝒅𝒙 =
= ∫(𝟑𝒙𝒆𝒙 + 𝟐)𝒅𝒙
𝒙(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏)= ∫
𝟑𝒆𝒙 +𝟐𝒙
(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙)𝟐 − 𝟏𝒅𝒙 =
𝒖=𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙
= ∫𝒅𝒖
𝒖𝟐 − 𝟏=𝟏
𝟐𝐥𝐨𝐠 |
𝒖 − 𝟏
𝒖 + 𝟏| + 𝑪 =
𝟏
𝟐𝐥𝐨𝐠 |
𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏
𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏| + 𝑪
Solution 2 by Ajetunmobi Abdulqoyyum-Nigeria
𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐
(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙𝒙 + 𝟑𝒙𝒆𝒙 + 𝒙)𝒅𝒙 =
= ∫𝟑𝒙𝒆𝒙 + 𝟐
𝒙(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =
= ∫𝟑𝒆𝒙 +
𝟐𝒙
(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =
𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒕
=𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒕
∫𝟏
(𝒕 − 𝟏)(𝒕 + 𝟏)𝒅𝒕 =
𝟏
𝟐∫𝒅𝒕
𝒕 − 𝟏−𝟏
𝟐∫𝒅𝒕
𝒕 + 𝟏=
=𝟏
𝟐𝐥𝐨𝐠 |
𝒕 − 𝟏
𝒕 + 𝟏| + 𝑪 =
𝟏
𝟐𝐥𝐨𝐠 |
𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏
𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏| + 𝑪
Solution 3 by Timson Azeez Folorunsho-Nigeria
𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐
(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙𝒙 + 𝟑𝒙𝒆𝒙 + 𝒙)𝒅𝒙 =
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= ∫𝟑𝒙𝒆𝒙 + 𝟐
𝒙(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =
= ∫𝟑𝒆𝒙 +
𝟐𝒙
(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =
𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒕
=𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒕
∫𝟏
(𝒕 − 𝟏)(𝒕 + 𝟏)𝒅𝒕 =
𝟏
𝟐∫𝒅𝒕
𝒕 − 𝟏−𝟏
𝟐∫𝒅𝒕
𝒕 + 𝟏=
=𝟏
𝟐𝐥𝐨𝐠 |
𝒕 − 𝟏
𝒕 + 𝟏| + 𝑪 =
𝟏
𝟐𝐥𝐨𝐠 |
𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 − 𝟏
𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏| + 𝑪
Solution 4 by Yen Tung Chung-Taichung-Taiwan
𝛀 = ∫𝟑𝒙𝒆𝒙 + 𝟐
(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙𝒙 + 𝟑𝒙𝒆𝒙 + 𝒙)𝒅𝒙 =
= ∫𝟑𝒙𝒆𝒙 + 𝟐
𝒙(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =
= ∫𝟑𝒆𝒙 +
𝟐𝒙
(𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏)(𝟐 𝐥𝐨𝐠𝒙 + 𝟑𝒆𝒙 + 𝟏)𝒅𝒙 =
𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒚
=𝟐 𝐥𝐨𝐠 𝒙+𝟑𝒆𝒙=𝒚
∫𝟏
(𝒚 − 𝟏)(𝒚 + 𝟏)𝒅𝒕 =
𝟏
𝟐∫
𝒅𝒕
𝒚 − 𝟏−𝟏
𝟐∫
𝒅𝒕
𝒚 + 𝟏=
=𝟏
𝟐𝐥𝐨𝐠 |
𝒚 − 𝟏
𝒚 + 𝟏| + 𝑪 =
𝟏
𝟐𝐥𝐨𝐠 |
𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 − 𝟏
𝟐 𝐥𝐨𝐠 𝒙 + 𝟑𝒆𝒙 + 𝟏| + 𝑪
1550. Prove that:
∫𝒅𝒙
𝟏 + √𝟏 + 𝐬𝐢𝐧𝟐 𝒙
𝝅𝟐
𝟎
=𝝅
𝟐⋅𝚪𝟒 (
𝟏𝟒) − 𝟖𝝅𝟐
(𝟐𝝅)𝟑𝟐𝚪𝟐 (
𝟏𝟒)=𝝅
𝟐(𝑮 −
𝟏
𝝅𝑮)
where 𝑮 =𝚪𝟐(
𝟏
𝟒)
(𝟐𝝅)𝟑𝟐
denotes Gauss Constant.
Proposed by Naren Bhandari-Bajura-Nepal
Solution by Kartick Chandra Betal-India
∫𝒅𝒙
𝟏 + √𝟏 + 𝐬𝐢𝐧𝟐 𝒙
𝝅𝟐
𝟎
= ∫√𝟏+ 𝐬𝐢𝐧𝟐 𝒙 − 𝟏
𝐬𝐢𝐧𝟐 𝒙𝒅𝒙
𝝅𝟐
𝟎
=
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= −(√𝟏 + 𝐬𝐢𝐧𝟐 𝒙 − 𝟏) 𝐜𝐨𝐭 𝒙|𝟎
𝝅𝟐+∫
𝟐 𝐬𝐢𝐧𝒙 𝐜𝐨𝐬 𝒙
𝟐√𝟏 + 𝐬𝐢𝐧𝟐 𝒙𝐜𝐨𝐭 𝒙
𝝅𝟐
𝟎
𝒅𝒙 =
= 𝐥𝐢𝐦𝒙→𝟎
[(√𝟏 + 𝐬𝐢𝐧𝟐 𝒙 − 𝟏)𝐜𝐨𝐭 𝒙] + ∫𝐜𝐨𝐬𝟐 𝒙
√𝟏+ 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙
𝝅𝟐
𝟎
=
= 𝐥𝐢𝐦𝒙→𝟎
𝐬𝐢𝐧𝒙 𝐜𝐨𝐬 𝒙
𝟏 + √𝟏 + 𝐬𝐢𝐧𝟐 𝒙+ ∫
𝟏 − 𝒙
√(𝟏 + 𝒙𝟐)(𝟏 − 𝒙𝟐)
𝟏
𝟎
𝒅𝒙 =
= −∫𝒙𝟐
√𝟏− 𝒙𝟒
𝟏
𝟎
𝒅𝒙 + ∫𝒅𝒙
√𝟏 − 𝒙𝟒
𝟏
𝟎
= −𝟏
𝟒∫𝒙𝟏𝟐+𝟏𝟒−𝟏
√𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 +𝟏
𝟒∫
𝒙𝟏𝟒−𝟏
√𝟏 − 𝒙
𝟏
𝟎
𝒅𝒙 =
= −𝟏
𝟒∫ 𝒙
𝟑𝟒−𝟏(𝟏 − 𝒙)
𝟏𝟐−𝟏
𝟏
𝟎
𝒅𝒙 +𝟏
𝟒∫ 𝒙
𝟏𝟒−𝟏(𝟏 − 𝒙)
𝟏𝟐−𝟏
𝟏
𝟎
𝒅𝒙 =
= −𝟏
𝟒⋅𝚪 (𝟑𝟒)𝚪(
𝟏𝟐)
𝚪 (𝟓𝟒)
+𝟏
𝟒⋅𝚪 (𝟏𝟒)𝚪 (
𝟏𝟐)
𝚪 (𝟑𝟒)
=
= −√𝝅
𝟒⋅√𝟐𝝅
𝟏𝟒𝚪𝟐 (
𝟏𝟒)+√𝝅
𝟒⋅𝚪𝟐 (
𝟏𝟒)
𝝅√𝟐=𝚪𝟒 (
𝟏𝟒) − 𝟖𝝅
𝟐
𝟒√𝟐𝝅 ⋅ 𝚪𝟐 (𝟏𝟒)=𝝅
𝟐⋅𝚪𝟒 (
𝟏𝟒) − 𝟖𝝅
𝟐
(𝟐𝝅)𝟑𝟐 ⋅ 𝚪𝟐 (
𝟏𝟒)=
=𝝅
𝟐⋅
{
𝚪𝟐 (𝟏𝟒)
(𝟐𝝅)𝟑𝟐
−𝟏
𝚪𝟐 (𝟏𝟒)
(𝟐𝝅)𝟑𝟐
⋅\𝒑𝒊}
=𝝅
𝟐(𝑮 −
𝟏
𝝅𝑮)
1551. Find:
𝛀 = ∫ 𝒙𝟐 ⋅ 𝐬𝐢𝐧−𝟏 𝒙 ⋅ 𝐬𝐢𝐧−𝟏(𝟒𝒙𝟑 − 𝟑𝒙)𝒅𝒙𝟏
𝟎
Proposed by Ty Halpen-Florida-USA
Solution by Asmat Qatea-Afghanistan
𝛀 = ∫ 𝒙𝟐 ⋅ 𝐬𝐢𝐧−𝟏 𝒙 ⋅ 𝐬𝐢𝐧−𝟏(𝟒𝒙𝟑 − 𝟑𝒙)𝒅𝒙𝟏
𝟎
=𝒙=𝐬𝐢𝐧𝒖
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= −∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ 𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧(𝟑𝒖)) 𝐜𝐨𝐬𝒖𝒅𝒖
𝝅𝟐
𝟎
=
= −∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ 𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧(𝟑𝒖)) 𝐜𝐨𝐬𝒖𝒅𝒖
𝝅𝟔
𝟎
−∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ 𝐬𝐢𝐧−𝟏(𝐬𝐢𝐧(𝟑𝒖)) 𝐜𝐨𝐬𝒖𝒅𝒖
𝝅𝟐
𝝅𝟔
= −∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ (𝟑𝒖) 𝐜𝐨𝐬𝒖𝒅𝒖
𝝅𝟔
𝟎
–∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖 ⋅ (−𝟑𝒖 + 𝝅) 𝐜𝐨𝐬𝒖𝒅𝒖
𝝅𝟐
𝝅𝟔
=
= −𝟑∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖𝟐 ⋅ 𝐜𝐨𝐬𝒖𝒅𝒖
𝝅𝟔
𝟎⏟ 𝑰𝟏
+ 𝟑∫ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝒖𝟐 ⋅ 𝐜𝐨𝐬𝒖𝒅𝒖
𝝅𝟐
𝝅𝟔⏟
𝑰𝟐
− 𝝅∫ 𝒖 ⋅ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝐜𝐨𝐬 𝒖
𝝅𝟐
𝝅𝟔
𝒅𝒖⏟
𝑰𝟑
∵ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝐜𝐨𝐬 𝒖 =𝟏
𝟒(𝐜𝐨𝐬𝒖 − 𝐜𝐨𝐬(𝟑𝒖)) ⇒ ∫𝐬𝐢𝐧𝟐 𝒖 𝐜𝐨𝐬𝒖𝒅𝒖
=𝟏
𝟒∫(𝐜𝐨𝐬𝒖 − 𝐜𝐨𝐬(𝟑𝒖))𝒅𝒖
𝑰𝟏 = −𝟑
𝟒[𝒖𝟐 (𝐬𝐢𝐧𝒖 −
𝐬𝐢𝐧(𝟑𝒖)
𝟑) − 𝟐𝒖(−𝐜𝐨𝐬 𝒖 +
𝐜𝐨𝐬(𝟑𝒖)
𝟗) + 𝟐(−𝐬𝐢𝐧 𝒖 +
𝐬𝐢𝐧(𝟑𝒖)
𝟐𝟕)]𝟎
𝝅𝟔
=
= −𝝅𝟐
𝟐𝟖𝟖−√𝟑𝝅
𝟖+𝟐𝟓
𝟑𝟔
𝑰𝟐 = −𝟑
𝟒[𝒖𝟐 (𝐬𝐢𝐧𝒖 −
𝐬𝐢𝐧(𝟑𝒖)
𝟑) − 𝟐𝒖(−𝐜𝐨𝐬 𝒖 +
𝐜𝐨𝐬(𝟑𝒖)
𝟗) + 𝟐(−𝐬𝐢𝐧 𝒖 +
𝐬𝐢𝐧(𝟑𝒖)
𝟐𝟕)]𝝅𝟔
𝝅𝟐
=𝟕𝟏𝝅𝟐
𝟐𝟖𝟖−√𝟑𝝅
𝟖−𝟑𝟏
𝟑𝟔
𝑰𝟏 + 𝑰𝟐 =𝟕𝟎𝝅𝟐
𝟐𝟖𝟖−√𝟑𝝅
𝟒−𝟏
𝟔
𝑰𝟑 = −𝝅∫ 𝒖 ⋅ 𝐬𝐢𝐧𝟐 𝒖 ⋅ 𝐜𝐨𝐬𝒖
𝝅𝟐
𝝅𝟔
𝒅𝒖 = −𝝅
𝟒∫ 𝒖(𝐜𝐨𝐬 𝒖 − 𝐜𝐨𝐬(𝟑𝒖))
𝝅𝟐
𝝅𝟔
𝒅𝒖 =
= −𝝅
𝟒[𝒖 (𝐬𝐢𝐧𝒖 −
𝐬𝐢𝐧(𝟑𝒖)
𝟑)—𝐜𝐨𝐬𝒖 +
𝐜𝐨𝐬(𝟑𝒖)
𝟗]𝝅𝟔
𝝅𝟐
= −𝟐𝟑𝝅𝟐
𝟏𝟒𝟒+√𝟑𝝅
𝟖
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𝛀 = 𝑰𝟏 + 𝑰𝟐 + 𝑰𝟑 =𝝅𝟐
𝟏𝟐−√𝟑𝝅
𝟖−𝟏
𝟔
1552. Prove that:
∫ 𝐥𝐨𝐠(𝒙𝟐 + 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)) 𝒅𝒙
𝝅𝟐
𝟎
= 𝝅 𝐥𝐨𝐠(𝐥𝐨𝐠 𝟐)
Proposed by Simon Peter-Madagascar
Solution by Luca Paes Barreto-Pernambuco-Brazil
It is well-known that:
∫𝐜𝐨𝐬 (𝒔 𝐭𝐚𝐧−𝟏 (
𝒙− 𝐥𝐨𝐠 𝐜𝐨𝐬 𝒙))
(𝒙𝟐 + 𝐥𝐨𝐠𝟐 𝐜𝐨𝐬 𝒙)𝒔𝟐
𝝅𝟐
𝟎
𝒅𝒙 =𝝅
𝟐⋅𝟏
𝐥𝐨𝐠𝒔 𝟐
Differentiable both sides w.r.t. 𝒔, we have:
𝝏𝒔(𝐜𝐨𝐬 (𝒔 𝐭𝐚𝐧−𝟏 (
𝒙− 𝐥𝐨𝐠 𝐜𝐨𝐬 𝒙))
(𝒙𝟐 + 𝐥𝐨𝐠𝟐 𝐜𝐨𝐬 𝒙)𝒔𝟐
)|
𝒔=𝟎
= −𝟏
𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝐥𝐨𝐠𝟐 𝐜𝐨𝐬 𝒙)
𝝏𝒔 (𝝅
𝟐⋅𝟏
𝐥𝐨𝐠𝒔 𝟐)|𝒔=𝟎
= −𝝅
𝟐⋅ 𝐥𝐨𝐠(𝐥𝐨𝐠𝟐) ⇒
−𝟏
𝟐∫ 𝐥𝐨𝐠(𝒙𝟐 + 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)) 𝒅𝒙
𝝅𝟐
𝟎
= −𝝅
𝟐⋅ 𝐥𝐨𝐠(𝐥𝐨𝐠𝟐)
Therefore,
∫ 𝐥𝐨𝐠(𝒙𝟐 + 𝐥𝐨𝐠𝟐(𝐜𝐨𝐬 𝒙)) 𝒅𝒙
𝝅𝟐
𝟎
= 𝝅 𝐥𝐨𝐠(𝐥𝐨𝐠𝟐)
1553. Find:
𝛀 = ∫𝑳𝒊𝟑(𝒙)
𝟏 − 𝒙𝐥𝐨𝐠 𝒙
𝟏
𝟎
𝒅𝒙
Proposed by Simon Peter-Madagascar
Solution 1 by Abdul Mukhtar-Nigeria
We know that:
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74 RMM-CALCULUS MARATHON 1501-1600
𝑳𝒊𝟑(𝒙)
𝟏 − 𝒙= ∑𝑯𝒏
(𝟑)𝒙𝒏∞
𝒏=𝟏
⇒ 𝛀 = ∫𝑳𝒊𝟑(𝒙)
𝟏 − 𝒙𝐥𝐨𝐠 𝒙
𝟏
𝟎
𝒅𝒙 = ∑𝑯𝒏(𝟑)∫ 𝒙𝒏 𝐥𝐨𝐠(𝒙)𝒅𝒙
𝟏
𝟎
∞
𝒏=𝟏
∵ ∫ 𝒙𝒏 𝐥𝐨𝐠𝒂(𝒙) 𝒅𝒙𝟏
𝟎
=(−𝟏)𝒂𝒂!
(𝒏 + 𝟏)𝒂+𝟏; 𝒇𝒐𝒓 𝒂 = 𝟏 ⇒
∫ 𝒙𝒏 𝐥𝐨𝐠(𝒙)𝒅𝒙𝟏
𝟎
=(−𝟏)𝟏𝟏!
(𝒏 + 𝟏)𝟏+𝟏= −
𝟏
(𝒏 + 𝟏)𝟐⇒∑𝑯𝒏
(𝟑)∫ 𝒙𝒏 𝐥𝐨𝐠(𝒙)𝒅𝒙𝟏
𝟎
∞
𝒏=𝟏
=
= ∑𝑯𝒏(𝟑) (−𝟏)
(𝒏 + 𝟏)𝟐
∞
𝒏=𝟏
= −∑𝑯𝒏(𝟑)
(𝒏 + 𝟏)𝟐
∞
𝒏=𝟏
= −∑𝑯𝒏−𝟏(𝟑)
𝒏𝟐
∞
𝒏=𝟏
We know that: 𝑯𝒏−𝟏(𝟑) = 𝑯𝒏
(𝟑) −𝟏
𝒏𝟑.
−∑𝑯𝒏−𝟏(𝟑)
𝒏𝟐
∞
𝒏=𝟏
= −∑𝟏
𝒏𝟐(𝑯𝒏
(𝟑)−𝟏
𝒏𝟑)
∞
𝒏=𝟏
= −∑𝑯𝒏(𝟑)
𝒏𝟐
∞
𝒏=𝟏
+∑𝟏
𝒏𝟓
∞
𝒏=𝟏
=
= −𝟏𝟏
𝟐𝜻(𝟓) + 𝟐𝜻(𝟐)𝜻(𝟑) + 𝜻(𝟓)
Solution 2 by Syed Shahabudeen-Kerala-India
𝛀 = ∫𝑳𝒊𝟑(𝒙)
𝟏 − 𝒙𝐥𝐨𝐠𝒙
𝟏
𝟎
𝒅𝒙 =𝝏
𝝏𝒂∫𝒙𝒂𝑳𝒊𝟑(𝒙)
𝟏 − 𝒙𝒅𝒙
𝟏
𝟎
=𝝏
𝝏𝒂∫ 𝑳𝒊𝟑(𝒙)∑𝒙𝒌
∞
𝒌=𝟎
𝒅𝒙𝟏
𝟎
=
=𝝏
𝝏𝒂∑∫ 𝒙𝒂+𝒌𝑳𝒊𝟑(𝒙)
𝟏
𝟎
𝒅𝒙
∞
𝒌=𝟎
=
= 𝐥𝐢𝐦𝒂→𝟎
𝝏
𝝏𝒂∑(
𝜻(𝟑)
𝒂 + 𝒌 + 𝟏−
𝜻(𝟐)
(𝒂 + 𝒌 + 𝟏)𝟐+
𝑯𝒂+𝒌+𝟏(𝒂 + 𝒌 + 𝟏)𝟑
)
∞
𝒌=𝟎
=
= 𝐥𝐢𝐦𝒂→𝟎
∑ (−𝜻(𝟑)
(𝒂 +𝒎)𝟐+
𝟐𝜻(𝟐)
(𝒂 +𝒎)𝟑+𝝏
𝝏𝒂
𝑯𝒂+𝒎(𝒂 +𝒎)𝟑
)
∞
𝒎=𝟎
=
= ∑ (−𝜻(𝟑)
𝒎𝟐+𝟐𝜻(𝟐)
𝒎𝟑)
∞
𝒎=𝟎
+ ∑ (𝝏
𝝏𝒂
𝑯𝒂+𝒎(𝒂 +𝒎)𝟑
)
∞
𝒎=𝟏
=
∵𝝏
𝝏𝒂
𝑯𝒂+𝒎(𝒂 + 𝒎)𝟑
= −𝟑𝑯𝒂+𝒎(𝒂 +𝒎)𝟒
+𝟏
(𝒂 +𝒎)𝟑(𝜻(𝟐) − 𝑯𝒂+𝒎
𝟐 )
⇒ 𝐥𝐢𝐦𝒂→𝟎
𝝏
𝝏𝒂
𝑯𝒂+𝒎(𝒂 +𝒎)𝟑
= −𝟑𝑯𝒂+𝒎(𝒂 +𝒎)𝟒
+𝟏
(𝒂 +𝒎)𝟑(𝜻(𝟐) − 𝑯𝒂+𝒎
𝟐 )
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75 RMM-CALCULUS MARATHON 1501-1600
⇒ 𝛀 = ∑ (−𝜻(𝟑)
𝒎𝟐+𝟐𝜻(𝟐)
𝒎𝟑)
∞
𝒎=𝟏
+ ∑ (−𝟑𝑯𝒂+𝒎(𝒂 +𝒎)𝟒
+𝟏
(𝒂 +𝒎)𝟑(𝜻(𝟐) − 𝑯𝒂+𝒎
𝟐 ))
∞
𝒎=𝟏
=
= 𝜻(𝟐)𝜻(𝟑) + ∑ (−𝟑𝑯𝒂+𝒎(𝒂 +𝒎)𝟒
+𝟏
(𝒂 +𝒎)𝟑(𝜻(𝟐) − 𝑯𝒂+𝒎
𝟐 ))
∞
𝒎=𝟏
∵ ∑𝑯𝒎𝒎𝟒
∞
𝒎=𝟏
= 𝟑𝜻(𝟓) − 𝜻(𝟐)𝜻(𝟑) 𝒂𝒏𝒅 ∑𝑯𝒎𝟐
𝒎𝟑
∞
𝒎=𝟏
= 𝟑𝜻(𝟐)𝜻(𝟑) −𝟗
𝟐𝜻(𝟓)
Therefore,
𝛀 = 𝟐𝜻(𝟐)𝜻(𝟑) + (−𝟗𝜻(𝟓) + 𝟒𝜻(𝟐)𝜻(𝟑) − 𝟑𝜻(𝟐)𝜻(𝟑) +𝟗
𝟐𝜻(𝟓)) = 𝟐𝜻(𝟐)𝜻(𝟑) −
𝟗
𝟐𝜻(𝟓)
𝛀 = ∫𝑳𝒊𝟑(𝒙)
𝟏 − 𝒙𝐥𝐨𝐠 𝒙
𝟏
𝟎
𝒅𝒙 = 𝟐𝜻(𝟐)𝜻(𝟑) −𝟗
𝟐𝜻(𝟓)𝜻(𝟑)
1554. Find:
𝛀 = ∫ ∫ ∫𝒚 𝐥𝐨𝐠 𝒙
(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)(𝟏 + 𝒛𝟐)
∞
𝟏
∞
𝟏
𝒅𝒙∞
𝟎
𝒅𝒚 𝒅𝒛
Proposed by Probal Chakraborty-India
Solution by Rana Ranino-Setif-Algerie
𝛀 = ∫ ∫ ∫𝒚 𝐥𝐨𝐠𝒙
(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)(𝟏 + 𝒛𝟐)
∞
𝟏
∞
𝟏
𝒅𝒙∞
𝟎
𝒅𝒚 𝒅𝒛 =𝝅
𝟐∫ ∫
𝒚 𝐥𝐨𝐠 𝒙
(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)
∞
𝟏
𝒅𝒙∞
𝟏
𝒅𝒚
=𝝅
𝟐∫ ∫
−𝒚 𝐥𝐨𝐠 𝒙
(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)
𝟏
𝟎
𝒅𝒙𝟏
𝟎
𝒅𝒚 = −𝝅
𝟐∫
𝟏
𝒚(𝟏 + 𝒚𝟐)
𝟏
𝟎
∫𝐥𝐨𝐠 𝒙
(𝟏 +𝒙𝒚)𝟐 𝒅𝒙
𝟏
𝟎
𝒅𝒚 =𝒕=𝒙𝒚
= −𝝅
𝟐∫
𝟏
𝟏 + 𝒚𝟐∫𝐥𝐨𝐠 𝒕 + 𝐥𝐨𝐠 𝒚
(𝟏 + 𝒕)𝟐𝒅𝒕
𝟏𝒚
𝟎
𝒅𝒚𝟏
𝟎
=
= −𝝅
𝟐∫
𝟏
𝟏 + 𝒚𝟐
𝟏
𝟎
∫𝐥𝐨𝐠 𝒕
(𝟏 + 𝒕)𝟐𝒅𝒕
𝟏𝒚
𝟎
𝒅𝒚 −𝝅
𝟐∫
𝐥𝐨𝐠 𝒚
(𝟏 + 𝒚𝟐)
𝟏
𝟎
∫𝟏
𝟏 + 𝒕𝟐
𝟏𝒚
𝟎
𝒅𝒕𝒅𝒚 =
=𝝅
𝟐∫
𝟏
𝟏 + 𝒚𝟐[𝐥𝐨𝐠(𝒕 + 𝟏) −
𝒕 𝐥𝐨𝐠 𝒕
𝒕 + 𝟏]𝟎
𝟏𝒚
𝒅𝒚𝟏
𝟎
+𝝅
𝟐∫
𝐥𝐨𝐠𝒚
𝟏 + 𝒚𝟐[
𝟏
(𝒚 + 𝟏)]𝟎
𝟏𝒚𝒅𝒚
𝟏
𝟎
=
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76 RMM-CALCULUS MARATHON 1501-1600
=𝝅
𝟐∫𝐥𝐨𝐠(𝟏 + 𝒚) − 𝐥𝐨𝐠 𝒚 +
𝐥𝐨𝐠 𝒚𝒚 + 𝟏
𝟏 + 𝒚𝟐
𝟏
𝟎
𝒅𝒚 −𝝅
𝟐∫
𝐥𝐨𝐠 𝒚
(𝟏 + 𝒚)(𝟏 + 𝒚𝟐)𝒅𝒚
𝟏
𝟎
=
=𝝅
𝟐∫𝐥𝐨𝐠(𝟏 + 𝒚)
𝟏 + 𝒚𝟐𝒅𝒚
𝟏
𝟎
−𝝅
𝟐∫
𝐥𝐨𝐠𝒚
𝟏 + 𝒚𝟐𝒅𝒚
𝟏
𝟎
=𝒚=𝐭𝐚𝐧 𝜽
=𝝅
𝟐∫ 𝐥𝐨𝐠(𝟏 + 𝐭𝐚𝐧𝜽)𝒅𝜽
𝝅𝟒
𝟎
−𝝅
𝟐∫ 𝐥𝐨𝐠(𝐭𝐚𝐧𝜽)𝒅𝜽
𝝅𝟒
𝟎⏟ −𝑮
∫ 𝐥𝐨𝐠(𝟏 + 𝐭𝐚𝐧 𝜽) 𝒅𝜽
𝝅𝟒
𝟎
= ∫ 𝐥𝐨𝐠 (√𝟐𝐜𝐨𝐬 (𝝅
𝟒− 𝜽))𝒅𝜽
𝝅𝟒
𝟎
−∫ 𝐥𝐨𝐠(𝐜𝐨𝐬 𝜽)𝒅𝜽
𝝅𝟒
𝟎
=
= ∫ 𝐥𝐨𝐠(√𝟐 𝐜𝐨𝐬 𝜽)𝒅𝜽 −
𝝅𝟒
𝟎
∫ 𝐥𝐨𝐠(𝐜𝐨𝐬 𝜽)𝒅𝜽
𝝅𝟒
𝟎
=𝝅
𝟖𝐥𝐨𝐠 𝟐
Therefore,
𝛀 = ∫ ∫ ∫𝒚 𝐥𝐨𝐠 𝒙
(𝒙 + 𝒚)𝟐(𝟏 + 𝒚𝟐)(𝟏 + 𝒛𝟐)
∞
𝟏
∞
𝟏
𝒅𝒙∞
𝟎
𝒅𝒚 𝒅𝒛 =𝝅
𝟐(𝑮 +
𝝅
𝟖𝐥𝐨𝐠 𝟐)
1555. If 𝐬𝐞𝐜𝝅
𝟕< 𝑎 ≤ 𝑏 then find:
𝛀(𝒂, 𝒃) = ∫ (𝐭𝐚𝐧−𝟏(𝒙
𝐬𝐞𝐜𝝅𝟕− 𝒙 𝐭𝐚𝐧
𝝅𝟕
) − 𝐭𝐚𝐧−𝟏 (𝒙 𝐬𝐞𝐜𝝅
𝟕− 𝐭𝐚𝐧
𝝅
𝟕))𝒅𝒙
𝒃
𝒂
Proposed by Daniel Sitaru-Romania
Solution 1 by Mohamed Rostami-Afghanistan
𝛀(𝒂, 𝒃) = ∫ (𝐭𝐚𝐧−𝟏 (𝒙
𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕
) − 𝐭𝐚𝐧−𝟏 (𝒙 𝐬𝐞𝐜𝝅
𝟕− 𝐭𝐚𝐧
𝝅
𝟕))𝒅𝒙
𝒃
𝒂
=
= ∫ (𝐭𝐚𝐧−𝟏(𝒙 𝐜𝐨𝐬
𝝅𝟕
𝟏 − 𝒙𝐬𝐢𝐧𝝅𝟕
) − 𝐭𝐚𝐧−𝟏 (𝒙 − 𝐬𝐢𝐧
𝝅𝟕
𝐜𝐨𝐬𝝅𝟕
))𝒅𝒙𝒃
𝒂
𝐭𝐚𝐧−𝟏(𝒙 𝐜𝐨𝐬
𝝅𝟕
𝟏 − 𝒙𝐬𝐢𝐧𝝅𝟕
) = 𝜶 ⇒ 𝐭𝐚𝐧𝜶 =𝒙 𝐜𝐨𝐬
𝝅𝟕
𝟏 − 𝒙 𝐬𝐢𝐧𝝅𝟕
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𝐭𝐚𝐧−𝟏(𝒙 − 𝐬𝐢𝐧
𝝅𝟕
𝐜𝐨𝐬𝝅𝟕
) = 𝜷 ⇒ 𝐭𝐚𝐧𝜷 =𝒙 − 𝐬𝐢𝐧
𝝅𝟕
𝐜𝐨𝐬𝝅𝟕
𝜶 − 𝜷 = 𝜸 ⇒𝐭𝐚𝐧𝜶 − 𝐭𝐚𝐧𝜷
𝟏 + 𝐭𝐚𝐧𝜶 𝐭𝐚𝐧𝜷= 𝐭𝐚𝐧 𝜸 ⇒
𝒙 𝐜𝐨𝐬𝝅𝟕
𝟏 − 𝒙𝐬𝐢𝐧𝝅𝟕
−𝒙 − 𝐬𝐢𝐧
𝝅𝟕
𝐜𝐨𝐬𝝅𝟕
𝟏 +𝒙 𝐜𝐨𝐬
𝝅𝟕
𝟏 − 𝒙𝐬𝐢𝐧𝝅𝟕
⋅𝒙 − 𝐬𝐢𝐧
𝝅𝟕
𝐜𝐨𝐬𝝅𝟕
= 𝐭𝐚𝐧𝜸 ⇒
𝐭𝐚𝐧 𝜸 =
𝒙(𝐜𝐨𝐬𝟐𝝅𝟕 − 𝐬𝐢𝐧
𝟐𝝅𝟕) − 𝒙 + 𝒙
𝟐 𝐬𝐢𝐧𝝅𝟕 + 𝐬𝐢𝐧
𝝅𝟕
𝐜𝐨𝐬𝝅𝟕 (𝟏 − 𝒙 𝐬𝐢𝐧
𝝅𝟕)
𝟏 − 𝒙𝐬𝐢𝐧𝝅𝟕 + 𝒙
𝟐 − 𝒙𝐬𝐢𝐧𝝅𝟕
𝟏 − 𝒙𝐬𝐢𝐧𝝅𝟕
=
=
𝟏
𝐜𝐨𝐬𝝅𝟕
[−𝒙 (𝟏 − 𝐜𝐨𝐬𝟐𝝅𝟕 ) + 𝒙
𝟐 𝐬𝐢𝐧𝝅𝟕 + 𝐬𝐢𝐧
𝝅𝟕]
𝟏 − 𝟐𝒙 𝐬𝐢𝐧𝝅𝟕 + 𝒙
𝟐
𝐭𝐚𝐧𝜸 =𝐭𝐚𝐧
𝝅𝟕 (−𝟐𝒙 𝐬𝐢𝐧
𝝅𝟕 + 𝒙
𝟐 + 𝟏)
𝟏 − 𝟐𝒙 𝐬𝐢𝐧𝝅𝟕 + 𝒙
𝟐= 𝐭𝐚𝐧
𝝅
𝟕⇒ 𝜸 = 𝜶− 𝜷 =
𝝅
𝟕
Therefore,
𝛀(𝒂, 𝒃) = ∫ (𝜶 − 𝜷)𝒃
𝒂
𝒅𝒙 = ∫𝝅
𝟕
𝒃
𝒂
𝒅𝒙 =𝝅
𝟕(𝒃 − 𝒂).
Solution 2 by Kamel Gandouli Rezgui-Tunisia
∵ 𝐭𝐚𝐧−𝟏 𝒙 ± 𝐭𝐚𝐧−𝟏 𝒚 = 𝐭𝐚𝐧−𝟏 (𝒙 ± 𝒚
𝟏 ∓ 𝒙𝒚) ⇒
𝐭𝐚𝐧−𝟏 (𝒙
𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕
) − 𝐭𝐚𝐧−𝟏 (𝒙𝐬𝐞𝐜𝝅
𝟕− 𝐭𝐚𝐧
𝝅
𝟕) = 𝐭𝐚𝐧−𝟏 (
𝒙
𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕
) =
=𝝅
𝟐− 𝐭𝐚𝐧−𝟏 (
𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕
𝒙)
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𝐭𝐚𝐧−𝟏 (𝒙 𝐬𝐞𝐜𝝅
𝟕− 𝐭𝐚𝐧
𝝅
𝟕) + 𝐭𝐚𝐧−𝟏 (
𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕
𝒙) =
= 𝐭𝐚𝐧−𝟏
(
𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕
𝒙 + 𝒙𝐬𝐞𝐜𝝅𝟕 − 𝐭𝐚𝐧
𝝅𝟕
𝟏 − (𝒙 𝐬𝐞𝐜𝝅𝟕 − 𝐭𝐚𝐧
𝝅𝟕) ⋅
𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕
𝒙 )
=
= 𝐭𝐚𝐧−𝟏 (𝐬𝐞𝐜
𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕 + 𝒙
𝟐 𝐬𝐞𝐜𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝝅𝟕
𝒙 − 𝒙𝐬𝐞𝐜𝟐𝝅𝟕 + 𝐬𝐞𝐜
𝝅𝟕 𝐭𝐚𝐧
𝝅𝟕 + 𝒙
𝟐 𝐬𝐞𝐜𝝅𝟕 𝐭𝐚𝐧
𝝅𝟕 − 𝒙 𝐭𝐚𝐧
𝟐 𝝅𝟕
) =
= 𝐭𝐚𝐧−𝟏 (𝐬𝐞𝐜
𝝅𝟕 − 𝟐𝒙 𝐭𝐚𝐧
𝝅𝟕 + 𝒙
𝟐 𝐬𝐞𝐜𝝅𝟕
𝐬𝐞𝐜𝝅𝟕 𝐭𝐚𝐧
𝝅𝟕 + 𝒙
𝟐 𝐬𝐞𝐜𝝅𝟕 𝐭𝐚𝐧
𝝅𝟕 − 𝟐𝒙 𝐭𝐚𝐧
𝟐 𝝅𝟕
) =(∗)
𝐬𝐞𝐜𝝅
𝟕𝐭𝐚𝐧
𝝅
𝟕+ 𝒙𝟐 𝐬𝐞𝐜
𝝅
𝟕𝐭𝐚𝐧
𝝅
𝟕− 𝟐𝒙 𝐭𝐚𝐧𝟐
𝝅
𝟕= 𝐭𝐚𝐧
𝝅
𝟕(𝐬𝐞𝐜
𝝅
𝟕− 𝟐𝒙 𝐭𝐚𝐧
𝝅
𝟕+ 𝒙𝟐 𝐬𝐞𝐜
𝝅
𝟕)
=(∗)𝐭𝐚𝐧−𝟏(
𝟏
𝐭𝐚𝐧𝝅𝟕
) =𝝅
𝟐−𝝅
𝟕
Therefore,
𝛀(𝒂, 𝒃) =𝝅
𝟕(𝒃 − 𝒂).
1556. If 𝟏 < 𝑎 ≤ 𝑏 then find:
𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
Proposed by Daniel Sitaru-Romania
Solution 1 by Serlea Kabay-Liberia
𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
=
= ∫ ∫ ∫ (𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 − 𝝅)𝒃
𝒂
𝒅𝒙𝒅𝒚𝒅𝒛𝒃
𝒂
𝒃
𝒂
=(∗)
(∵ ∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙 = 𝒙 𝐭𝐚𝐧−𝟏 𝒙 − 𝐥𝐨𝐠(√𝟏 + 𝒙𝟐) )
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=(∗)∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
+∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 𝒚𝒅𝒙𝒅𝒚𝒅𝒛𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
+ ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 𝒛𝒅𝒙𝒅𝒚𝒅𝒛𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
−𝝅(𝒃 − 𝒂)𝟐 =
= (𝒃 − 𝒂)𝟐 (∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝒃
𝒂
+∫ 𝐭𝐚𝐧−𝟏 𝒚𝒅𝒚𝒃
𝒂
+∫ 𝐭𝐚𝐧−𝟏 𝒛𝒃
𝒂
𝒅𝒛) − 𝝅(𝒃 − 𝒂)𝟐 =
= 𝟑(𝒃 − 𝒂)𝟐 (𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏
𝟐(𝒂𝟐 + 𝟏
𝒃𝟐 + 𝟏)) − 𝝅(𝒃 − 𝒂)𝟐
Solution 2 by Remus Florin Stanca-Romania
𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙) = 𝐭𝐚𝐧−𝟏 (
𝒚 + 𝒛 − 𝒙(𝟏 − 𝒚𝒛)
𝟏 − 𝒙(𝒚 + 𝒛) − 𝒚𝒛) =
= 𝐭𝐚𝐧−𝟏 (
𝒚 + 𝒛𝟏 − 𝒚𝒛 + 𝒙
𝟏 + 𝒙𝒚 + 𝒛𝟏 − 𝒚𝒛
) = 𝐭𝐚𝐧−𝟏 (𝒚 + 𝒛
𝟏 − 𝒚𝒛) + 𝐭𝐚𝐧−𝟏 𝒛 − 𝝅 =
= 𝐭𝐚𝐧−𝟏 (𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒚) + 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒛)
𝟏 − 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒚) ⋅ 𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒛)) + 𝐭𝐚𝐧−𝟏 𝒙 − 𝝅 =
= 𝐭𝐚𝐧−𝟏(𝐭𝐚𝐧(𝐭𝐚𝐧−𝟏 𝒚) + 𝐭𝐚𝐧−𝟏 𝒛) + 𝐭𝐚𝐧−𝟏 𝒙 − 𝝅 =
= 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 − 𝝅
∫ 𝐭𝐚𝐧−𝟏 𝒙𝒃
𝒂
𝒅𝒙 = [𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 −𝟏
𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝟏)]
𝒂
𝒃
=
= 𝒃𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 −𝟏
𝟐𝐥𝐨𝐠(𝒃𝟐 + 𝟏) +
𝟏
𝟐𝐥𝐨𝐠(𝒂𝟐 + 𝟏)
𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
=
= ∫ ∫ (𝒃 𝐭𝐚𝐧−𝟏 𝒃 −𝟏
𝟐𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +
𝟏
𝟐𝐥𝐨𝐠(𝒂𝟐 + 𝟏) + (𝒃 − 𝒂) 𝐭𝐚𝐧−𝟏 𝒚
𝒃
𝒂
𝒃
𝒂
+ (𝒃 − 𝒂) 𝐭𝐚𝐧−𝟏 𝒛)𝒅𝒚𝒅𝒛 − 𝝅(𝒃 − 𝒂)𝟐 =
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= ∫ (𝒃(𝒃 − 𝒂) 𝐭𝐚𝐧−𝟏 𝒃 −𝟏
𝟐(𝒃 − 𝒂) 𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝒂 𝐭𝐚𝐧−𝟏 𝒂(𝒃 − 𝒂)
𝒃
𝒂
+𝟏
𝟐𝐥𝐨𝐠(𝒂𝟐 + 𝟏)(𝒃 − 𝒂)
+ (𝒃 − 𝒂) (𝒃 𝐭𝐚𝐧−𝟏 𝒃 −𝟏
𝟐𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +
𝟏
𝟐𝐥𝐨𝐠(𝒂𝟐 + 𝟏))
+ (𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒛)𝒅𝒛 − 𝝅(𝒃 − 𝒂)𝟐 =
= 𝟑𝒃(𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒃 −𝟑
𝟐(𝒃 − 𝒂)𝟐 𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝟑𝒂(𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒂
+𝟑
𝟐(𝒃 − 𝒂)𝟐 𝐥𝐨𝐠(𝒂𝟐 + 𝟏) − 𝝅(𝒃 − 𝒂)𝟐
Therefore,
𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
=
= 𝟑𝒃(𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒃 −𝟑
𝟐(𝒃 − 𝒂)𝟐 𝐥𝐨𝐠(𝒃𝟐 + 𝟏) − 𝟑𝒂(𝒃 − 𝒂)𝟐 𝐭𝐚𝐧−𝟏 𝒂
+𝟑
𝟐(𝒃 − 𝒂)𝟐 𝐥𝐨𝐠(𝒂𝟐 + 𝟏) − 𝝅(𝒃 − 𝒂)𝟐
Solution 3 by Asmat Qatea-Afghanistan
𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 =
{
𝐭𝐚𝐧−𝟏 (
𝒙 + 𝒚
𝟏 − 𝒙𝒚) , 𝒙𝒚 < 1
𝝅 + 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚
𝟏 − 𝒙𝒚) , 𝒙 > 0, 𝑦 > 0(𝒙𝒚 > 1)
−𝝅 + 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚
𝟏− 𝒙𝒚) , 𝒙 < 0, 𝑦 < 0(𝑥𝑦 > 1)
𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 = 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚
𝟏− 𝒙𝒚) ⇒ 𝒙 ∈ (𝟏,∞), 𝒚 ∈ (𝟏,∞)
𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 = 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚
𝟏 − 𝒙𝒚) + 𝐭𝐚𝐧−𝟏 𝒛
{𝒙𝒚 > 1 ⇒
𝒙 + 𝒚
𝟏 − 𝒙𝒚< 0
𝒛 ∈ (𝟏,∞)
𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛 = 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)
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∫ ∫ ∫ (𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒚 + 𝐭𝐚𝐧−𝟏 𝒛)𝒃
𝒂
𝒅𝒙𝒅𝒚𝒅𝒛𝒃
𝒂
𝒃
𝒂
=
= 𝝅(𝒃 − 𝒂)𝟐 +∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
𝟑∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 𝒙𝒃
𝒂
𝒅𝒙𝒅𝒚𝒅𝒛𝒃
𝒂
𝒃
𝒂
= 𝝅(𝒃 − 𝒂)𝟐 +∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
𝟑(𝒃 − 𝒂)𝟐 [𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 −𝟏
𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝟏)]
𝒂
𝒃
=
= 𝝅(𝒃 − 𝒂)𝟑 +∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
Therefore,
𝛀(𝒂, 𝒃) = ∫ ∫ ∫ 𝐭𝐚𝐧−𝟏 (𝒙 + 𝒚 + 𝒛 − 𝒙𝒚𝒛
𝟏 − 𝒙𝒚 − 𝒚𝒛 − 𝒛𝒙)𝒅𝒙𝒅𝒚𝒅𝒛
𝒃
𝒂
𝒃
𝒂
𝒃
𝒂
=
= 𝟑(𝒃 − 𝒂)𝟐 (𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏
𝟐(𝒂𝟐 + 𝟏
𝒃𝟐 + 𝟏)) − 𝝅(𝒃 − 𝒂)𝟑
1557. Prove that:
∫ 𝒆−𝒙𝟐𝑯𝟐𝒏(𝜶𝒙)
∞
−∞
𝒅𝒙 = √𝝅(𝟐𝒏)!
𝒏!(𝜶𝟐 − 𝟏)𝒏, 𝒏 > 0
where, 𝑯𝟐𝒏 −Hermite polynomials.
Proposed by Tobi Joshua-Nigeria
Solution by Serlea Kabay-Liberia
Using Hermite polynomial as a special case of the Laguerre polynomial,
𝑯𝟐𝒏(𝜶𝒙) = (−𝟒)𝒏𝒏! 𝑳𝒏
−𝟏𝟐(𝒂𝟐𝒙𝟐) ⇒ ∫ 𝒆−𝒙
𝟐𝑯𝟐𝒏(𝜶𝒙)
∞
−∞
𝒅𝒙
= (−𝟒)𝒏𝒏!∫ 𝒆−𝒙𝟐𝑳𝒏−𝟏𝟐(𝒂𝟐𝒙𝟐)
∞
−∞
𝒅𝒙
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𝑰 = ∫ 𝒆−𝒙𝟐𝑯𝟐𝒏(𝜶𝒙)
∞
−∞
𝒅𝒙 = (−𝟒)𝒏𝒏! ∫ 𝒆−𝒙𝟐𝑳𝒏−𝟏𝟐(𝒂𝟐𝒙𝟐)
∞
−∞
𝒅𝒙
𝑳𝒏(𝒂)(𝒙) = ∑(−𝟏)𝒌 (
𝒏 + 𝒂
𝒏 − 𝒌)𝒙𝟐𝒌
𝒌!
𝒏
𝒌=𝟎
⇒ 𝑳𝒏(−𝟏𝟐)(𝒂𝟐𝒙𝟐) = ∑(−𝟏)𝒌(
𝒏 −𝟏𝟐
𝒏 − 𝒌)𝒂𝟐𝒌𝒙𝟐𝒌
𝒌!
𝒏
𝒌=𝟎
∵ 𝑰 = (−𝟒)𝒏𝒏!∫ 𝒆−𝒙𝟐∑(−𝟏)𝒌(
𝒏 −𝟏𝟐
𝒏 − 𝒌)𝒂𝟐𝒌𝒙𝟐𝒌
𝒌!
𝒏
𝒌=𝟎
𝒅𝒙 =∞
−∞
= 𝟐(−𝟒)𝒏𝒏!∫ 𝒆−𝒙𝟐∑(−𝟏)𝒌 (
𝒏 −𝟏𝟐
𝒏 − 𝒌)𝒂𝟐𝒌𝒙𝟐𝒌
𝒌!
𝒏
𝒌=𝟎
𝒅𝒙∞
𝟎
Using Dominated convergence theorem:
𝑰 = 𝟐𝟐𝒏+𝟏(−𝟏)𝒏∑(−𝟏)𝒌(𝒏 −
𝟏𝟐
𝒏 − 𝒌)𝒂𝟐𝒌
𝒌!
𝒏
𝒌=𝟎
∫ 𝒆−𝒙𝟐𝒙𝟐𝒌
∞
𝟎
𝒅𝒙 =
= 𝟐𝟐𝒏(−𝟏)𝒏∑(−𝟏)𝒌(𝒏 −
𝟏𝟐
𝒏 − 𝒌)𝒂𝟐𝒌𝚪 (𝒌 +
𝟏𝟐)
𝒌!
𝒏
𝒌=𝟎
(𝒏 −
𝟏𝟐
𝒏 − 𝒌) =
𝚪 (𝒏 +𝟏𝟐)
𝚪(𝒏 + 𝟏 − 𝒌)𝚪(𝒌 +𝟏𝟐)=
(𝟐𝒏)!√𝝅
𝟒𝒏𝒏! 𝚪(𝒏 + 𝟏 − 𝒌)𝚪(𝒌 +𝟏𝟐)
𝑰 = (𝟐𝒏)! √𝝅(−𝟏)𝒏∑(−𝟏)𝒌𝒏
𝒌=𝟎
𝒂𝟐𝒌𝚪(𝒌 +𝟏𝟐)
𝒌! 𝚪(𝒏 + 𝟏 − 𝒌)𝚪(𝒌 +𝟏𝟐)=
= (𝟐𝒏)!√𝝅(−𝟏)𝒏∑(−𝟏)𝒌𝒂𝟐𝒌
𝒌! (𝒏 − 𝒌)!
𝒏
𝒌=𝟎
⇒ 𝑰 =(𝟐𝒏)! √𝝅(−𝟏)𝒏
𝒏!∑(−𝟏)𝒌
𝒏! 𝒂𝟐𝒌
𝒌! (𝒏 − 𝒌)!𝒂𝟐𝒌
𝒏
𝒌=𝟎
=
=(𝟐𝒏)! √𝝅(−𝟏)𝒏
𝒏!∑(
𝒏
𝒌) (−𝒂𝟐)𝒌
𝒏
𝒌=𝟎
𝑰 =(𝟐𝒏)! √𝝅(−𝟏)𝒏
𝒏!(𝟏 − 𝒂𝟐)𝒏 =
√𝝅 ⋅ (𝟐𝒏)! ⋅ (𝒂𝟐 − 𝟏)𝒏
𝒏!
www.ssmrmh.ro
83 RMM-CALCULUS MARATHON 1501-1600
Therefore,
∫ 𝒆−𝒙𝟐𝑯𝟐𝒏(𝜶𝒙)
∞
−∞
𝒅𝒙 = √𝝅(𝟐𝒏)!
𝒏!(𝜶𝟐 − 𝟏)𝒏, 𝒏 > 0
𝑳𝒏𝒂(⋅) −Laguerre polynomial.
1558. Prove that:
𝛀 = ∫𝐭𝐚𝐧−𝟏(𝒂𝒙)
𝐬𝐢𝐧𝐡(𝒃𝒙)
∞
𝟎
𝒅𝒙 =𝝅
𝒃(𝐥𝐨𝐠(
𝚪 (𝒃𝟐𝝅𝒂
)
𝚪 (𝟏𝟐+
𝒃𝟐𝝅𝒂
)) +
𝟏
𝟐𝐥𝐨𝐠 (
𝒃
𝟐𝝅𝒂))
Proposed by Ose Favour-Nigeria
Solution 1 by Felix Marin-Romania
∫𝐭𝐚𝐧−𝟏(𝒂𝒙)
𝐬𝐢𝐧𝐡(𝒃𝒙)𝒅𝒙
∞
𝟎
= ∫𝒂𝒙
𝐬𝐢𝐧𝐡(𝒃𝒙)
∞
𝟎
∫𝒅𝒚
𝒚𝟐 + (𝒂𝒙)𝟐
∞
𝟏
𝒅𝒙 =
𝒔𝒈𝒏 (𝒃)∫𝒂𝒙
𝒚𝟐 + (𝒂𝒙)𝟐𝟏
𝐬𝐢𝐧(|𝒃|𝒙)
∞
𝟏
𝒅𝒙𝒅𝒚 =
|𝒃|𝒙→𝝅𝒙
𝒗=𝝅𝒂|𝒃| 𝝅
𝒃∫ ∫
𝒗𝒙
𝒚𝟐 + (𝒗𝒙)𝟐𝟏
𝐬𝐢𝐧𝐡(𝝅𝒙)𝒅𝒙𝒅𝒚
∞
𝟎
∞
𝟏
=
=𝝅
𝒃∫ [𝒊∫
(𝒚 + 𝒗𝒙𝒊)−𝟏 − (𝒚 − 𝒗𝒙𝒊)−𝟏
𝟐 𝐬𝐢𝐧𝐡(𝝅𝒙)𝒅𝒙
∞
𝟎
] 𝒅𝒚∞
𝟏
=
=𝝅
𝒃∫ [∑(−𝟏)𝒏(𝒚 + 𝟐𝒏𝒗)−𝟏 −
𝟏
𝟐(𝒚 + 𝒗𝒙)−𝟏|𝒙=𝟎
∞
𝒏=𝟎
] 𝒅𝒚∞
𝟏
=
=𝝅
𝒃∫ {∑[
𝟏
𝒚 + 𝟐𝒏𝒗−
𝟏
𝒚 + (𝟐𝒏 + 𝟏)𝒗] −
𝟏
𝟐𝒚
∞
𝒏=𝟎
}∞
𝟏
𝒅𝒚 =
=𝝅
𝒃∫ {
𝟏
𝟐𝒗∑[
𝟏
𝒏 +𝒚𝟐𝒗
−𝟏
𝒏 +𝟏𝟐 +
𝒚𝟐𝒗
]
∞
𝒏=𝟎
−𝟏
𝟐𝒚}𝒅𝒚
∞
𝟏
=
=𝝅
𝒃∫ {
𝟏
𝟐𝒗[𝚿(
𝟏
𝟐+𝒚
𝟐𝒗) −𝚿(
𝒚
𝟐𝒗)] −
𝟏
𝟐𝒚} 𝒅𝒚
∞
𝟏
=
=𝝅
𝒃[𝐥𝐨𝐠(
𝚪(𝟏𝟐+
𝒚[𝟐𝒗]
)
𝚪 (𝒚[𝟐𝒗]
)) −
𝟏
𝟐𝐥𝐨𝐠𝒚]
𝟏
∞
=𝝅
𝒃[𝟏
𝟐𝐥𝐨𝐠 (
𝟏
𝟐𝒗) − 𝐥𝐨𝐠(
𝚪 ([𝟏 + 𝒗][𝟐𝒗]
)
𝚪 (𝟏[𝟐𝒗]
))]
www.ssmrmh.ro
84 RMM-CALCULUS MARATHON 1501-1600
Therefore,
𝛀 = ∫𝐭𝐚𝐧−𝟏(𝒂𝒙)
𝐬𝐢𝐧𝐡(𝒃𝒙)
∞
𝟎
𝒅𝒙 =𝝅
𝒃(𝐥𝐨𝐠(
𝚪 (𝒃𝟐𝝅𝒂)
𝚪 (𝟏𝟐 +
𝒃𝟐𝝅𝒂)
) +𝟏
𝟐𝐥𝐨𝐠 (
𝒃
𝟐𝝅𝒂))
Solution 2 by Abdul Mukhtar-Nigeria
𝐋𝐞𝐭 𝑰(𝒂, 𝒃) = ∫𝟏
𝐬𝐢𝐧𝐡(𝒃𝒙)𝐭𝐚𝐧−𝟏 (
𝒙
𝒂)𝒅𝒙
∞
𝟎
Using Feynmann parametrization technique for integrating 𝐬𝐢𝐧𝒙
𝒙 , we have:
𝑰(𝒂, 𝒃) = ∫𝟏
𝐬𝐢𝐧𝐡(𝒃𝒙)
∞
𝟎
∫𝐬𝐢𝐧 𝒕𝒙
𝒕
∞
𝟎
𝒆−𝒂𝒕𝒅𝒕𝒅𝒙 =
= ∫𝒆−𝒂𝒕
𝒕
∞
𝟎
∫𝐬𝐢𝐧 𝒕𝒙
𝐬𝐢𝐧𝐡(𝒃𝒙)
∞
𝟎
𝒅𝒙𝒅𝒕 = ∫ 𝒆−𝒕𝒙 𝐬𝐢𝐧 𝒕𝒙𝒅𝒙∞
𝟎
=𝒕
𝒕𝟐 + 𝒏𝟐; (∵
𝟏
𝐬𝐢𝐧𝐡 𝒙=
𝟐𝒆−𝒙
𝟏 − 𝒆−𝟐𝒙)
∫𝐬𝐢𝐧 𝒕𝒙
𝐬𝐢𝐧𝐡(𝒃𝒙)𝒅𝒙
∞
𝟎
= 𝟐∑∫ 𝒆−𝒙(𝟐𝒃𝒏+𝒏) 𝐬𝐢𝐧 𝒕𝒙𝒅𝒙∞
𝟎
∞
𝒏=𝟎
= 𝟐𝒕∑𝟏
𝒕𝟐 + (𝟐𝒃𝒏 + 𝒃)𝟐
∞
𝒏=𝟎
Again, using Weierstrass product of 𝐜𝐨𝐬𝐡 𝒙:
𝐜𝐨𝐬𝐡 (𝝅𝒙
𝒃) =∏(𝟏+
𝒙𝟐
(𝟐𝒌𝒏+ 𝒌)𝟐)
𝒌≥𝟎
Taking logarithmic differention w.r.t. 𝒙, we get:
𝝅
𝟐𝒃𝐭𝐚𝐧𝐡 (
𝝅𝒙
𝟐𝒃) = 𝟐𝒙∑
𝟏
𝒙𝟐 + (𝟐𝒌𝒏 + 𝒌)𝒌≥𝟎
𝑰(𝒂, 𝒃) =𝝅
𝟐𝒃∫
𝒆−𝒂𝒕
𝒕∫
𝐬𝐢𝐧 𝒕𝒙
𝐬𝐢𝐧𝐡(𝒃𝒙)𝒅𝒙𝒅𝒕
∞
𝟎
∞
𝟎
=𝝅
𝟐𝒃∫
𝒆−𝒂𝒕
𝒕𝐭𝐚𝐧𝐡 (
𝝅𝒕
𝟐𝒃)
∞
𝟎
𝒅𝒕
Now, using Leibniz rule w.r.t. 𝒂, we get
𝑰(𝒂′, 𝒃) = −𝝅
𝟐𝒃∫ 𝒆−𝒂𝒕 𝐭𝐚𝐧𝐡 (
𝝅𝒕
𝟐𝒃)
∞
𝟎
𝒅𝒕 = −𝝅
𝟐𝒃∫
𝟏 − 𝒆𝝅𝒕𝒃
𝟏 + 𝒆𝝅𝒕𝒃
𝒆−𝒂𝒕∞
𝟎
𝒅𝒕 =
= −𝝅
𝟐𝒃∫ [−𝟏 + 𝟐∑(−𝟏)𝒏𝒆
𝝅𝒕𝒏𝒃
𝒏≥𝟎
] 𝒆−𝒂𝒕∞
𝟎
𝒅𝒕 =
=𝝅
𝟐𝒃− 𝝅∑(−𝟏)𝒏
𝟏
𝒏𝝅 + 𝒂𝒏≥𝟎
=𝝅
𝟐𝒃𝒂−𝟏
𝝅[𝝍(
𝒂 + 𝝅
𝟐𝝅) − 𝝍(
𝒂
𝟐𝝅)]
www.ssmrmh.ro
85 RMM-CALCULUS MARATHON 1501-1600
Therefore,
𝛀 = ∫𝐭𝐚𝐧−𝟏(𝒂𝒙)
𝐬𝐢𝐧𝐡(𝒃𝒙)
∞
𝟎
𝒅𝒙 =𝝅
𝒃(𝐥𝐨𝐠(
𝚪 (𝒃𝟐𝝅𝒂)
𝚪 (𝟏𝟐 +
𝒃𝟐𝝅𝒂)
) +𝟏
𝟐𝐥𝐨𝐠 (
𝒃
𝟐𝝅𝒂))
Solution 3 by Syed Shahabudeen-India
𝛀 = ∫𝐭𝐚𝐧−𝟏(𝒂𝒙)
𝐬𝐢𝐧𝐡(𝒃𝒙)
∞
𝟎
𝒅𝒙 = ∫𝟏
𝐬𝐢𝐧𝐡(𝒃𝒙)𝑳 {𝐬𝐢𝐧(𝒙𝒕)
𝒕}𝒔=𝟏𝒂
∞
𝟎
𝒅𝒙 =
= ∫𝒆−𝒕𝒂
𝒕
∞
𝟎
∫𝐬𝐢𝐧(𝒙𝒕)𝒆𝒃𝒙
𝒆𝟐𝒃𝒙 − 𝟏
∞
𝟎
𝒅𝒙𝒅𝒕 = ∫𝒆−𝒕𝒂
𝒕
∞
𝟎
∫ 𝐬𝐢𝐧(𝒙𝒕)𝒆−𝒃𝒙∞
𝟎
∑𝒆−𝟐𝒃𝒙𝒌∞
𝒌=𝟎
𝒅𝒙𝒅𝒕 =
= ∫𝒆−𝒕𝒂
𝒕∑∫ 𝒆(−𝟐𝒃𝒌+𝒃)𝒙 𝐬𝐢𝐧(𝒙𝒕)
∞
𝟎
∞
𝒌=𝟎
𝒅𝒙𝒅𝒕∞
𝟎
= ∫𝒆−𝒕𝒂
𝒕∑𝑳{𝐬𝐢𝐧(𝒙𝒕)}𝒔=(𝟐𝒃𝒌+𝒃)
∞
𝒌=𝟎
∞
𝟎
𝒅𝒕 =
= ∫𝒆−𝒕𝒂
𝒕∑
𝒕
(𝟐𝒃𝒌 + 𝒃)𝟐 + 𝒕𝟐
∞
𝒌=𝟎
𝒅𝒕∞
𝟎
= ∫ 𝒆−𝒕𝒂∑
𝟏
(𝟐𝒃𝒌 + 𝒃)𝟐 + 𝒕𝟐
∞
𝒌=𝟎
𝒅𝒕∞
𝟎
∑𝟏
(𝟐𝒃𝒌 + 𝒃)𝟐 + 𝒕𝟐
∞
𝒌=𝟎
=𝟏
𝒃𝟐∑
𝟏
(𝟐𝒌 + 𝟏)𝟐 +𝒕𝟐
𝒃𝟐
∞
𝒌=𝟎
=𝝅
𝟐𝒃𝒕𝐭𝐚𝐧𝐡 (
𝝅𝒕
𝟐𝒃)
𝛀 =𝝅
𝟐𝒃∫ 𝒆−
𝒕𝒂𝐭𝐚𝐧𝐡 (
𝝅𝒕𝟐𝒃)
𝒕
∞
𝟎
𝒅𝒕 = (𝝅
𝟐𝒃)𝟐
∫ 𝒆−𝒕𝒂𝐭𝐚𝐧𝐡 (
𝝅𝒕𝟐𝒃)
𝝅𝒕𝟐𝒃
𝒅𝒕∞
𝟎
=𝒙=𝝅𝒕𝟐𝒃
=𝝅
𝟐𝒃∫ 𝒆−
𝟐𝒃𝝅𝒂𝒙
∞
𝟎
𝐭𝐚𝐧𝐡𝒙
𝒙𝒅𝒙 =
𝝅
𝟐𝒃𝑳 {𝐭𝐚𝐧𝐡𝒙
𝒙}𝒔=𝟐𝒃𝝅𝒂
It is well-know that:
𝑳 {𝐭𝐚𝐧𝐡 𝒙
𝒙}𝒔=𝟐𝒃𝝅𝒂
= 𝟐 𝐥𝐨𝐠(√𝒔𝚪 (
𝒔𝟒)
𝟐𝚪(𝒔 + 𝟐𝟒 )
) ⇒
𝛀 =𝝅
𝒃𝐥𝐨𝐠
(
√𝟐𝒃𝝅𝒂𝚪 (
𝒃𝟐𝝅𝒂)
𝟐𝚪(
𝒃𝝅𝒂 + 𝟏
𝟐)
)
=𝝅
𝒃(𝐥𝐨𝐠(
𝚪 (𝒃𝟐𝝅𝒂
)
𝚪(𝟏𝟐 +
𝒃𝟐𝝅𝒂)
) +𝟏
𝟐𝐥𝐨𝐠 (
𝒃
𝟐𝝅𝒂)
www.ssmrmh.ro
86 RMM-CALCULUS MARATHON 1501-1600
Therefore,
𝛀 = ∫𝐭𝐚𝐧−𝟏(𝒂𝒙)
𝐬𝐢𝐧𝐡(𝒃𝒙)
∞
𝟎
𝒅𝒙 =𝝅
𝒃(𝐥𝐨𝐠(
𝚪 (𝒃𝟐𝝅𝒂)
𝚪 (𝟏𝟐 +
𝒃𝟐𝝅𝒂)
) +𝟏
𝟐𝐥𝐨𝐠 (
𝒃
𝟐𝝅𝒂))
1559. If 𝟏
√𝟑𝟏< 𝑎 ≤ 𝑏 then find:
𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟑𝟎𝒙𝟑 − 𝟏𝟎𝒙
𝟑𝟏𝒙𝟐 − 𝟏)
𝒃
𝒂
𝒅𝒙
Proposed by Daniel Sitaru-Romania
Solution by Rana Ranino-Setif-Algerie
𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟑𝟎𝒙𝟑 − 𝟏𝟎𝒙
𝟑𝟏𝒙𝟐 − 𝟏)
𝒃
𝒂
𝒅𝒙 = ∫ 𝐭𝐚𝐧−𝟏 (𝟏𝟎𝒙 − 𝟑𝟎𝒙𝟑
𝟏 − 𝟑𝟏𝒙𝟐)
𝒃
𝒂
𝒅𝒙
𝐭𝐚𝐧−𝟏 (𝟏𝟎𝒙 − 𝟑𝟎𝒙𝟑
𝟏 − 𝟑𝟏𝒙𝟐) = 𝐭𝐚𝐧−𝟏 (
𝟓𝒙 + 𝟓𝒙(𝟏 − 𝟔𝒙𝟐)
𝟏 − 𝟑𝟏𝒙𝟐) = 𝐭𝐚𝐧−𝟏 (
𝟓𝒙𝟏− 𝟔𝒙𝟐
+ 𝟓𝒙
𝟏 −𝟐𝟓𝒙𝟐
𝟏 − 𝟔𝒙𝟐
)
𝑭𝒐𝒓 𝒙 ≥𝟏
√𝟑𝟏⇒
𝟐𝟓𝒙𝟐
𝟏 − 𝟔𝒙𝟐≥ 𝟏 ⇒ 𝐭𝐚𝐧−𝟏 (
𝟏𝟎𝒙 − 𝟑𝟎𝒙𝟑
𝟏 − 𝟑𝟏𝒙𝟐) =
= 𝐭𝐚𝐧−𝟏 (𝟓𝒙
𝟏 − 𝟔𝒙𝟐) + 𝐭𝐚𝐧−𝟏(𝟓𝒙) − 𝝅
𝐭𝐚𝐧−𝟏 (𝟓𝒙
𝟏 − 𝟔𝒙𝟐) = 𝐭𝐚𝐧−𝟏 (
𝟐𝒙 + 𝟑𝒙
𝟏 − 𝟔𝒙𝟐) = 𝐭𝐚𝐧−𝟏(𝟐) + 𝐭𝐚𝐧−𝟏(𝟑𝒙)
𝐭𝐚𝐧−𝟏 (𝟏𝟎𝒙 − 𝟑𝟎𝒙𝟑
𝟏 − 𝟑𝟏𝒙𝟐) = 𝐭𝐚𝐧−𝟏(𝟐𝒙) + 𝐭𝐚𝐧−𝟏(𝟑𝒙) + 𝐭𝐚𝐧−𝟏(𝟓𝒙) − 𝝅
𝛀(𝒂, 𝒃) = ∫ (−𝝅 + 𝐭𝐚𝐧−(𝟐𝒙) + 𝐭𝐚𝐧−𝟏(𝟑𝒙) + 𝐭𝐚𝐧−𝟏(𝟓𝒙))𝒃
𝒂
𝒅𝒙 =
= [−𝝅𝒙 + 𝒙 𝐭𝐚𝐧−𝟏(𝟐𝒙) + 𝒙 𝐭𝐚𝐧−𝟏(𝟑𝒙) + 𝒙 𝐭𝐚𝐧−𝟏(𝟓𝒙) −𝟏
𝟒𝐥𝐨𝐠(𝟒𝒙𝟐 + 𝟏) −
𝟏
𝟔(𝟗𝒙𝟐 + 𝟏)
−𝟏
𝟏𝟎(𝟐𝟓𝒙𝟐 + 𝟏)]
𝒂
𝒃
=
= [𝒙 𝐭𝐚𝐧−𝟏 (𝟑𝟎𝒙𝟑 − 𝟏𝟎𝒙
𝟑𝟏𝒙𝟐 − 𝟏) −
𝟏
𝟒𝐥𝐨𝐠(𝟒𝒙𝟐 + 𝟏) −
𝟏
𝟔(𝟗𝒙𝟐 + 𝟏) −
𝟏
𝟏𝟎(𝟐𝟓𝒙𝟐 + 𝟏)]
𝒂
𝒃
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Therefore,
𝛀(𝒂, 𝒃) = 𝒃 𝐭𝐚𝐧−𝟏 (𝟑𝟎𝒃𝟑 − 𝟏𝟎𝒃
𝟑𝟏𝒃𝟐 − 𝟏) − 𝒂 𝐭𝐚𝐧−𝟏 (
𝟑𝟎𝒂𝟑 − 𝟏𝟎𝒂
𝟑𝟏𝒂𝟐 − 𝟏) −
𝟏
𝟒𝐥𝐨𝐠(
𝟒𝒃𝟐 + 𝟏
𝟒𝒂𝟐 + 𝟏) −
−𝟏
𝟔𝐥𝐨𝐠(
𝟗𝒃𝟐 + 𝟏
𝟗𝒂𝟐 + 𝟏) −
𝟏
𝟏𝟎𝐥𝐨𝐠 (
𝟐𝟓𝒃𝟐 + 𝟏
𝟐𝟓𝒂𝟐 + 𝟏)
1560. If 𝟎 < 𝑎 ≤ 𝑏 <𝝅
𝟔 then find:
𝛀(𝒂, 𝒃) = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐
𝐜𝐨𝐬𝟐 𝒙 − 𝟑 𝐬𝐢𝐧𝟐 𝒙
𝒃
𝒂
𝒅𝒙
Proposed by Daniel Sitaru-Romania
Solution 1 by Asmat Qatea-Afghanistan
𝛀(𝒂, 𝒃) = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐
𝐜𝐨𝐬𝟐 𝒙 − 𝟑𝐬𝐢𝐧𝟐 𝒙
𝒃
𝒂
𝒅𝒙 = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐
𝐜𝐨𝐬𝟐 𝒙 (𝟏 + 𝟑 𝐭𝐚𝐧𝟐 𝒙)
𝒃
𝒂
𝒅𝒙 ⇒𝐭𝐚𝐧 𝒙=𝒕
𝛀 = ∫(𝟏 + 𝒕𝟐)𝟐
𝟏 − 𝟑𝒕𝟐𝒅𝒕 = ∫
𝟏 + 𝟐𝒕𝟐 + 𝒕𝟒
𝟏 − 𝟑𝒕𝟐𝒅𝒕 = ∫(−
𝟏
𝟑𝒕𝟐 −
𝟕
𝟗+
𝟏𝟔𝟗
𝟏 − 𝟑𝒕𝟐)𝒅𝒕 =
= −𝟏
𝟗𝒕𝟑 −
𝟕
𝟗𝒕 +𝟏𝟔
𝟐𝟕∫
𝟏
𝟏𝟑− 𝒕𝟐
𝒅𝒕 = −𝟏
𝟗𝒕𝟑 −
𝟕
𝟗𝒕 +
𝟏𝟔
𝟐𝟕⋅√𝟑
𝟐𝐥𝐨𝐠(
𝟏
√𝟑+ 𝒕
𝟏
√𝟑− 𝒕)+ 𝑪 =
= −𝟏
𝟗𝒕𝟑 −
𝟕
𝟗𝒕 +
𝟏𝟔√𝟑
𝟐𝟕⋅𝟏
𝟐𝐥𝐨𝐠 (
𝟏 + √𝟑𝒕
𝟏 − √𝟑𝒕) + 𝑪 =
= −𝟏
𝟗𝒕𝟑 −
𝟕
𝟗𝒕 +
𝟏𝟔√𝟑
𝟐𝟕𝐭𝐚𝐧𝐡−𝟏(√𝟑𝒕) + 𝑪
𝛀(𝒂, 𝒃) = [−𝟏
𝟗𝐭𝐚𝐧𝟑 𝒙 −
𝟕
𝟗𝐭𝐚𝐧𝒙 +
𝟏𝟔√𝟑
𝟐𝟕𝐭𝐚𝐧𝐡−𝟏(√𝟑 𝐭𝐚𝐧 𝒙)]
𝒂
𝒃
𝛀(𝒂, 𝒃) = −𝟏
𝟗𝐭𝐚𝐧𝟑 𝒃 −
𝟕
𝟗𝐭𝐚𝐧 𝒃 +
𝟏𝟔√𝟑
𝟐𝟕𝐭𝐚𝐧𝐡−𝟏(√𝟑 𝐭𝐚𝐧 𝒃) +
𝟏
𝟗𝐭𝐚𝐧𝟑 𝒂 +
𝟕
𝟗𝐭𝐚𝐧𝒂 −
−𝟏𝟔√𝟑
𝟐𝟕𝐭𝐚𝐧𝐡−𝟏(√𝟑 𝐭𝐚𝐧 𝒂)
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Solution 2 by Remus Florin Stanca-Romania
𝛀(𝒂, 𝒃) = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐
𝐜𝐨𝐬𝟐 𝒙 − 𝟑𝐬𝐢𝐧𝟐 𝒙
𝒃
𝒂
𝒅𝒙 = ∫𝟏
𝐜𝐨𝐬𝟐 𝒙⋅(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐
𝟏 − 𝟑 𝐭𝐚𝐧𝟐 𝒙
𝒃
𝒂
𝒅𝒙 =𝐭𝐚𝐧 𝒙=𝒕
= ∫(𝟏 + 𝒕𝟐)𝟐
𝟏 − 𝟑𝒕𝟐
𝐭𝐚𝐧 𝒃
𝐭𝐚𝐧 𝒂
𝒅𝒕 = ∫𝟏 + 𝒕𝟒 + 𝟐𝒕𝟐
𝟏 − 𝟑𝒕𝟐
𝐭𝐚𝐧 𝒃
𝐭𝐚𝐧 𝒂
𝒅𝒕 =𝟏
𝟑∫
𝟑𝒕𝟒 + 𝟔𝒕𝟐 + 𝟑
𝟏 − 𝟑𝒕𝟐
𝐭𝐚𝐧 𝒃
𝐭𝐚𝐧 𝒂
𝒅𝒕 =
=𝟏
𝟑∫
𝟑𝒕𝟒 − 𝒕𝟐 + 𝟕𝒕𝟐 + 𝟑
𝟏 − 𝟑𝒕𝟐
𝐭𝐚𝐧 𝒃
𝐭𝐚𝐧 𝒂
𝒅𝒕 =𝟏
𝟑∫ (−𝒕𝟐 +
𝟕𝒕𝟐 + 𝟑
𝟏 − 𝟑𝒕𝟐)𝒅𝒕
𝐭𝐚𝐧 𝒃
𝐭𝐚𝐧 𝒂
=
=𝟏
𝟑∫ (−𝒕𝟐 +
𝟏
𝟑(−𝟕 +
𝟏𝟔
𝟏 − 𝟑𝒕𝟐))
𝐭𝐚𝐧 𝒃
𝐭𝐚𝐧 𝒂
𝒅𝒕 = [−𝒕𝟑
𝟗−𝟕𝒕
𝟗−𝟏𝟔
𝟐𝟕⋅√𝟑
𝟐𝐥𝐨𝐠 |
𝒕 −𝟏
√𝟑
𝒕 +𝟏
√𝟑
|]
𝐭𝐚𝐧 𝒂
𝐭𝐚𝐧 𝒃
𝛀(𝒂, 𝒃) = −𝐭𝐚𝐧𝟑 𝒃
𝟗−𝟕 𝐭𝐚𝐧 𝒃
𝟗−𝟖√𝟑
𝟐𝟕𝐥𝐨𝐠 |
√𝟑 𝐭𝐚𝐧 𝒃 − 𝟏
√𝟑 𝐭𝐚𝐧 𝒃 + 𝟏| +𝐭𝐚𝐧𝟑 𝒂
𝟗
+𝟕 𝐭𝐚𝐧𝒂
𝟗𝐥𝐨𝐠 |
√𝟑 𝐭𝐚𝐧 𝒂 − 𝟏
√𝟑 𝐭𝐚𝐧 𝒂 + 𝟏|
Solution 3 by Ghuiam Shah Naseri-Afghanistan
𝛀(𝒂, 𝒃) = ∫(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐
𝐜𝐨𝐬𝟐 𝒙 − 𝟑𝐬𝐢𝐧𝟐 𝒙
𝒃
𝒂
𝒅𝒙 = ∫𝟏
𝐜𝐨𝐬𝟐 𝒙⋅(𝟏 + 𝐭𝐚𝐧𝟐 𝒙)𝟐
𝟏 − 𝟑 𝐭𝐚𝐧𝟐 𝒙
𝒃
𝒂
𝒅𝒙 ⇒𝐭𝐚𝐧 𝒙=𝒖
𝛀 = ∫(𝟏 + 𝒖𝟐)𝟐
𝟏 − 𝟑𝒖𝟐𝒅𝒖 = ∫
𝒖𝟒 + 𝟐𝒖𝟐 + 𝟏
−𝟑𝒖𝟐 + 𝟏𝒅𝒖
= ∫(𝟖√𝟑
𝟐𝟕⋅
𝟏
𝒖 +√𝟑𝟑
−𝟖√𝟑
𝟐𝟕⋅
𝟏
𝒖 −√𝟑𝟑
−𝟏
𝟑𝒖𝟐 −
𝟕
𝟗)𝒅𝒖 =
= ∫𝟖√𝟑
𝟐𝟕⋅
𝟏
𝒖 +√𝟑𝟑
𝒅𝒖 −∫𝟖√𝟑
𝟐𝟕⋅
𝟏
𝒖 −√𝟑𝟑
𝒅𝒖 −∫𝟏
𝟑𝒖𝟐𝒅𝒖 − ∫
𝟕
𝟗𝒅𝒖 =
=𝟖√𝟑
𝟐𝟕𝐥𝐨𝐠 (𝒖 +
√𝟑
𝟑) −
𝟖√𝟑
𝟑𝐥𝐨𝐠 (𝒖 −
√𝟑
𝟑) −
𝟏
𝟑𝒖𝟑 −
𝟕
𝟗𝒖
𝛀(𝒂, 𝒃) = [𝟖√𝟑
𝟐𝟕𝐥𝐨𝐠(𝐭𝐚𝐧 𝒙 +
√𝟑
𝟑) −
𝟖√𝟑
𝟑𝐥𝐨𝐠 (𝐭𝐚𝐧𝒙 −
√𝟑
𝟑) −
𝟏
𝟑𝐭𝐚𝐧𝟑 𝒙 −
𝟕
𝟗𝐭𝐚𝐧𝒙]
𝒂
𝒃
=
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89 RMM-CALCULUS MARATHON 1501-1600
=𝟖√𝟑
𝟐𝟕𝐥𝐨𝐠 (𝐭𝐚𝐧𝒃 +
√𝟑
𝟑) −
𝟖√𝟑
𝟑𝐥𝐨𝐠 (𝐭𝐚𝐧𝒃 −
√𝟑
𝟑) −
𝟏
𝟑𝐭𝐚𝐧𝟑 𝒃 −
𝟕
𝟗𝐭𝐚𝐧 𝒃 −
−(𝟖√𝟑
𝟐𝟕𝐥𝐨𝐠 (𝐭𝐚𝐧 𝒂 +
√𝟑
𝟑) −
𝟖√𝟑
𝟑𝐥𝐨𝐠 (𝐭𝐚𝐧𝒂 −
√𝟑
𝟑) −
𝟏
𝟑𝐭𝐚𝐧𝟑 𝒂 −
𝟕
𝟗𝐭𝐚𝐧 𝒂) =
=𝟖√𝟑
𝟐𝟕𝐥𝐨𝐠(
𝐭𝐚𝐧𝒃 +√𝟑𝟑
𝐭𝐚𝐧𝒃 −√𝟑𝟑
) −𝟏
𝟑𝐭𝐚𝐧𝟑 𝒃 −
𝟕
𝟗𝐭𝐚𝐧 𝒃 −
𝟖√𝟑
𝟐𝟕𝐥𝐨𝐠(
𝐭𝐚𝐧𝒂 +√𝟑𝟑
𝐭𝐚𝐧𝒂 −√𝟑𝟑
)+𝟏
𝟑𝐭𝐚𝐧𝟑 𝒂
+𝟕
𝟗𝐭𝐚𝐧𝒂
1561. Find a closed form:
𝛀 = ∫ 𝒙−𝟐 ⋅ 𝒆−𝟒𝒙 ⋅ 𝐬𝐢𝐧𝟐(𝟐𝒙)∞
𝟎
𝒅𝒙
Proposed by Abdul Mukhtar-Nigeria
Solution 1 by Ty Halpen-Florida-USA
We will parametrize the integral:
𝑰(𝒂) = ∫ 𝒙−𝟐 ⋅ 𝒆−𝒂𝒙 ⋅ 𝐬𝐢𝐧𝟐(𝟐𝒙)∞
𝟎
𝒅𝒙
𝝏𝟐𝑰(𝒂)
𝝏𝒂𝟐= ∫ 𝒆−𝟐𝒂 𝐬𝐢𝐧𝟐(𝟐𝒙)𝒅𝒙
∞
𝟎
=
= [𝒆−𝒂𝒙(𝒂𝟐 𝐜𝐨𝐬(𝟒𝒙) − 𝒂𝟐 − 𝟒𝒂 𝐬𝐢𝐧(𝟒𝒙) − 𝟏𝟔)
𝟐𝒂(𝒂𝟐 + 𝟏𝟔)]𝒙=𝟎
𝒙=∞
=𝟖
𝒂(𝒂𝟐 + 𝟏𝟔)
𝝏𝑰(𝒂)
𝝏𝒂= ∫
𝟖
𝒂(𝒂𝟐 + 𝟏𝟔)𝒅𝒂 =
𝟏
𝟐𝐥𝐨𝐠𝒂 −
𝟏
𝟒𝐥𝐨𝐠(𝒂𝟐 + 𝟏𝟔) + 𝑪𝟏
𝑰(𝒂) = ∫(𝟏
𝟐𝐥𝐨𝐠𝒂 −
𝟏
𝟒𝐥𝐨𝐠(𝒂𝟐 + 𝟏𝟔) + 𝑪𝟏)𝒅𝒂 =
𝑰𝑩𝑷
= −𝒂
𝟒𝐥𝐨𝐠(𝒂𝟐 + 𝟏𝟔) +
𝒂
𝟐𝐥𝐨𝐠 𝒂 − 𝟐 𝐭𝐚𝐧−𝟏 (
𝒂
𝟒) + 𝒂𝑪𝟏 + 𝑪𝟐
Now, notice that 𝐥𝐢𝐦𝒂→∞
𝑰(𝒂) = 𝟎 and 𝑰(𝒂) = 𝟎 from the famous Dirichlet integral:
𝑰(𝟎) = 𝝅 = 𝑪𝟐
𝐥𝐢𝐦𝒂→∞
𝑰(𝟎) = 𝟎 = −𝟐 (𝝅
𝟐) + 𝒂𝑪𝟏 + 𝝅 ⇒ 𝑪𝟏 = 𝟎
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Then,
𝑰(𝟒) =𝝅
𝟐− 𝐥𝐨𝐠 𝟐
Solution 2 by Rana Ranino-Setif-Algerie
𝛀 = ∫ 𝒙−𝟐 ⋅ 𝒆−𝟒𝒙 ⋅ 𝐬𝐢𝐧𝟐(𝟐𝒙)∞
𝟎
𝒅𝒙 = ∫𝒆−𝟒𝒙 𝐬𝐢𝐧𝟐(𝟐𝒙)
𝒙𝟐𝒅𝒙
∞
𝟎
=
= ∫ 𝒆−𝟒𝒙 𝐬𝐢𝐧𝟐(𝟐𝒙)(∫ 𝒚𝒆−𝒙𝒚𝒅𝒚∞
𝟎
)𝒅𝒙∞
𝟎
= ∫ ∫ 𝒚 𝐬𝐢𝐧𝟐(𝟐𝒙)𝒆−(𝒚+𝟒)𝒙∞
𝟎
𝒅𝒙𝒅𝒚∞
𝟎
=
=𝟏
𝟐∫ ∫ 𝒚(𝟏 − 𝐜𝐨𝐬(𝟒𝒙))𝒆−(𝒚+𝟒)𝒙
∞
𝟎
∞
𝟎
𝒅𝒙𝒅𝒕 =𝟏
𝟐∫ (
𝒚
𝒚 + 𝟒−
𝒚(𝒚 + 𝟒)
(𝒚 + 𝟏)𝟐 + 𝟏𝟔)
∞
𝟎
𝒅𝒚 =
= ∫ (𝟐𝒚 + 𝟖
𝒚𝟐 + 𝟖𝒚 + 𝟑𝟐−
𝟐
𝒚 + 𝟒+
𝟖
(𝒚 + 𝟒)𝟐 + 𝟏𝟔)
∞
𝟎
𝒅𝒚 =
= [𝐥𝐨𝐠 (𝒚𝟐 + 𝟖𝒚 + 𝟑𝟐
𝒚𝟐 + 𝟖𝒚 + 𝟏𝟔) + 𝟐 𝐭𝐚𝐧−𝟏 (
𝒚 + 𝟒
𝟒)]𝟎
∞
=𝝅
𝟐− 𝐥𝐨𝐠 𝟐
Solution 3 by Yen Tung Chung-Taichung-Taiwan
𝛀 = ∫ 𝒙−𝟐 ⋅ 𝒆−𝟒𝒙 ⋅ 𝐬𝐢𝐧𝟐(𝟐𝒙)∞
𝟎
𝒅𝒙 = ∫ 𝒙−𝟐 ⋅ 𝒆−𝟒𝒙 ⋅𝟏 − 𝐜𝐨𝐬(𝟒𝒙)
𝟐
∞
𝟎
𝒅𝒙 =
=𝟏
𝟐∫ 𝒆−𝟒𝒙 ⋅
𝟏 − 𝐜𝐨𝐬(𝟒𝒙)
𝒙𝟐
∞
𝟎
𝒅𝒙 =𝟏
𝟐𝑳(𝟏 − 𝐜𝐨𝐬(𝟒𝒕)
𝒕𝟐)|𝒔=𝟒
=
=𝟏
𝟐(∫ ∫ 𝑳(𝟏 − 𝐜𝐨𝐬(𝟒𝒕))𝒅𝒔𝒅𝒔
∞
𝒔
∞
𝒔
)|𝒔=𝟒
=𝟏
𝟐(∫ ∫ (
𝟏
𝒔−
𝒔
𝒔𝟐 + 𝟏𝟔)𝒅𝒔𝒅𝒔
∞
𝒔
∞
𝒔
)|𝒔=𝟒
=
=𝟏
𝟐(∫ (𝐥𝐨𝐠 𝒔 −
𝟏
𝟐𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔))
𝒔
∞∞
𝒔
)|𝒔=𝟒
=𝟏
𝟐(∫ (
𝟏
𝟐𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) − 𝐥𝐨𝐠 𝒔)𝒅𝒙
∞
𝒔
)|𝒔=𝟒
=
=𝟏
𝟐((𝟏
𝟐𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) + 𝟒 𝐭𝐚𝐧−𝟏 (
𝒔
𝟒) − 𝒔 𝐥𝐨𝐠 𝒔)|
𝒔
∞
)|𝒔=𝟒
=
=𝟏
𝟐(𝟐𝝅 − (
𝟏
𝟐𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) + 𝟒 𝐭𝐚𝐧−𝟏 (
𝒔
𝟒) − 𝒔 𝐥𝐨𝐠 𝒔)|
𝒔=𝟒=
=𝟏
𝟐(𝟐𝝅 − (𝟏𝟎 𝐥𝐨𝐠 𝟐 + 𝝅 − 𝟖 𝐥𝐨𝐠 𝟐)) =
𝟏
𝟐(𝝅 − 𝟐 𝐥𝐨𝐠 𝟐) =
𝝅
𝟐− 𝐥𝐨𝐠𝟐.
Where,
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91 RMM-CALCULUS MARATHON 1501-1600
𝒊) ∫ 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔)𝒅𝒔 = 𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) − 𝟐∫𝒔𝟐
𝒔𝟐 + 𝟏𝟔𝒅𝒔 =
𝒖=𝐥𝐨𝐠(𝒔𝟐+𝟏𝟔)
= 𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) − 𝟐∫(𝟏 −𝟏𝟔
𝒔𝟐 + 𝟏𝟔)𝒅𝒔 = 𝒔 𝐥𝐨𝐠(𝒔𝟐 + 𝟏𝟔) − 𝟐𝒔 + 𝟖 𝐭𝐚𝐧−𝟏 (
𝒔
𝟒) + 𝑪
𝒊𝒊) ∫ 𝐥𝐨𝐠 𝒔 𝒅𝒔 = 𝒔 𝐥𝐨𝐠 𝒔 − 𝒔 + 𝑪.
1562. Find a closed form:
𝛀(𝒂) = ∫𝒙√𝒙
(𝒙𝟐 + 𝟏)(𝟏 + 𝒂𝟐𝒙𝟐)
∞
𝟎
𝒅𝒙, 𝒂 > 𝟎
Proposed by Vasile Mircea Popa-Romania
Solution by Rana Ranino-Setif-Algerie
𝛀(𝒂) = ∫𝒙√𝒙
(𝒙𝟐 + 𝟏)(𝟏 + 𝒂𝟐𝒙𝟐)
∞
𝟎
𝒅𝒙 = ∫√𝒙
(𝟏 + 𝒙𝟐)(𝒂𝟐 + 𝒙𝟐)𝒅𝒙
∞
𝟎
=
=𝟏
𝒂𝟐 − 𝟏(∫
√𝒙
𝟏 + 𝒙𝟐
∞
𝟎
𝒅𝒙⏟
𝑨
− ∫√𝒙
𝒂𝟐 + 𝒙𝟐
∞
𝟎
𝒅𝒙⏟
𝑩
)
𝑨 = ∫√𝒙
𝟏 + 𝒙𝟐
∞
𝟎
𝒅𝒙 =𝝅
𝟐𝐬𝐢𝐧 (𝟑𝝅𝟒 )
=𝝅
√𝟒
𝑩 =𝒙=𝒂𝒚 𝟏
√𝒂∫
√𝒚
𝟏 + 𝒚𝟐
∞
𝟎
𝒅𝒚 =𝝅
√𝟐𝒂
𝛀 =𝝅
(𝒂𝟐 − 𝟏)√𝟐(𝟏 −
𝟏
√𝒂) =
𝝅
(√𝒂 + 𝟏)(𝒂 + 𝟏)√𝟐𝒂
Therefore,
𝛀(𝒂) = ∫𝒙√𝒙
(𝒙𝟐 + 𝟏)(𝟏 + 𝒂𝟐𝒙𝟐)
∞
𝟎
𝒅𝒙 =𝝅
(√𝒂 + 𝟏)(𝒂 + 𝟏)√𝟐𝒂
1563. If 𝟎 < 𝒂 ≤ 𝒃 <𝝅
𝟖 then find:
𝛀(𝒂, 𝒃) = ∫ ∫(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧 (
𝝅𝟒− 𝒙 − 𝒚))
𝟏 + 𝐭𝐚𝐧 𝒙 ⋅ 𝐭𝐚𝐧 𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒− 𝒙 − 𝒚)
𝒅𝒙𝒃
𝒂
𝒅𝒚𝒃
𝒂
Proposed by Daniel Sitaru-Romania
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92 RMM-CALCULUS MARATHON 1501-1600
Solution 1 by Kamel Gandouli Rezgui-Tunisia
𝐭𝐚𝐧 (𝝅
𝟒− 𝒙 − 𝒚) =
𝟏 − 𝐭𝐚𝐧(𝒙 + 𝒚)
𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)⇒ 𝟏 + 𝐭𝐚𝐧 (
𝝅
𝟒− 𝒙 − 𝒚) =
𝟐
𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)
(𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)) (𝐭𝐚𝐧(𝝅
𝟒− (𝒙 + 𝒚)) = 𝟏 − 𝐭𝐚𝐧(𝒙 + 𝒚)
= [𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)] [𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧 (𝝅
𝟒− (𝒙 + 𝒚))] =
= 𝟏 + 𝐭𝐚𝐧𝒙 𝐭𝐚𝐧𝒚 + 𝐭𝐚𝐧(𝒙 + 𝒚) − 𝐭𝐚𝐧𝒙 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧(𝒙 + 𝒚) =
= 𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧𝒚 + 𝐭𝐚𝐧 𝒙 + 𝐭𝐚𝐧 𝒚
Hence,
(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚))
𝟏 + 𝐭𝐚𝐧𝒙 ⋅ 𝐭𝐚𝐧 𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚)
=𝟐(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚)
𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧 (𝝅𝟒 −
(𝒙 + 𝒚))=
=𝟐(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧𝒚)
𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚=𝟐(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧𝒚)
(𝟏 + 𝐭𝐚𝐧 𝒚)(𝟏 + 𝐭𝐚𝐧 𝒙)= 𝟐
Therefore,
𝛀(𝒂, 𝒃) = ∫ ∫(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧 (
𝝅𝟒 − 𝒙 − 𝒚))
𝟏 + 𝐭𝐚𝐧 𝒙 ⋅ 𝐭𝐚𝐧𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚)
𝒅𝒙𝒃
𝒂
𝒅𝒚𝒃
𝒂
= 𝟐(𝒃 − 𝒂)𝟐
Solution 2 by Remus Florin Stanca-Romania
(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧𝒚) (𝟏 + 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚))
𝟏 + 𝐭𝐚𝐧𝒙 ⋅ 𝐭𝐚𝐧𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚)
=
=(𝟏 + 𝐭𝐚𝐧𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) ⋅
𝟐 − 𝟐 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 + 𝐭𝐚𝐧𝒙 + 𝐭𝐚𝐧𝒚
𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 ⋅𝟏 − 𝐭𝐚𝐧(𝒙 + 𝒚)𝟏 + 𝐭𝐚𝐧(𝒙 + 𝒚)
=
= 𝟐 ⋅(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) ⋅
𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚𝟏 − 𝐭𝐚𝐧𝒙 𝐭𝐚𝐧 𝒚 + 𝐭𝐚𝐧𝒙 + 𝐭𝐚𝐧 𝒚
𝟏 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 ⋅𝟏 −
𝐭𝐚𝐧𝒙 + 𝐭𝐚𝐧𝒚𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚
𝟏 +𝐭𝐚𝐧𝒙 + 𝐭𝐚𝐧𝒚𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚
=
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93 RMM-CALCULUS MARATHON 1501-1600
= 𝟐 ⋅(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚)(𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚)
𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 + 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧 𝒚 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝟐 𝒚 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧 𝒚 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧𝟐 𝒚=
= 𝟐 ⋅𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧𝒚 + 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧𝟐 𝒚 + 𝐭𝐚𝐧 𝒙 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝒚 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝟐 𝒚
𝟏 − 𝐭𝐚𝐧 𝒙 𝐭𝐚𝐧𝒚 + 𝐭𝐚𝐧 𝒚 𝐭𝐚𝐧𝟐 𝒚 + 𝐭𝐚𝐧 𝒙 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝒚 − 𝐭𝐚𝐧𝟐 𝒙 𝐭𝐚𝐧𝟐 𝒚= 𝟐
Therefore,
𝛀(𝒂, 𝒃) = ∫ ∫(𝟏 + 𝐭𝐚𝐧 𝒙)(𝟏 + 𝐭𝐚𝐧 𝒚) (𝟏 + 𝐭𝐚𝐧 (
𝝅𝟒 − 𝒙 − 𝒚))
𝟏 + 𝐭𝐚𝐧 𝒙 ⋅ 𝐭𝐚𝐧𝒚 ⋅ 𝐭𝐚𝐧 (𝝅𝟒 − 𝒙 − 𝒚)
𝒅𝒙𝒃
𝒂
𝒅𝒚𝒃
𝒂
= 𝟐(𝒃 − 𝒂)𝟐
1564.
𝑨 = ∫𝐥𝐨𝐠 𝒙
𝟏 + 𝒆𝒙 + 𝒆𝟐𝒙 + 𝒆𝟑𝒙𝒅𝒙
∞
𝟎
, 𝑩 = ∫𝒙 𝐥𝐨𝐠(𝟏 + 𝒙)
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
Prove that:
𝛀 = 𝑨+ 𝑩 =𝝅𝟐
𝟗𝟔−𝟏
𝟐𝐥𝐨𝐠𝟐 𝟐 −
𝝅
𝟖𝐥𝐨𝐠𝑹 and hence find the value of 𝑹.
Proposed by Ajetunmobi Abdulqoyyum-Nigeria
Solution by Rana Ranino-Setif-Algerie
𝑨 = ∫𝐥𝐨𝐠𝒙
𝟏 + 𝒆𝒙 + 𝒆𝟐𝒙 + 𝒆𝟑𝒙𝒅𝒙
∞
𝟎
=𝟏
𝟐(∫
𝐥𝐨𝐠𝒙
𝒆𝒙 + 𝟏𝒅𝒙
∞
𝟎⏟ 𝑨𝟏
+∫𝐥𝐨𝐠𝒙
𝒆𝟐𝒙 + 𝟏𝒅𝒙
∞
𝟎⏟ 𝑨𝟐
−∫𝒆𝒙 𝐥𝐨𝐠𝒙
𝒆𝟐𝒙 + 𝟏𝒅𝒙
∞
𝟎⏟ 𝑨𝟑
)
𝑨𝟏 = ∫𝐥𝐨𝐠 𝒙
𝒆𝒙 + 𝟏𝒅𝒙
∞
𝟎
=𝝏
𝝏𝒔{∫
𝒙𝒔−𝟏
𝒆𝒙 + 𝟏𝒅𝒙
∞
𝟎
}𝒔=𝟏
=𝝏
𝝏𝒔{𝜼(𝒔)𝚪(𝒔)}𝒔=𝟏 =
= 𝜼′(𝟏) + 𝜼(𝟏)𝚪(𝟏)𝝍(𝟏) = −𝟏
𝟐𝐥𝐨𝐠𝟐 𝟐
𝑨𝟐 =𝟏
𝟐∫
𝐥𝐨𝐠 𝒙 − 𝐥𝐨𝐠𝟐
𝒆𝒙 + 𝟏𝒅𝒙
∞
𝟎
=𝟏
𝟐∫
𝐥𝐨𝐠 𝒙
𝒆𝒙 + 𝟏𝒅𝒙
∞
𝟎⏟ 𝑨𝟏
−𝟏
𝟐𝐥𝐨𝐠𝟐∫
𝒅𝒙
𝒆𝒙 + 𝟏
∞
𝟎⏟ 𝜼(𝟏)=𝐥𝐨𝐠 𝟐
= −𝟑
𝟒𝐥𝐨𝐠𝟐 𝟐
𝑨𝟑 = ∫𝒆−𝒙 𝐥𝐨𝐠 𝒙
𝟏 + 𝒆−𝟐𝒙𝒅𝒙
∞
𝟎
=𝒕=𝒆−𝒙
∫𝐥𝐨𝐠 (𝐥𝐨𝐠 (
𝟏𝒕))
𝟏 + 𝒕𝟐𝒅𝒕
𝟏
𝟎
Using Malmsten’s integral:
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94 RMM-CALCULUS MARATHON 1501-1600
𝑰(𝝋) = ∫𝐥𝐨𝐠 (𝐥𝐨𝐠
𝟏𝒙)
𝟏 + 𝟐𝒙 𝐜𝐨𝐬𝝋 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
=𝝅
𝟐𝐬𝐢𝐧𝝋𝐥𝐨𝐠 {
(𝟐𝝅)𝝋𝝅 (𝟏𝟐 +
𝝋𝟐𝝅)
𝚪 (𝟏𝟐 −
𝝋𝟐𝝅)
}
𝑨𝟑 = 𝑰 (𝝅
𝟐) =
𝝅
𝟐𝐥𝐨𝐠{
√𝟐𝝅𝚪 (𝟑𝟒)
𝚪 (𝟏𝟒)
} =𝝅
𝟐𝐥𝐨𝐠 {
𝟐𝝅𝟑𝟐
𝚪𝟐 (𝟏𝟒)} =
𝝅
𝟒𝐥𝐨𝐠{
𝟒𝝅𝟑
𝚪𝟒 (𝟏𝟒)}
𝑨 = −𝟓
𝟖𝐥𝐨𝐠𝟐 𝟐 −
𝝅
𝟖𝐥𝐨𝐠(
𝟒𝝅𝟑
𝚪 (𝟏𝟒))
𝑩 = ∫𝒙 𝐥𝐨𝐠(𝟏 + 𝒙)
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
=∬𝒙𝟐
(𝟏 + 𝒙𝟐)(𝟏 + 𝒙𝒚)𝒅𝒙𝒅𝒚
𝟏
𝟎
=
= ∫𝟏
𝟏 + 𝒚𝟐
𝟏
𝟎
∫ (𝒙𝒚 − 𝟏
𝟏 + 𝒙𝟐+
𝟏
𝟏 + 𝒙𝒚)
𝟏
𝟎
𝒅𝒙𝒅𝒚 =
= ∫𝟏
𝟏 + 𝒚𝟐{𝟏
𝟐𝒚 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − 𝐭𝐚𝐧−𝟏 𝒙 +
𝟏
𝒚𝐥𝐨𝐠(𝟏 + 𝒙𝒚)}
𝟎
𝟏
𝒅𝒚𝟏
𝟎
=
= ∫𝟏
𝟏 + 𝒚𝟐(𝒚
𝟐𝐥𝐨𝐠 𝟐 −
𝝅
𝟒+𝐥𝐨𝐠(𝟏 + 𝒚)
𝒚)𝒅𝒚
𝟏
𝟎
=
=𝟏
𝟐𝐥𝐨𝐠 𝟐∫
𝒚
𝟏 + 𝒚𝟐𝒅𝒚
𝟏
𝟎
−𝝅
𝟒∫
𝒅𝒚
𝟏 + 𝒚𝟐
𝟏
𝟎
+∫𝐥𝐨𝐠(𝟏 + 𝒚)
𝒚𝒅𝒚
𝟏
𝟎
− 𝑩 =
=𝟏
𝟒𝐥𝐨𝐠𝟐 𝟐 −
𝝅𝟐
𝟏𝟔− 𝑳𝒊𝟐(−𝟏) − 𝑩 =
𝟏
𝟖𝐥𝐨𝐠𝟐 𝟐 −
𝝅𝟐
𝟑𝟐+𝝅𝟐
𝟐𝟒=𝟏
𝟖𝐥𝐨𝐠𝟐 𝟐 +
𝝅𝟐
𝟗𝟔
𝑨 + 𝑩 =𝟓
𝟖𝐥𝐨𝐠𝟐 𝟐 −
𝝅
𝟖𝐥𝐨𝐠(
𝟒𝝅𝟑
𝚪 (𝟏𝟒)) +
𝟏
𝟖𝐥𝐨𝐠𝟐 𝟐 +
𝝅𝟐
𝟗𝟔=
=𝝅𝟐
𝟗𝟔−𝟏
𝟐𝐥𝐨𝐠𝟐 𝟐 −
𝝅
𝟖𝐥𝐨𝐠(
𝟒𝝅𝟑
𝚪𝟒 (𝟏𝟒)) ⇒ 𝑹 =
𝟒𝝅𝟑
𝚪𝟒 (𝟏𝟒)
1565. If 𝟓 < 𝒂 ≤ 𝒃 then find:
𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑
𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏)𝒅𝒙
𝒃
𝒂
Proposed by Daniel Sitaru-Romania
www.ssmrmh.ro
95 RMM-CALCULUS MARATHON 1501-1600
Solution 1 by Amrit Awasthi-India
𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑
𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏)𝒅𝒙
𝒃
𝒂
=𝒙=𝐭𝐚𝐧 𝒕
= ∫ 𝐬𝐞𝐜𝟐 𝒕 ⋅ 𝐭𝐚𝐧−𝟏 (𝟒 𝐭𝐚𝐧 𝒕 − 𝟒 𝐭𝐚𝐧𝟑 𝒕
𝐭𝐚𝐧𝟒 𝒕 − 𝟔 𝐭𝐚𝐧𝟐 𝒕 + 𝟏)
𝐭𝐚𝐧−𝟏 𝒃
𝐭𝐚𝐧−𝟏 𝒂
𝒅𝒕 =
= ∫ 𝐬𝐞𝐜𝟐 𝒕 ⋅ 𝐭𝐚𝐧−𝟏(𝐭𝐚𝐧𝟒𝒕) 𝒅𝒕𝐭𝐚𝐧−𝟏 𝒃
𝐭𝐚𝐧−𝟏 𝒂
= 𝟒∫ 𝒕 ⋅ 𝐬𝐞𝐜𝟐 𝒕𝐭𝐚𝐧−𝟏 𝒃
𝐭𝐚𝐧−𝟏 𝒂
𝒅𝒕 =
= [𝟒𝒕 ⋅ 𝐭𝐚𝐧 𝒕 + 𝟒 𝐥𝐨𝐠|𝐜𝐨𝐬 𝒕|]𝐭𝐚𝐧−𝟏 𝒂𝐭𝐚𝐧−𝟏 𝒃 − 𝟐𝝅(𝒃 − 𝒂) =
= 𝟒𝒃 ⋅ 𝐭𝐚𝐧−𝟏 𝒃 − 𝟒𝒂 ⋅ 𝐭𝐚𝐧−𝟏 𝒂 + 𝟒 𝐥𝐨𝐠𝐜𝐨𝐬(𝐭𝐚𝐧−𝟏 𝒃)
𝐜𝐨𝐬(𝐭𝐚𝐧−𝟏 𝒂)− 𝟐𝝅(𝒃 − 𝒂) =
= 𝟒 [𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏
𝟐𝐥𝐨𝐠 (
𝒂𝟐 + 𝟏
𝒃𝟐 + 𝟏)] − 𝟐𝝅(𝒃 − 𝒂)
Solution 2 by Asmat Qatea-Afghanistan
∵ 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝒙 = 𝝅 + 𝐭𝐚𝐧−𝟏 (𝟐𝒙
𝟏 − 𝒙𝟐) , 𝒙 ∈ (𝟏,∞)
𝟒 𝐭𝐚𝐧−𝟏 𝒙 = 𝟐𝝅 + 𝐭𝐚𝐧−𝟏 (𝟐𝒙
𝟏 − 𝒙𝟐) + 𝐭𝐚𝐧−𝟏 (
𝟐𝒙
𝟏 − 𝒙𝟐)
𝟒 𝐭𝐚𝐧−𝟏 𝒙 = 𝟐𝝅 + 𝐭𝐚𝐧−𝟏 (
𝟒𝒙𝟏− 𝒙𝟐
𝟏 −𝟒𝒙𝟐
(𝟏 − 𝒙𝟐)𝟐
)
𝟒 𝐭𝐚𝐧−𝟏 𝒙 = 𝟐𝝅 + 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑
𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏) , ∀𝒙 ∈ (𝟏,∞)
∫ 𝟒 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝒃
𝒂
= ∫ 𝟐𝝅𝒅𝒙𝒃
𝒂
+𝛀
𝛀 = 𝟒 [𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝟏
𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝟏)]
𝒂
𝒃
− 𝟐𝝅(𝒃 − 𝒂)
Therefore,
𝛀(𝒂, 𝒃) = 𝟒 [𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏
𝟐𝐥𝐨𝐠 (
𝒂𝟐 + 𝟏
𝒃𝟐 + 𝟏)] − 𝟐𝝅(𝒃 − 𝒂)
www.ssmrmh.ro
96 RMM-CALCULUS MARATHON 1501-1600
Solution 3 by Ajetunmobi Abdulquoyyum-Nigeria
𝛀(𝒂, 𝒃) = ∫ 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑
𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏)𝒅𝒙
𝒃
𝒂
− 𝟐𝝅∫ 𝒅𝒙𝒃
𝒂
𝑨 = ∫ 𝐭𝐚𝐧−𝟏 (𝟒𝒙 − 𝟒𝒙𝟑
𝒙𝟒 − 𝟔𝒙𝟐 + 𝟏)𝒅𝒙 = ∫
𝟒𝒙(𝟏 − 𝒙𝟐)
(𝒙𝟐 − 𝟐𝒙 − 𝟏)(𝒙𝟐 + 𝟐𝒙 − 𝟏)𝒅𝒙 =
= ∫𝐭𝐚𝐧−𝟏(𝟏 − 𝒙𝟐) (𝟒𝒙
(𝒙𝟐 − 𝟐𝒙 − 𝟏)(𝒙𝟐 + 𝟐𝒙 − 𝟏))𝒅𝒙 =
= ∫𝐭𝐚𝐧−𝟏(𝟏 − 𝒙𝟐) (𝟏
𝒙𝟐 + 𝟐𝒙 − 𝟏−
𝟏
𝒙𝟐 + 𝟐𝒙 − 𝟏)𝒅𝒙 =
= ∫𝐭𝐚𝐧−𝟏 (𝟏 − 𝒙𝟐
𝟏 − 𝟐𝒙 − 𝒙𝟐−
𝟏 − 𝒙𝟐
𝟏 + 𝟐𝒙 − 𝒙𝟐)𝒅𝒙 =
= ∫𝐭𝐚𝐧−𝟏(𝟏
𝟏 −𝟐𝒙
𝟏 − 𝒙𝟐
−𝟏
𝟏 +𝟐𝒙
𝟏 − 𝒙𝟐
)𝒅𝒙 =𝐭𝐚𝐧 𝒕=𝒙
= ∫𝐬𝐞𝐜𝟐 𝒕 𝐭𝐚𝐧−𝟏 (𝟏
𝟏 −𝟐 𝐭𝐚𝐧 𝒕𝟏 − 𝐭𝐚𝐧𝟐 𝒕
−𝟏
𝟏 +𝟐 𝐭𝐚𝐧 𝒕𝟏 − 𝐭𝐚𝐧𝟐 𝒕
)𝒅𝒕 =
= ∫𝐬𝐞𝐜𝟐 𝒕 𝐭𝐚𝐧−𝟏 (𝟏
𝟏 − 𝐭𝐚𝐧 𝟐𝒕−
𝟏
𝟏 + 𝐭𝐚𝐧𝟐𝒕)𝒅𝒕 =
= ∫𝐬𝐞𝐜𝟐 𝒕 𝐭𝐚𝐧−𝟏𝟏 + 𝐭𝐚𝐧 𝟐𝒕 − 𝟏 + 𝐭𝐚𝐧𝟐𝒕
(𝟏 − 𝐭𝐚𝐧 𝟐𝒕)(𝟏 + 𝐭𝐚𝐧𝟐𝒕)𝒅𝒕 =
= ∫𝐬𝐞𝐜𝟐 𝒕 𝐭𝐚𝐧−𝟏(𝐭𝐚𝐧𝟒𝒕) 𝒅𝒕 = 𝟒∫𝒕 ⋅ 𝐬𝐞𝐜𝟐 𝒕 𝒅𝒕 =𝑰𝑩𝑷
= 𝟒(𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝟏
𝟐𝐥𝐨𝐠(𝟏 + 𝒙𝟐)) + 𝑪
𝛀(𝒂, 𝒃) = 𝟒 [𝒃 𝐭𝐚𝐧−𝟏 𝒃 − 𝒂 𝐭𝐚𝐧−𝟏 𝒂 +𝟏
𝟐𝐥𝐨𝐠 (
𝒂𝟐 + 𝟏
𝒃𝟐 + 𝟏)] − 𝟐𝝅(𝒃 − 𝒂)
1566. If 𝟎 < 𝒂 ≤ 𝒃 <𝝅
𝟐 then find:
𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬 𝟒𝒙
𝟏 − 𝐜𝐨𝐬 𝟒𝒙
𝒃
𝒂
𝒅𝒙
Proposed by Daniel Sitaru-Romania
www.ssmrmh.ro
97 RMM-CALCULUS MARATHON 1501-1600
Solution 1 by Amrit Awasthi-India
𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬𝟒𝒙
𝟏 − 𝐜𝐨𝐬𝟒𝒙
𝒃
𝒂
𝒅𝒙 = 𝟒∫𝒅𝒙
𝟐𝐬𝐢𝐧𝟐 𝟐𝒙
𝒃
𝒂
−∫ 𝒅𝒙𝒃
𝒂
=
= 𝟐∫ 𝐜𝐬𝐜𝟐 𝟐𝒙𝒃
𝒂
𝒅𝒙 − (𝒃 − 𝒂) = −𝐜𝐨𝐭𝟐𝒙|𝒂𝒃 − (𝒃 − 𝒂) = 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭 𝟐𝒃 − (𝒃 − 𝒂)
Solution 2 by Adrian Popa-Romania
𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬 𝟒𝒙
𝟏 − 𝐜𝐨𝐬 𝟒𝒙
𝒃
𝒂
𝒅𝒙 = −∫−𝟑 − 𝐜𝐨𝐬 𝟒𝒙
𝟏 − 𝐜𝐨𝐬𝟒𝒙
𝒃
𝒂
𝒅𝒙 = −∫𝟏 − 𝐜𝐨𝐬 𝟒𝒙 − 𝟒
𝟏 − 𝐜𝐨𝐬 𝟒𝒙𝒅𝒙
𝒃
𝒂
=
= −∫ 𝒅𝒙𝒃
𝒂
+ ∫𝟒
𝟏 − 𝐜𝐨𝐬 𝟒𝒙𝒅𝒙
𝒃
𝒂
= (𝒂 − 𝒃) +∫𝟐
𝐬𝐢𝐧𝟐 𝟐𝒙𝒅𝒙
𝒃
𝒂
=
= 𝒂 − 𝒃 − 𝐜𝐨𝐭𝟐𝒙|𝒂𝒃 = 𝒂 − 𝒃 + 𝐜𝐨𝐭𝟐𝒂 − 𝐜𝐨𝐭 𝟐𝒃
Solution 3 by Mohammad Hamed Nasery-Afghanistan
𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬𝟒𝒙
𝟏 − 𝐜𝐨𝐬𝟒𝒙
𝒃
𝒂
𝒅𝒙 = −∫−𝟑 − 𝐜𝐨𝐬 𝟒𝒙
𝟏 − 𝐜𝐨𝐬𝟒𝒙𝒅𝒙
𝒃
𝒂
=
= −∫𝟏 − 𝐜𝐨𝐬𝟒𝒙 − 𝟑 − 𝟏
𝟏 − 𝐜𝐨𝐬𝟒𝒙𝒅𝒙
𝒃
𝒂
= −∫𝟏 − 𝐜𝐨𝐬𝟒𝒙 − 𝟒
𝟏 − 𝐜𝐨𝐬𝟒𝒙𝒅𝒙
𝒃
𝒂
=
= 𝟒∫𝟏
𝟏 − 𝐜𝐨𝐬 𝟒𝒙𝒅𝒙
𝒃
𝒂
−∫ 𝒅𝒙𝒃
𝒂
= 𝟐∫𝟏
𝐬𝐢𝐧𝟐 𝒙
𝒃
𝒂
𝒅𝒙 − (𝒃 − 𝒂) =
= 𝒂 − 𝒃 + 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭 𝟐𝒃
Solution 4 by Hussain Reza Zadah-Afghanistan
𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬 𝟒𝒙
𝟏 − 𝐜𝐨𝐬 𝟒𝒙
𝒃
𝒂
𝒅𝒙 =𝒕=𝟒𝒙 𝟏
𝟒∫
𝟑 + 𝐜𝐨𝐬 𝒕
𝟏 − 𝐜𝐨𝐬 𝒕
𝟒𝒃
𝟒𝒂
𝒅𝒕 =𝐭𝐚𝐧
𝒕𝟐=𝒖
=𝟏
𝟒∫
𝟑 +𝟏 − 𝒖𝟐
𝟏 + 𝒖𝟐
𝟏 −𝟏 − 𝒖𝟐
𝟏 + 𝒖𝟐
⋅𝟐𝒅𝒖
𝟏 + 𝒖𝟐
𝐭𝐚𝐧 𝟐𝒃
𝐭𝐚𝐧 𝟐𝒂
=
=𝟏
𝟐∫
𝒖𝟐 + 𝟐
𝒖𝟐(𝟏 + 𝒖𝟐)
𝐭𝐚𝐧 𝟐𝒃
𝐭𝐚𝐧 𝟐𝒂
𝒅𝒖 =𝟏
𝟐∫
𝟐
𝒖𝟐
𝐭𝐚𝐧 𝟐𝒃
𝐭𝐚𝐧 𝟐𝒂
𝒅𝒖 − ∫𝟏
𝒖𝟐 + 𝟏𝒅𝒖
𝐭𝐚𝐧 𝟐𝒃
𝐭𝐚𝐧 𝟐𝒂
=
= [−𝟐
𝒖−𝟏
𝟐𝐭𝐚𝐧−𝟏 𝒖]
𝐭𝐚𝐧 𝟐𝒂
𝐭𝐚𝐧 𝟐𝒃
= 𝒂 − 𝒃 + 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭 𝟐𝒃
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98 RMM-CALCULUS MARATHON 1501-1600
Solution 5 by Ajetunmobi Abdulqoyyum-Nigeria
𝛀 = ∫𝟑 + 𝐜𝐨𝐬𝟒𝒙
𝟏 − 𝐜𝐨𝐬𝟒𝒙𝒅𝒙 =
𝟏
𝟒∫𝟑 + 𝐜𝐨𝐬𝒙
𝟏 − 𝐜𝐨𝐬 𝒙𝒅𝒙 =
𝐭𝐚𝐧𝒙𝟐=𝒕
=𝟏
𝟒∫𝟑 +
𝟏 − 𝒕𝟐
𝟏 + 𝒕𝟐
𝟏 −𝟏 − 𝒕𝟐
𝟏 + 𝒕𝟐
⋅𝟐𝒅𝒕
𝟏 + 𝒕𝟐=𝟏
𝟐∫
𝟐𝒕𝟐 + 𝟒
𝟐𝒕𝟐(𝟏 + 𝒕𝟐)𝒅𝒕 =
𝟏
𝟐∫
𝒕𝟐 + 𝟐
𝒕𝟐(𝟏 + 𝒕𝟐)𝒅𝒕 =
=𝟏
𝟐∫𝟏
𝒕𝟐𝒅𝒕 +∫
𝟏
𝒕𝟐(𝒕𝟐 + 𝟏)𝒅𝒕 = ∫
𝟏
𝒕𝟐𝒅𝒕 − ∫
𝟏
𝒕𝟐 + 𝟏𝒅𝒕 =
= −𝟏
𝒕−𝟏
𝟐𝐭𝐚𝐧−𝟏 𝒕 = −
𝟏
𝐭𝐚𝐧 𝟐𝒙−𝟏
𝟐𝐭𝐚𝐧−𝟏(𝐭𝐚𝐧𝟐𝒙) = −𝐜𝐨𝐭𝟐𝒙 − 𝒙
Therefore,
𝛀(𝒂, 𝒃) = 𝒂 − 𝒃 + 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭𝟐𝒃
Solution 6 by Satyam Roy-India
By generalization:
𝛀 = ∫𝒎+ 𝐜𝐨𝐬 𝟒𝒙
𝒏 − 𝐜𝐨𝐬 𝟒𝒙𝒅𝒙 = ∫
𝒑
𝒏 − 𝐜𝐨𝐬𝟒𝒙
𝒃
𝒂
𝒅𝒙 + 𝒂 − 𝒃
If 𝒎,𝒏 ∈ ℕ,𝒑 = 𝒎 + 𝒏. Here 𝒎 = 𝟑,𝒏 = 𝟏,𝒑 = 𝟒
∫𝟒
𝟏 + 𝐜𝐨𝐬𝟒𝒙
𝒃
𝒂
𝒅𝒙 = ∫𝟐
𝐬𝐢𝐧𝟐 𝟐𝒙
𝒃
𝒂
𝒅𝒙 = 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭𝟐𝒃
Therefore,
𝛀(𝒂, 𝒃) = 𝒂 − 𝒃 + 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭𝟐𝒃
Solution 7 by Sujit Bhowmick-India
𝛀(𝒂, 𝒃) = ∫𝟑 + 𝐜𝐨𝐬𝟒𝒙
𝟏 − 𝐜𝐨𝐬𝟒𝒙
𝒃
𝒂
𝒅𝒙 = ∫𝟑(𝟏 + 𝐭𝐚𝐧𝟐 𝟐𝒙) + 𝟏 − 𝐭𝐚𝐧𝟐 𝟐𝒙
𝟏 + 𝐭𝐚𝐧𝟐 𝟐𝒙 − (𝟏 − 𝐭𝐚𝐧𝟐 𝟐𝒙)𝒅𝒙
𝒃
𝒂
=
= ∫𝟏 + 𝟐 𝐭𝐚𝐧𝟐 𝟐𝒙
𝟐 𝐭𝐚𝐧𝟐 𝟐𝒙
𝒃
𝒂
𝒅𝒙 = ∫𝟏 + 𝐬𝐞𝐜𝟐 𝟐𝒙
𝐭𝐚𝐧𝟐 𝟐𝒙𝒅𝒙
𝒃
𝒂
= 𝟐∫ 𝐜𝐬𝐜𝟐 𝟐𝒙𝒅𝒙𝒃
𝒂
−∫ 𝒅𝒙𝒃
𝒂
=
= 𝒂 − 𝒃 + 𝐜𝐨𝐭 𝟐𝒂 − 𝐜𝐨𝐭 𝟐𝒃
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99 RMM-CALCULUS MARATHON 1501-1600
1567. If 𝟎 < 𝒂 ≤ 𝒃 find a closed form:
𝛀(𝒂, 𝒃) = ∫
(
𝒙
𝟏 +𝒙𝟐
𝟑 +𝒙𝟐
𝟓 +𝒙𝟐
𝟕 +⋯)
𝒃
𝒂
𝒅𝒙
Proposed by Daniel Sitaru-Romania
Solution by Naren Bhandari-Bajura-Nepal
Due to Lambert continued fraction (particular case of Gauss continued fraction)
𝐭𝐚𝐧𝒙 =𝒙
𝟏 + 𝕂𝒏=𝟏∞ −𝒙𝟐
𝟐𝒏 + 𝟏
Now we replace 𝒙 by 𝒊𝒙 giving us
𝐭𝐚𝐧𝐡 𝒙 =𝒙
𝟏 +𝕂𝒏=𝟏∞ 𝒙𝟐
𝟐𝒏 + 𝟏
So, we need to integrate 𝑰 + ∫ 𝐭𝐚𝐧𝐡 𝒙𝒅𝒙𝒃
𝒂 which is easy to see
𝑰 = ∫𝒅
𝒅𝒙𝐥𝐨𝐠(𝐜𝐨𝐬𝐡𝒙)
𝒃
𝒂
𝒅𝒙 = 𝐥𝐨𝐠 (𝐜𝐨𝐬𝐡𝒃
𝐜𝐨𝐬𝐡𝒂)
1568. Prove that:
∫ 𝐬𝐢𝐧 (𝒙
𝟐) 𝐭𝐚𝐧𝐡−𝟏(𝐬𝐢𝐧 𝟐𝒙)
𝝅𝟐
𝟎
𝒅𝒙 =
= 𝐥𝐨𝐠((𝟐√𝟐 − √𝟐 + 𝟐√𝟐 − 𝟏)
√𝟐+√𝟐
(𝟏 + 𝟐√𝟐 − 𝟐√𝟐 − √𝟐)
√𝟐−√𝟐
)
Proposed by Naren Bhandari-Bajura-Nepal
Solution by proposer
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100 RMM-CALCULUS MARATHON 1501-1600
𝑰 = ∫ 𝐬𝐢𝐧 (𝒙
𝟐) 𝐭𝐚𝐧𝐡−𝟏(𝐬𝐢𝐧𝟐𝒙)
𝝅𝟐
𝟎
𝒅𝒙 =𝑰𝑩𝑷𝟒∫
𝐜𝐨𝐬 (𝒙𝟐)
𝐜𝐨𝐬𝟐𝒙
𝝅𝟐
𝟎
𝒅𝒙 = 𝟒∫𝐜𝐨𝐬 (
𝒙𝟐)
𝟏 − 𝟐𝐬𝐢𝐧𝟐 𝒙
𝝅𝟐
𝟎
𝒅𝒙
By the use of compound angle formula and subbing 𝐬𝐢𝐧 (𝒙
𝟐) = 𝒖 leads us to
∫𝟖𝒅𝒖
𝟏 − 𝟖𝒖𝟐(𝟏 − 𝒖𝟐)
𝟏
√𝟐
𝟎
= ∫𝟖𝒅𝒖
𝟖𝒖𝟒 − 𝟖𝒖𝟐 + 𝟏
𝟏
√𝟐
𝟎
= 𝐥𝐢𝐦𝒂→
𝟏
√𝟐
∫𝟖𝒅𝒖
𝑷(𝒖)
𝒂
𝟎
Note that the polynomial 𝑷(𝒖) is reducible over ℝ or
𝟐𝑷(𝒖) = (𝟒𝒖𝟐 − √𝟐 − 𝟐)(𝟒𝒖𝟐 + √𝟐 − 𝟐), with positive factors (√𝟐 + √𝟐⏟ 𝒓𝟏
, √𝟐 − √𝟐⏟ 𝒓𝟐
).
Therefore,
𝑰 = 𝟒√𝟐∫ (𝒅𝒖
𝟒𝒖𝟐 − √𝟐− 𝟐−
𝒅𝒖
𝟒𝒖𝟐 + √𝟐)
𝟏
√𝟐
𝟎
= −𝒓𝟐 𝐥𝐨𝐠 (𝒓𝟐 + 𝟏
𝟏 − 𝒓𝟐) + 𝒓𝟏 𝐥𝐨𝐠 (
𝒓𝟏 + 𝟏
𝒓𝟏 − 𝟏)
By putting the roots we have the result however, to get the desired result as presented in final form we observe that:
(𝒓𝟏 + 𝟏
𝒓𝟏 − 𝟏,𝒓𝟐 + 𝟏
𝟏 − 𝒓𝟐) =𝒓𝒂𝒕𝒊𝒐𝒏𝒂𝒍𝒊𝒛𝒂𝒕𝒊𝒐𝒏
(𝟐√𝟐− 𝟏 + 𝟐√𝟐 + √𝟐,𝟏
𝟏 + 𝟐√𝟐 − 𝟐√𝟐 − √𝟐)
Hence,
∫ 𝐬𝐢𝐧 (𝒙
𝟐) 𝐭𝐚𝐧𝐡−𝟏(𝐬𝐢𝐧𝟐𝒙)
𝝅𝟐
𝟎
𝒅𝒙 =
= 𝐥𝐨𝐠((𝟐√𝟐− √𝟐 + 𝟐√𝟐− 𝟏)
√𝟐+√𝟐
(𝟏 + 𝟐√𝟐− 𝟐√𝟐− √𝟐)
√𝟐−√𝟐
)
1569. Prove that:
𝑰𝟐(𝒌) = ∫𝒙 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙
√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙
𝝅𝟐
𝟎
𝒅𝒙 = −𝝅
𝟐⋅𝒌′
𝒌𝟐+𝑬(𝒌)
𝒌𝟐
Proposed by Onikoyi Adeboye-Nigeria
Solution 1 by Kamel Gandouli Rezgui-Tunisia
𝑰𝟐(𝒌) = ∫𝒙 𝐬𝐢𝐧𝒙 𝐜𝐨𝐬 𝒙
√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙
𝝅𝟐
𝟎
𝒅𝒙 =
𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙=𝒖
𝒙=𝐬𝐢𝐧−𝟏√𝒖
𝒌𝟐 𝟏
𝟐𝒌𝟐∫ 𝐬𝐢𝐧−𝟏√
𝒖
𝒌𝟐𝒅𝒖
√𝟏 − 𝒌𝟐
𝒌𝟐
𝟎
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101 RMM-CALCULUS MARATHON 1501-1600
∫ 𝐬𝐢𝐧−𝟏√𝒖
𝒌𝟐𝒅𝒖
√𝟏 − 𝒌𝟐
𝒌𝟐
𝟎
= 𝑱(𝒌) ⇒
𝑱′(𝒌) = (∫ 𝐬𝐢𝐧−𝟏√𝒖
𝒌𝟐𝒅𝒖
√𝟏 − 𝒌𝟐
𝒌𝟐
𝟎
)
′
=𝐬𝐢𝐧−𝟏 𝟏
√𝟏 − 𝒌𝟐⋅ 𝟐𝒌 =
𝒌𝝅
√𝟏 − 𝒌𝟐
⇒ 𝑱(𝒌) = −𝝅√𝟏 − 𝒌𝟐 + 𝟐𝑬(𝒌) ⇒
𝑰𝟐(𝒌) =𝟏
𝟐𝒌𝟐(−𝝅√𝟏 − 𝒌𝟐 + 𝟐𝑬(𝒌)) = −
𝝅
𝟐⋅𝒌′
𝒌𝟐+𝑬(𝒌)
𝒌𝟐⇒ 𝒌′ = √𝟏 − 𝒌𝟐
For 𝒌 = 𝟏 ⇒ 𝑬(𝒙) = ∫ 𝒙 𝐬𝐢𝐧𝒙𝒅𝒙𝝅
𝟐𝟎
= 𝟏
Solution 2 by Sediqakbar Restheen-Afghanistan
𝑰𝟐(𝒌) = ∫𝒙𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙
√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙
𝝅𝟐
𝟎
𝒅𝒙 = −𝟏
𝒌𝟐∫−𝟐𝒌𝟐 𝐬𝐢𝐧 𝒙 𝐜𝐨𝐬 𝒙
𝟐√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙
𝝅𝟐
𝟎
𝒅𝒙 =
= −𝟏
𝒌𝟐[𝒙√𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙]
𝟎
𝝅𝟐+𝟏
𝒌𝟐∫ √𝟏 − 𝒌𝟐 𝐬𝐢𝐧𝟐 𝒙𝒅𝒙
𝝅𝟐
𝟎
= −𝝅√𝟏 − 𝒌𝟐
𝟐𝒌𝟐+𝑬(𝒌)
𝒌𝟐
Let 𝒌′ = √𝟏 − 𝒌𝟐 ⇒ 𝑰𝟐(𝒌) = −𝝅
𝟐⋅𝒌′
𝒌𝟐+𝑬(𝒌)
𝒌𝟐
1570. Find:
𝛀 = ∫ ∫ ∫𝒙𝒚𝒛
(𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙)
𝟏
𝟎
𝒅𝒙𝒅𝒚𝒅𝒛𝟏
𝟎
𝟏
𝟎
Proposed by Asmat Qatea-Afghanistan
Solution by proposer
∵ (𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙) = ∑(𝒙𝟐𝒚 + 𝒙𝟐𝒛)
𝒄𝒚𝒄
+ 𝟐𝒙𝒚𝒛
𝑰 = ∫ ∫ ∫𝒙𝟐𝒚 + 𝒙𝟐𝒛 +
𝟐𝟑𝒙𝒚𝒛
(𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙)
𝟏
𝟎
𝒅𝒙𝒅𝒚𝒅𝒛𝟏
𝟎
𝟏
𝟎
=𝟏
𝟑
∫ ∫ ∫𝒙𝟐
(𝒙 + 𝒚)(𝒙 + 𝒛)
𝟏
𝟎
𝒅𝒙𝒅𝒚𝒅𝒛𝟏
𝟎
𝟏
𝟎
+𝟐
𝟑∫ ∫ ∫
𝒙𝒚𝒛
(𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙)
𝟏
𝟎
𝒅𝒙𝒅𝒚𝒅𝒛𝟏
𝟎
𝟏
𝟎
=𝟏
𝟑
∫ ∫𝒙𝟐
𝒙 + 𝒚𝐥𝐨𝐠 (
𝒙 + 𝟏
𝒙)𝒅𝒙𝒅𝒚
𝟏
𝟎
𝟏
𝟎
+𝟐
𝟑𝛀 =
𝟏
𝟑
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102 RMM-CALCULUS MARATHON 1501-1600
∫ 𝒙𝟐 𝐥𝐨𝐠𝟐 (𝒙 + 𝟏
𝒙)𝒅𝒙
𝟏
𝟎
+𝟐
𝟑𝛀 =
𝟏
𝟑
∫ 𝒙𝟐 𝐥𝐨𝐠𝟐(𝒙 + 𝟏)𝒅𝒙𝟏
𝟎
+∫ 𝒙𝟐 𝐥𝐨𝐠𝟐 𝒙𝟏
𝟎
𝒅𝒙 − 𝟐∫ 𝒙𝟐 𝐥𝐨𝐠(𝒙 + 𝟏) 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
+𝟐
𝟑𝛀 =
𝟏
𝟑
𝑰𝟏 + 𝑰𝟐 − 𝟐𝑰𝟑 +𝟐
𝟑𝛀 =
𝟏
𝟑; (𝑰), where
𝑰𝟏 = ∫ 𝒙𝟐 𝐥𝐨𝐠𝟐(𝒙 + 𝟏)𝒅𝒙𝟏
𝟎
=𝒙+𝟏=𝒆𝒚
∫ (𝟐𝟐𝒚 − 𝟐𝒆𝒚 + 𝟏)𝒚𝟐𝒆𝒚𝐥𝐨𝐠 𝟐
𝟎
𝒅𝒚 =
= ∫ (𝒆𝟑𝒚 − 𝟐𝒆𝟐𝒚 + 𝒆𝒚)𝒚𝟐𝐥𝐨𝐠 𝟐
𝟎
𝒅𝒚 =𝟐
𝟑𝐥𝐨𝐠𝟐 𝟐 −
𝟏𝟔
𝟗𝐥𝐨𝐠 𝟐 +
𝟓𝟓
𝟓𝟒
𝑰𝟐 = ∫ 𝒙𝟐 𝐥𝐨𝐠𝟐 𝒙𝟏
𝟎
𝒅𝒙 =𝒙=𝒆𝒚
∫ 𝒆𝟑𝒚𝒚𝟐𝒅𝒚𝟎
−∞
=𝟐
𝟐𝟕
𝑰𝟑 = ∫ 𝒙𝟐 𝐥𝐨𝐠(𝒙 + 𝟏) 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
= −∫ 𝒙𝟐 𝐥𝐨𝐠 𝒙∑(−𝟏)𝒌𝒙𝒌
𝒌
∞
𝒌=𝟏
𝒅𝒙𝟏
𝟎
=
= −∑(−𝟏)𝒌
𝒌∫ 𝒙𝒌+𝟐 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
∞
𝒌=𝟏
=∑(−𝟏)𝒌
𝒌(𝒌 + 𝟑)𝟐
∞
𝒌=𝟏
= ∑(−𝟏)𝒌 (𝑨𝟏𝒌+
𝑨𝟐𝒌 + 𝟑
+𝑨𝟑
(𝒌 + 𝟑)𝟐)
∞
𝒌=𝟏
𝑨𝟏 =𝟏
𝟗, 𝑨𝟐 = −
𝟏
𝟑, 𝑨𝟑 = −
𝟏
𝟗
Hence,
𝑰𝟑 =𝟏
𝟗(− 𝐥𝐨𝐠𝟐 − (𝟏 −
𝟏
𝟐+𝟏
𝟑+∑
(−𝟏)𝒌−𝟏
𝒌
∞
𝒌=𝟒
) −𝟏
𝟑(𝟏
𝟏𝟐−𝟏
𝟐𝟐+𝟏
𝟑𝟐+∑
(−𝟏)𝒌
(𝒌 + 𝟑)𝟐
∞
𝒌=𝟏
) +𝟒𝟏
𝟏𝟎𝟖
= −𝟐
𝟗𝐥𝐨𝐠 𝟐 −
𝝅𝟐
𝟑𝟔+𝟒𝟏
𝟏𝟎𝟖
From (𝑰): 𝑰𝟏 + 𝑰𝟐 − 𝟐𝑰𝟑 +𝟐
𝟑𝛀 =
𝟏
𝟑 it follows:
𝟐
𝟑𝐥𝐨𝐠𝟐 𝟐 −
𝟏𝟔
𝟗𝐥𝐨𝐠𝟐 +
𝟓𝟓
𝟓𝟒+𝟐
𝟐𝟕− 𝟐 (−
𝟐
𝟗𝐥𝐨𝐠 𝟐 −
𝝅𝟐
𝟑𝟔+𝟒𝟏
𝟏𝟎𝟖) +
𝟐
𝟑𝛀 =
𝟏
𝟑
𝟐
𝟑𝐥𝐨𝐠𝟐 𝟐 −
𝟏𝟐
𝟗𝐥𝐨𝐠𝟐 +
𝝅𝟐
𝟏𝟖+𝟐
𝟑𝛀 = 𝟎 ⇔
𝟐 𝐥𝐨𝐠𝟐 𝟐 −𝟏𝟐
𝟑𝐥𝐨𝐠 𝟐 +
𝝅𝟐
𝟔+ 𝟐𝛀 = 𝟎 ⇔
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103 RMM-CALCULUS MARATHON 1501-1600
𝛀 = 𝟐 𝐥𝐨𝐠𝟐 − 𝐥𝐨𝐠𝟐 𝟐 −𝝅𝟐
𝟏𝟐
Therefore,
𝛀 = ∫ ∫ ∫𝒙𝒚𝒛
(𝒙 + 𝒚)(𝒚 + 𝒛)(𝒛 + 𝒙)
𝟏
𝟎
𝒅𝒙𝒅𝒚𝒅𝒛𝟏
𝟎
𝟏
𝟎
= 𝟐 𝐥𝐨𝐠 𝟐 − 𝐥𝐨𝐠𝟐 𝟐 −𝝅𝟐
𝟏𝟐
1571. Find a closed form:
𝛀 = ∫𝒙 𝐭𝐚𝐧−𝟏 𝒙
𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙
∞
𝟎
Proposed by Vasile Mircea Popa-Romania
Solution 1 by Asmat Qatea-Afghanistan
𝛀 = ∫𝒙 𝐭𝐚𝐧−𝟏 𝒙
𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙
∞
𝟎
=𝒙=𝟏𝒙= ∫
𝒙 𝐭𝐚𝐧−𝟏 (𝟏𝒙)
𝟏 − 𝒙𝟐 + 𝒙𝟒
∞
𝟎
𝒅𝒙
∵ 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 (𝟏
𝒙) =
𝝅
𝟐
𝟐𝛀 =𝝅
𝟐∫
𝒙
𝟏 − 𝒙𝟐 + 𝒙𝟒
∞
𝟎
𝒅𝒙 =𝒙𝟐=𝒙 𝝅
𝟒∫
𝟏
𝟏 − 𝒙 + 𝒙𝟐
∞
𝟎
𝒅𝒙
𝛀 =𝝅
𝟖∫
𝟏
(𝒙 −𝟏𝟐)𝟐
+ (√𝟑𝟐)
𝟐
∞
𝟎
𝒅𝒙 =𝝅
𝟒√𝟑[𝐭𝐚𝐧−𝟏(
𝒙 −𝟏𝟐
√𝟑𝟐
)]
𝟎
∞
=𝝅
𝟒√𝟑(𝝅
𝟐+𝝅
𝟔)
𝛀 = ∫𝒙 𝐭𝐚𝐧−𝟏 𝒙
𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙
∞
𝟎
=𝝅𝟐
𝟔√𝟑
Solution 2 by Sesiqakbar Restheen-Afghanistan
∵ 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 (𝟏
𝒙) =
𝝅
𝟐,∀𝒙 ∈ ℝ
𝛀 = ∫𝒙 𝐭𝐚𝐧−𝟏 𝒙
𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙
∞
𝟎
=𝒖=𝟏𝒙∫
𝟏𝒖 𝐭𝐚𝐧
−𝟏 (𝟏𝒖)
𝟏𝒖𝟒−𝟏𝒖𝟐+ 𝟏
∞
𝟎
(−𝒅𝒖
𝒖𝟐) =
= ∫𝒖(𝝅𝟐 − 𝐭𝐚𝐧
−𝟏 𝒖)
𝒖𝟒 − 𝒖𝟐 + 𝟏𝒅𝒖
∞
𝟎
=𝝅
𝟒∫
𝒖
𝒖𝟒 − 𝒖𝟐 + 𝟏
∞
𝟎
𝒅𝒖
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104 RMM-CALCULUS MARATHON 1501-1600
∫𝒖
𝒖𝟒 − 𝒖𝟐 + 𝟏𝒅𝒖 =
𝟏
√𝟑𝐭𝐚𝐧−𝟏 [
𝟏
√𝟑(𝟐𝒙𝟐 − 𝟏)] + 𝑪
∫𝒖
𝒖𝟒 − 𝒖𝟐 + 𝟏
∞
𝟎
𝒅𝒖 =𝟏
√𝟑[𝝅
𝟐− 𝐭𝐚𝐧−𝟏 (−
𝟏
√𝟑)] =
𝟏
√𝟑[𝝅
𝟐—𝝅
𝟔) =
𝟐𝝅
𝟑√𝟑
Therefore,
𝛀 = ∫𝒙 𝐭𝐚𝐧−𝟏 𝒙
𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙
∞
𝟎
=𝝅𝟐
𝟔√𝟑
Solution 3 by Ghuiam Naseri-Afghanistan
∵ 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 (𝟏
𝒙) =
𝝅
𝟐,∀𝒙 ∈ ℝ
𝛀 = ∫𝒙 𝐭𝐚𝐧−𝟏 𝒙
𝒙𝟒 − 𝒙𝟐 + 𝟏𝒅𝒙
∞
𝟎
=𝒙=𝟏𝒙∫
𝒙𝟐 𝐭𝐚𝐧−𝟏 (𝟏𝒙)
𝒙𝟒 − 𝒙𝟐 + 𝟏
∞
𝟎
𝒅𝒙
𝝅
𝟐∫
𝒙
𝒙𝟒 − 𝒙𝟐 + 𝟏
∞
𝟎
𝒅𝒙 =𝒙𝟐=𝒙 𝝅
𝟒∫
𝟏
𝒙𝟐 − 𝒙 + 𝟏𝒅𝒙
∞
𝟎
=
=𝝅
𝟒⋅𝟐
√𝟑𝐭𝐚𝐧−𝟏 (
𝟐𝒙 − 𝟏
√𝟑) =𝒙=𝒙𝟐 𝝅
𝟒⋅𝟐
√𝟑[𝐭𝐚𝐧−𝟏 (
𝟐𝒙𝟐 − 𝟏
√𝟑)]𝟎
∞
=𝝅𝟐
𝟔√𝟑
1572.
𝛀(𝒏) = ∫𝒙𝒏−𝟏(𝒙 − 𝒏) 𝐥𝐨𝐠 𝒙
𝒆𝒙𝒅𝒙
∞
𝟎
, 𝒏 ≥ 𝟏
Find:
𝛀 =∑𝟏
𝛀(𝒏)
∞
𝒏=𝟏
Proposed by Daniel Sitaru-Romania
Solution 1 by Ajetunmobi Abdulqoyyum-Nigeria
𝛀(𝒏) = ∫ 𝒆−𝒙𝒙𝒏 𝐥𝐨𝐠 𝒙∞
𝟎
𝒅𝒙 − 𝒏∫ 𝒙𝒏−𝟏𝒆−𝒙 𝐥𝐨𝐠 𝒙∞
𝟎
𝒅𝒙 =
=𝝏
𝝏𝒔|𝒔=𝟎(∫ 𝒆−𝒙𝒙(𝒏+𝒔+𝟏)−𝟏
∞
𝟎
𝒅𝒙 − 𝒏∫ 𝒙𝒏+𝒔−𝟏𝒆−𝒙∞
𝟎
𝒅𝒙) =
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105 RMM-CALCULUS MARATHON 1501-1600
=𝝏
𝝏𝒔|𝒔=𝟎
(𝚪(𝒏 + 𝒔 + 𝟏) − 𝒏𝚪(𝒏 + 𝒔)) = 𝚪′(𝒏 + 𝟏) − 𝒏𝚪′(𝒏) =
= ∑𝟏
𝚪(𝒏 + 𝟏)𝝍(𝒏+ 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏)
∞
𝒏=𝟏
= ∑𝟏
𝚪(𝒏 + 𝟏)𝝍(𝒏+ 𝟏) − 𝚪(𝒏 + 𝟏)𝝍(𝒏)
∞
𝒏=𝟏
=
= ∑𝟏
𝚪(𝒏 + 𝟏)(𝝍(𝒏 + 𝟏) − 𝝍(𝒏))
∞
𝒏=𝟏
= ∑𝟏
𝚪(𝒏 + 𝟏) (𝟏𝒏)
∞
𝒏=𝟏
= ∑𝒏
𝚪(𝒏 + 𝟏)
∞
𝒏=𝟏
= ∑𝟏
𝚪(𝒏)
∞
𝒏=𝟏
= 𝒆
Therefore,
𝛀 = ∑𝟏
𝛀(𝒏)
∞
𝒏=𝟏
= 𝒆
Solution 2 by Amrit Awasthi-India
𝝍(𝒛) =𝟏
𝚪(𝒛)∫ 𝒙𝒛−𝟏𝒆−𝒙 𝐥𝐨𝐠 𝒙𝒅𝒙∞
𝟎
𝛀(𝒏) = ∫ 𝒙𝒏+𝟏𝒆−𝒙 𝐥𝐨𝐠 𝒙 𝒅𝒙∞
𝟎
− 𝒏∫ 𝒙𝒏−𝟏𝒆−𝒙 𝐥𝐨𝐠𝒙𝒅𝒙∞
𝟎
=
= 𝚪(𝒏 + 𝟏)𝝍(𝒏 + 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏) = 𝒏! [𝝍(𝒏+ !) − 𝝍(𝒏)] =
= 𝒏! (𝟏
𝒏) = (𝒏 − 𝟏)!
Therefore,
𝛀 = ∑𝟏
𝛀(𝒏)
∞
𝒏=𝟏
= ∑𝟏
(𝒏 − 𝟏)!
∞
𝒏−𝟏
= 𝒆
Solution 3 by Santiago Alvarez-Mexico
𝛀(𝒏) = ∫𝒙𝒏−𝟏(𝒙 − 𝒏) 𝐥𝐨𝐠𝒙
𝒆𝒙𝒅𝒙
∞
𝟎
= (𝝏
𝝏𝒔∫ 𝒙𝒏−𝟏+𝒔(𝒙 − 𝒏)𝒆−𝒙𝒅𝒙∞
𝟎
)𝒔=𝟎
=
= (𝝏
𝝏𝒔∫ 𝒙𝒏+𝒔𝒆−𝒙𝒅𝒙∞
𝟎
− 𝒏𝝏
𝝏𝒔∫ 𝒙𝒏+𝒔−𝟏𝒆−𝒙𝒅𝒙∞
𝟎
)𝒔=𝟎
=
= (𝝏
𝝏𝒔𝚪(𝒏 + 𝒔 + 𝟏) − 𝒏
𝝏
𝝏𝒔𝚪(𝒏 + 𝒔))
𝒔=𝟎
=
= (𝚪(𝒏 + 𝒔 + 𝟏)𝝍(𝒏 + 𝒔 + 𝟏) − 𝒏𝚪(𝒏 + 𝒔)𝝍(𝒏 + 𝒔))𝒔=𝟎
=
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106 RMM-CALCULUS MARATHON 1501-1600
= 𝚪(𝒏 + 𝟏)𝝍(𝒏+ 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏)
𝛀 = ∑𝟏
𝚪(𝒏 + 𝟏)𝝍(𝒏 + 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏)
∞
𝒏=𝟏
= ∑𝟏
𝒏𝚪(𝒏)𝝍(𝒏 + 𝟏) − 𝒏𝚪(𝒏)𝝍(𝒏)
∞
𝒏=𝟏
=
= ∑𝟏
𝒏𝚪(𝒏) (𝟏𝒏)
∞
𝒏=𝟏
= ∑𝟏
(𝒏 − 𝟏)!
∞
𝒏=𝟏
= ∑𝟏
𝒏!
∞
𝒏=𝟎
= 𝒆
1573. Find:
𝛀 = ∫𝒙𝟒
𝒙𝟒 𝐥𝐨𝐠𝟒 𝟒 + 𝟒(𝒙𝟑 𝐥𝐨𝐠𝟑 𝟒 + 𝟑𝒙𝟐 𝐥𝐨𝐠𝟐 𝟒 + 𝟔𝒙 𝐥𝐨𝐠 𝟒 + 𝟔 + 𝟔 ⋅ 𝟒𝒙)𝒅𝒙
Proposed by Daniel Sitaru-Romania
Solution by Almas Babirov-Azerbaijan
𝛀 = ∫𝒙𝟒
𝒙𝟒 𝐥𝐨𝐠𝟒 𝟒 + 𝟒(𝒙𝟑 𝐥𝐨𝐠𝟑 𝟒 + 𝟑𝒙𝟐 𝐥𝐨𝐠𝟐 𝟒 + 𝟔𝒙 𝐥𝐨𝐠 𝟒 + 𝟔 + 𝟔 ⋅ 𝟒𝒙)𝒅𝒙 =
= ∫𝒙𝟒
𝒙𝟒 𝐥𝐨𝐠𝟒 𝟒 + 𝟒𝒙𝟑 𝐥𝐨𝐠𝟑 𝟒 + 𝟏𝟐𝒙𝟐 𝐥𝐨𝐠𝟐 𝟒 + 𝟐𝟒𝒙 𝐥𝐨𝐠 𝟒 + 𝟐𝟒 + 𝟐𝟒 ⋅ 𝟒𝒙𝒅𝒙 = (∗)
∵ 𝐥𝐨𝐠 𝟒𝒙 = 𝒕 ⇒ 𝟒𝒙 = 𝒆𝒕 ⇒ 𝒙 = 𝐥𝐨𝐠𝟒 𝒆𝒕 = 𝒕 𝐥𝐨𝐠𝟒 𝒆
(∗) = ∫𝒕𝟒 𝐥𝐨𝐠𝟒
𝟒 𝒆
𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕𝒅(𝒕 𝐥𝐨𝐠𝟒 𝒆) =
= 𝐥𝐨𝐠𝟒𝟓 𝒆 ⋅ ∫(
𝒕𝟒 − (𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕)
𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕+ 𝟏)𝒅𝒕 =
= 𝒕 𝐥𝐨𝐠𝟒𝟓 𝒆 − 𝐥𝐨𝐠𝟒
𝟓 𝒆 ⋅ ∫𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕
𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕𝒅𝒕 =
=𝒕
𝐥𝐨𝐠𝟓 𝟒−
𝟏
𝐥𝐨𝐠𝟓 𝟒⋅ ∫𝒅(𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕)
𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕=
=𝒕
𝐥𝐨𝐠𝟓 𝟒−
𝟏
𝐥𝐨𝐠𝟓 𝟒⋅ 𝐥𝐨𝐠(𝒕𝟒 + 𝟒𝒕𝟑 + 𝟏𝟐𝒕𝟐 + 𝟐𝟒𝒕 + 𝟐𝟒 + 𝟐𝟒𝒆𝒕) + 𝑪 =
=𝒙
𝐥𝐨𝐠𝟒 𝟒−
𝟏
𝐥𝐨𝐠𝟓 𝟒⋅ 𝐥𝐨𝐠(𝒙𝟒 𝐥𝐨𝐠𝟒 𝟒 + 𝟒𝒙𝟑 𝐥𝐨𝐠𝟑 𝟒 + 𝟏𝟐𝒙𝟐 𝐥𝐨𝐠𝟐 𝟒 + 𝟐𝟒𝒙 𝐥𝐨𝐠𝟒 + 𝟐𝟒 + 𝟐𝟒 ⋅ 𝟒𝒙) + 𝑪
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107 RMM-CALCULUS MARATHON 1501-1600
1574. If 𝟎 < 𝑎 < 4𝒃 then find:
𝛀 = ∫𝐬𝐢𝐧−𝟏(√𝒙)
𝒂𝒙𝟐 − 𝒂𝒙 + 𝒃
𝟏
𝟎
𝒅𝒙
Proposed by Marin Chirciu-Romania
Solution by Marian Ursărescu-Romania
𝛀 = ∫𝐬𝐢𝐧−𝟏(√𝒙)
𝒂𝒙𝟐 − 𝒂𝒙 + 𝒃
𝟏
𝟎
𝒅𝒙 =𝒙=𝟏−𝒕
∫𝐬𝐢𝐧−𝟏(√𝟏 − 𝒕)
𝒂(𝟏 − 𝒕)𝟐 − 𝒂(𝟏 − 𝒕) + 𝒃
𝟎
𝟏
(−𝒅𝒕) =
= ∫𝐬𝐢𝐧−𝟏(√𝟏 − 𝒕)
𝒂 − 𝟐𝒂𝒕 + 𝒂𝒕𝟐 − 𝒂 + 𝒂𝒕 + 𝒃𝒅𝒕
𝟏
𝟎
= ∫𝐬𝐢𝐧−𝟏(√𝟏 − 𝒕)
𝒂𝒕𝟐 − 𝒂𝒕 + 𝒃𝒅𝒕
𝟏
𝟎
𝚪 = ∫𝐬𝐢𝐧−𝟏(√𝟏 − 𝒕)
𝒂𝒕𝟐 − 𝒂𝒕 + 𝒃𝒅𝒕
𝟏
𝟎
⇒ 𝛀 = 𝚪.
But 𝐬𝐢𝐧−𝟏(√𝒙) + 𝐬𝐢𝐧−𝟏(√𝟏 − 𝒙) =𝝅
𝟐; ∀𝒙 ∈ [𝟎, 𝟏], because
𝒇(𝒙) = 𝐬𝐢𝐧−𝟏(√𝒙) + 𝐬𝐢𝐧−𝟏(√𝟏 − 𝒙) ; 𝒇′(𝒙) = 𝟎 ⇒ 𝒇(𝒙) = 𝒄 =𝝅
𝟐⇒
𝛀 + 𝚪 = ∫𝐬𝐢𝐧−𝟏(√𝒙) + 𝐬𝐢𝐧−𝟏(√𝟏 − 𝒙)
𝒂𝒙𝟐 − 𝒂𝒙 + 𝒃𝒅𝒙
𝟏
𝟎
=𝝅
𝟐∫
𝒅𝒙
𝒂𝒙𝟐 − 𝒂𝒙 + 𝒃
𝟏
𝟎
=
=𝝅
𝟐𝒂∫
𝒅𝒙
𝒙𝟐 − 𝒙 +𝒃𝒂
𝟏
𝟎
=𝝅
𝟐𝒂∫
𝒅𝒙
(𝒙 −𝟏𝟐)𝟐
+𝟒𝒃 − 𝒂𝟒𝒂
𝟏
𝟎
=𝝅
𝟐𝒂∫
𝒅𝒙
(𝒙 −𝟏𝟐)𝟐
+ (√𝟒𝒃− 𝒂𝟒𝒂 )
𝟐
𝟏
𝟎
𝝅
𝟐𝒂⋅ 𝟐√
𝒂
𝟒𝒃 − 𝒂⋅ 𝐭𝐚𝐧−𝟏
𝒙 −𝟏𝟐
√𝟒𝒃 − 𝒂𝒂
||
𝟎
𝟏
=𝝅
√𝒂(𝟒𝒃 − 𝒂)⋅ 𝟐 𝐭𝐚𝐧−𝟏√
𝒂
𝟒𝒃 − 𝒂; (𝟐)
From (1),(2) it follows that:
𝛀 =𝝅
√𝒂(𝟒𝒃 − 𝒂)⋅ 𝟐 𝐭𝐚𝐧−𝟏√
𝒂
𝟒𝒃 − 𝒂
1575.
𝒇 ∈ 𝑪𝟏([𝟎, 𝟑]), 𝒇(𝟎) = 𝟒, 𝒇(𝟑) = 𝒌, 𝒇′(𝒙) = 𝐭𝐚𝐧−𝟏 𝒙
Find:
𝛀(𝒌) = ∫ 𝒙𝟐𝒇(𝒙)𝟑
𝟎
𝒅𝒙
Proposed by Abdul Mukhtar-Nigeria
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108 RMM-CALCULUS MARATHON 1501-1600
Solution 1 by Adrian Popa-Romania
𝛀(𝒌) = ∫ 𝒙𝟐𝒇(𝒙)𝟑
𝟎
𝒅𝒙 =𝒙𝟑
𝟑𝒇(𝒙)|
𝟎
𝟑
−𝟏
𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑
𝟎
𝒅𝒙 = 𝟗𝒇(𝟑) −𝟏
𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑
𝟎
𝒅𝒙 =
= 𝟗𝒌 −𝟏
𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑
𝟎
𝒅𝒙
𝛀(𝒌) = 𝟗𝒌 −𝒙𝟒 𝐭𝐚𝐧−𝟏 𝒙
𝟏𝟐|𝟎
𝟑
+𝟏
𝟏𝟐∫
𝒙𝟒
𝟏 + 𝒙𝟐𝒅𝒙
𝟑
𝟎
=
= 𝟗𝒌 −𝟐𝟕
𝟒𝐭𝐚𝐧−𝟏 𝟑 +
𝟏
𝟏𝟐∫ (𝒙𝟐 − 𝟏 +
𝟏
𝒙𝟐 + 𝟏)𝒅𝒙
𝟑
𝟎
=
= 𝟗𝒌−𝟐𝟕
𝟒𝐭𝐚𝐧−𝟏 𝟑 +
𝟏
𝟏𝟐⋅𝒙𝟑
𝟑|𝟎
𝟑
−𝟏
𝟏𝟐⋅ 𝒙|
𝟎
𝟑
+𝟏
𝟏𝟐𝐭𝐚𝐧−𝟏 𝒙|
𝟎
𝟑
=
= 𝟗𝒌 −𝟐𝟕
𝟒𝐭𝐚𝐧−𝟏 𝟑 +
𝟑
𝟒−𝟏
𝟒+𝟏
𝟏𝟐𝐭𝐚𝐧−𝟏 𝟑 = 𝟗𝒌 −
𝟐𝟎
𝟑𝐭𝐚𝐧−𝟏 𝟑 +
𝟏
𝟐
Solution 2 by Fayssal Abdelli-Bejaia-Algerie
𝛀(𝒌) = ∫ 𝒙𝟐𝒇(𝒙)𝟑
𝟎
𝒅𝒙 =𝒙𝟑
𝟑𝒇(𝒙)|
𝟎
𝟑
−𝟏
𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑
𝟎
𝒅𝒙 =
𝟗𝒌𝟐 −𝟏
𝟑𝚪;
𝚪 = ∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝟑
𝟎
𝒅𝒙 =𝒙𝟒
𝟒𝐭𝐚𝐧−𝟏 𝒙|
𝟎
𝟑
−𝟏
𝟒∫
𝒙𝟒
𝟏 + 𝒙𝟐𝒅𝒙
𝟑
𝟎
=
=𝟖𝟏
𝟒𝐭𝐚𝐧−𝟏 𝒙 −
𝟏
𝟒∫(𝒙𝟐 − 𝟏)(𝒙𝟐 + 𝟏)
𝒙𝟐 + 𝟏𝒅𝒙
𝟑
𝟎
−∫𝒅𝒙
𝟏 + 𝒙𝟐
𝟑
𝟎
=
=𝟖𝟏
𝟒𝐭𝐚𝐧−𝟏 𝟑 −
𝟏
𝟒𝐭𝐚𝐧−𝟏 𝒙|
𝟎
𝟑
−𝟏
𝟒⋅ (𝒙𝟑
𝟑− 𝒙)|
𝟎
𝟑
= 𝟐𝟎 𝐭𝐚𝐧−𝟏 𝟑 −𝟑
𝟐
Therefore,
𝛀(𝒌) = 𝟗𝒌 −𝟐𝟎
𝟑𝐭𝐚𝐧−𝟏 𝟑 +
𝟏
𝟐
Solution 3 by Soumitra Mandal-Chandar Nagore-India
𝒇′(𝒙) = 𝐭𝐚𝐧−𝟏 𝒙 ⇒ ∫ 𝒇′(𝒙)𝟑
𝟎
𝒅𝒙 = ∫ 𝐭𝐚𝐧−𝟏 𝒙𝟑
𝟎
𝒅𝒙 ⇒
𝒇(𝟑) − 𝒇(𝟎) = 𝒙 𝐭𝐚𝐧−𝟏 𝒙|𝟎
𝟑−∫
𝒙
𝟏 + 𝒙𝟐𝒅𝒙
𝟑
𝟎
= 𝟑 𝐭𝐚𝐧−𝟏 𝟑 −𝟏
𝟐𝐥𝐨𝐠(𝟏 + 𝒙𝟐)|
𝟎
𝟑
=
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109 RMM-CALCULUS MARATHON 1501-1600
= 𝟑 𝐭𝐚𝐧−𝟏 𝟑 −𝟏
𝟐⇒ 𝒌 − 𝟒 = 𝟑 𝐭𝐚𝐧−𝟏 𝟑 −
𝟏
𝟐⇒ 𝒌 =
𝟕
𝟐+ 𝟑 𝐭𝐚𝐧−𝟏 𝟑
𝛀(𝒌) = ∫ 𝒙𝟐𝒇(𝒙)𝟑
𝟎
𝒅𝒙 = 𝟗𝒇(𝟑) −𝟏
𝟑∫ 𝒙𝟑 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝟑
𝟎
=
= 𝟗𝒇(𝟑) −𝟏
𝟏𝟐𝒙𝟒 𝐭𝐚𝐧−𝟏 𝒙|
𝟎
𝟑
+𝟏
𝟏𝟐∫
𝒙𝟒
𝟏 + 𝒙𝟐𝒅𝒙
𝟑
𝟎
=
= 𝟗𝒌 −𝟐𝟕
𝟒𝐭𝐚𝐧−𝟏 𝟑 +
𝟏
𝟏𝟐∫ 𝒙𝟐 (𝟏 −
𝟏
𝟏 + 𝒙𝟐)
𝟑
𝟎
𝒅𝒙 =
= 𝟗𝒌−𝟐𝟕
𝟒𝐭𝐚𝐧−𝟏 𝟑 +
𝟏
𝟑𝟔(𝟑𝟑 − 𝟎) −
𝟏
𝟏𝟐∫
𝒙𝟐
𝟏 + 𝒙𝟐𝒅𝒙
𝟑
𝟎
=
= 𝟗(𝟕
𝟐+ 𝟑 𝐭𝐚𝐧−𝟏 𝟑) −
𝟐𝟕
𝟒𝐭𝐚𝐧−𝟏 𝟑 +
𝟑
𝟒−𝟏
𝟏𝟐∫ (𝟏 −
𝟏
𝟏 + 𝒙𝟐)
𝟑
𝟎
𝒅𝒙 =
=𝟔𝟑
𝟐+𝟖𝟏
𝟒𝐭𝐚𝐧−𝟏 𝟑 +
𝟑
𝟒−𝟏
𝟒+𝟏
𝟏𝟐𝐭𝐚𝐧−𝟏 𝒙|
𝟎
𝟑
=
= 𝟑𝟐 +𝟏𝟐𝟏
𝟔𝐭𝐚𝐧−𝟏 𝟑
1576. If 𝟎 < 𝒂 ≤ 𝒃 <𝝅
𝟐 then:
∫𝐭𝐚𝐧 (
𝝅 − 𝟐𝒙𝟒
) (𝟏 + 𝐬𝐢𝐧 𝒙)
𝐬𝐢𝐧 𝒙𝒅𝒙
𝒃
𝒂
≤𝐬𝐢𝐧 𝒃 − 𝐬𝐢𝐧 𝒂
𝐬𝐢𝐧 𝒂
Proposed by Daniel Sitaru-Romania
Solution by Ravi Prakash-New Delhi-India
𝐭𝐚𝐧 (𝝅− 𝟐𝒙𝟒
)(𝟏 + 𝐬𝐢𝐧𝒙)
𝐬𝐢𝐧 𝒙=
𝟏 − 𝐭𝐚𝐧𝒙𝟐
𝟏 + 𝐭𝐚𝐧𝒙𝟐
(𝐜𝐨𝐬𝒙𝟐 + 𝐬𝐢𝐧
𝒙𝟐)𝟐
𝐬𝐢𝐧𝒙=
=
𝐜𝐨𝐬𝒙𝟐 − 𝐬𝐢𝐧
𝒙𝟐
𝐜𝐨𝐬𝒙𝟐 + 𝐬𝐢𝐧
𝒙𝟐
(𝐜𝐨𝐬𝒙𝟐 + 𝐬𝐢𝐧
𝒙𝟐)𝟐
𝐬𝐢𝐧 𝒙=(𝐜𝐨𝐬
𝒙𝟐 − 𝐬𝐢𝐧
𝒙𝟐) (𝐜𝐨𝐬
𝒙𝟐 + 𝐬𝐢𝐧
𝒙𝟐)
𝐬𝐢𝐧 𝒙=
=𝐜𝐨𝐬𝟐
𝒙𝟐 − 𝐬𝐢𝐧
𝟐 𝒙𝟐
𝐬𝐢𝐧 𝒙=𝐜𝐨𝐬 𝒙
𝐬𝐢𝐧𝒙
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Hence,
∫𝐭𝐚𝐧 (
𝝅− 𝟐𝒙𝟒 )(𝟏 + 𝐬𝐢𝐧𝒙)
𝐬𝐢𝐧 𝒙𝒅𝒙
𝒃
𝒂
= 𝐥𝐨𝐠(𝐬𝐢𝐧 𝒙)|𝒂𝒃 = 𝐥𝐨𝐠 (
𝐬𝐢𝐧𝒃
𝐬𝐢𝐧𝒂) = 𝐥𝐨𝐠 (𝟏 −
𝐬𝐢𝐧𝒃 − 𝐬𝐢𝐧𝒂
𝐬𝐢𝐧𝒂)
≤
≤𝐬𝐢𝐧 𝒃 − 𝐬𝐢𝐧 𝒂
𝐬𝐢𝐧𝒂; (∵ 𝐥𝐨𝐠(𝟏 + 𝒙) ≤ 𝒙, ∀𝒙 ≥ 𝟎)
1577.
𝛀(𝒂) = ∫ 𝐥𝐨𝐠(𝟏 + 𝒙) ⋅ 𝐭𝐚𝐧−𝟏(√𝒙)𝒅𝒙𝒂
𝟎
, 𝒂 > 𝟎
Prove that:
𝛀(𝒂) + 𝛀(𝒃) + 𝛀(𝒄) < (𝒂 + 𝒃 + 𝒄) (𝒂 + 𝒃 + 𝒄 +𝟏
𝟐) , ∀𝒂, 𝒃, 𝒄 > 𝟎
Proposed by Florică Anastase-Romania
Solution 1 by Ruxandra Daniela Tonilă-Romania
We have: 𝒆𝒙 ≥ 𝒙 + 𝟏,∀𝒙 > 𝟎 ⇔ 𝐥𝐨𝐠(𝟏 + 𝒙) ≤ 𝒙, ∀𝒙 > 𝟎 and 𝐭𝐚𝐧−𝟏(√𝒙) ≤𝝅
𝟐, ∀𝒙 > 𝟎.
Therefore, 𝛀(𝒂) ≤𝝅
𝟐∫ 𝒙𝒃
𝒂𝒅𝒙 ⇔ 𝛀(𝒂) ≤
𝝅
𝟒𝒂𝟐. Hence,
𝛀(𝒂) + 𝛀(𝒃) + 𝛀(𝒄) ≤𝝅
𝟒(𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐) < 𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐 <
< (𝒂+ 𝒃 + 𝒄)𝟐 < (𝒂 + 𝒃 + 𝒄)𝟐 +𝟏
𝟐(𝒂 + 𝒃 + 𝒄) = (𝒂 + 𝒃 + 𝒄) (𝒂 + 𝒃 + 𝒄 +
𝟏
𝟐)
Solution 2 by Adrian Popa-Romania
First, we prove that: 𝐥𝐨𝐠(𝟏 + 𝒙) 𝐭𝐚𝐧−𝟏(√𝒙) < 𝟐𝒙 +𝟏
𝟐.
If 𝒇(𝒙) = 𝐥𝐨𝐠(𝟏 + 𝒙) 𝐭𝐚𝐧−𝟏(√𝒙) − 𝟐𝒙 −𝟏
𝟐, then 𝒇′(𝒙) =
𝐭𝐚𝐧−𝟏(√𝒙)
𝟏+𝒙+𝐥𝐨𝐠(𝟏+𝒙)
𝟐(𝟏+𝒙)√𝒙− 𝟐
𝐭𝐚𝐧−𝟏(√𝒙) < √𝒙 < 𝟐√𝒙 < 𝟏 + 𝒙 ⇒𝐭𝐚𝐧−𝟏(√𝒙)
𝟏 + 𝒙< 𝟏; (𝟏)
𝐥𝐨𝐠(𝟏 + 𝒙) < 𝒙 <(∗)
𝟐(𝟏 + 𝒙)√𝒙 ⇔ √𝒙 < 𝟐(𝟏 + 𝒙)
√𝒙 ≤𝒙+𝟏
𝟐< 𝟐(𝒙 + 𝟏) ⇔ (∗) true. Thus,
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111 RMM-CALCULUS MARATHON 1501-1600
𝐥𝐨𝐠(𝟏 + 𝒙)
𝟐(𝟏 + 𝒙)√𝒙< 𝟏; (𝟐)
From (1),(2) we have 𝒇′(𝒙) < 𝟎 ⇒ 𝒇(𝒙) < 𝒇(𝟎) = −𝟏
𝟐< 𝟎.
Now, integrating the inequality: 𝐥𝐨𝐠(𝟏 + 𝒙) 𝐭𝐚𝐧−𝟏(√𝒙) < 𝟐𝒙 +𝟏
𝟐, we get:
𝛀(𝒂) < ∫ (𝟐𝒙 +𝟏
𝟐)𝒅𝒙
𝒂
𝟎
= 𝒂𝟐 +𝟏
𝟐𝒂
Therefore,
𝛀(𝒂) + 𝛀(𝒃) + 𝛀(𝒄) < 𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐 +𝟏
𝟐(𝒂 + 𝒃 + 𝒄) <
< (𝒂 + 𝒃 + 𝒄)𝟐 +𝟏
𝟐(𝒂 + 𝒃 + 𝒄) = (𝒂 + 𝒃 + 𝒄) (𝒂 + 𝒃 + 𝒄 +
𝟏
𝟐)
Solution 3 by proposer
𝛀(𝒂) = ∫ 𝐥𝐨𝐠(𝟏 + 𝒙) ⋅𝒂
𝟎
𝐭𝐚𝐧−𝟏(√𝒙)𝒅𝒙 = ∫𝐥𝐨𝐠(𝟏 + 𝒙)
√𝒙
𝒂
𝟎
⋅ √𝒙 𝐭𝐚𝐧−𝟏(√𝒙)𝒅𝒙
Let be the function 𝒇: ℝ → ℝ, 𝒇(𝒙) = 𝐭𝐚𝐧−𝟏 𝒙 −𝟏
𝒙𝟐+𝟏⋅ 𝐭𝐚𝐧−𝟏 𝒙
𝒇′(𝒙) =𝟐𝒙⋅𝐭𝐚𝐧−𝟏 𝒙+𝒙𝟐
(𝒙𝟐+𝟏)𝟐 ≥ 𝟎 ⇒ 𝒇 −increasing on [𝟎,∞) ⇒ 𝒇(𝒙) ≥ 𝒇(𝟎) = 𝟎
𝟎 <𝒙𝟐
𝒙𝟐 + 𝟏⋅ 𝐭𝐚𝐧−𝟏 𝒙 < 𝟏 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 < 𝒙;∀𝒙 ≥ 𝟎 ⇒
𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙
𝒙𝟐 + 𝟏< 𝟏 ⟺
𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 < 𝒙𝟐 + 𝟏 ⟺ √𝒙 ⋅ 𝐭𝐚𝐧−𝟏 √𝒙 < 𝒙 + 𝟏;∀𝒙 > 𝟎; (𝟏)
It is well-known: 𝐥𝐨𝐠(𝟏 + 𝒙) ≤𝒙
√𝟏+𝒙; ∀𝒙 ≥ 𝟎 ⇒
𝐥𝐨𝐠(𝟏 + 𝒙)
√𝒙≤
√𝒙
√𝒙 + 𝟏;∀𝒙 ≥ 𝟎; (𝟐)
From (1),(2) it follows that:
𝛀(𝒂) = ∫𝐥𝐨𝐠(𝟏 + 𝒙)
√𝒙
𝒂
𝟎
⋅ √𝒙 𝐭𝐚𝐧−𝟏(√𝒙)𝒅𝒙 < ∫√𝒙
√𝒙 + 𝟏⋅ (𝒙 + 𝟏)
𝒂
𝟎
𝒅𝒙 =
= ∫ √𝒙(𝒙 + 𝟏)𝒂
𝟎
𝒅𝒙 ≤𝑨𝑴−𝑮𝑴
∫ (𝒙 +𝟏
𝟐)𝒅𝒙
𝒂
𝟎
=𝟏
𝟐(𝒂𝟐 + 𝒂)
Therefore,
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112 RMM-CALCULUS MARATHON 1501-1600
𝛀(𝒂) + 𝛀(𝒃) + 𝛀(𝒄) <𝟏
𝟐(𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐) +
𝟏
𝟐(𝒂 + 𝒃 + 𝒄) < (𝒂 + 𝒃 + 𝒄)𝟐 +
𝟏
𝟐(𝒂 + 𝒃 + 𝒄)
= (𝒂 + 𝒃 + 𝒄) (𝒂 + 𝒃 + 𝒄 +𝟏
𝟐)
1578. If 𝟎 < 𝒂 ≤ 𝒃 then:
∫𝟏
√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (
𝒌𝝅
𝟐𝒏 + 𝟏)
[𝒙]
𝒌=𝟏
𝒅𝒙𝒃
𝒂
≥𝟏
𝟐𝒂−𝟏
𝟐𝒃, [∗] − 𝑮𝑰𝑭.
Proposed by Daniel Sitaru-Romania
Solution by Asmat Qatea-Afghanistan
∫𝟏
√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (
𝒌𝝅
𝟐𝒏 + 𝟏)
[𝒙]
𝒌=𝟏
𝒅𝒙𝒃
𝒂
≥𝟏
𝟐𝒂−𝟏
𝟐𝒃⇔
∫𝟏
√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (
𝒌𝝅
𝟐𝒏 + 𝟏)
[𝒙]
𝒌=𝟏
𝒅𝒙𝒃
𝒂
≥(?)
𝐥𝐨𝐠𝟐∫ 𝟐−𝒙𝒃
𝒂
𝒅𝒙
𝟏
√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (
𝒌𝝅
𝟐𝒏 + 𝟏)
[𝒙]
𝒌=𝟏
≥(?)
𝟐−𝒙 𝐥𝐨𝐠 𝟐 ⇒ 𝒙 ∈ [𝟎,∞)
Case 1. If 𝒙 ∈ [𝟎, 𝟏) then:
𝟏
√𝟐 ⋅ 𝟎 + 𝟏⋅∏𝐬𝐢𝐧 (
𝒌𝝅
𝟐𝒏 + 𝟏)
𝟎
𝒌=𝟏
≥(?)
𝟐−𝒙 𝐥𝐨𝐠 𝟐 ⇒ 𝟏 ≥ 𝟐−𝒙 𝐥𝐨𝐠𝟐 ⇒ 𝟐𝒙 ≥ 𝐥𝐨𝐠 𝟐 − 𝐭𝐫𝐮𝐞.
Case 2. If 𝒙 ∈ [𝒏, 𝒏 + 𝟏), 𝒏 ∈ ℕ then:
𝟏
√𝟐 ⋅ 𝒏 + 𝟏⋅∏𝐬𝐢𝐧 (
𝒌𝝅
𝟐𝒏 + 𝟏)
𝒏
𝒌=𝟏
≥(?)
𝟐−𝒙 𝐥𝐨𝐠 𝟐
∵∏𝐬𝐢𝐧 (𝒌𝝅
𝟐𝒏 + 𝟏)
𝒏
𝒌=𝟏
=√𝟐𝒏 + 𝟏
𝟐𝒏
𝟏
𝟐𝒏≥(?)
𝟐−𝒙 𝐥𝐨𝐠𝟐 ⇒ 𝟐𝒙−𝒏 ≥ 𝐥𝐨𝐠𝟐 − 𝐭𝐫𝐮𝐞, 𝐛𝐞𝐜𝐚𝐮𝐬𝐞 𝒙 − 𝒏 ≥ 𝟎
Therefore,
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113 RMM-CALCULUS MARATHON 1501-1600
∫𝟏
√𝟐[𝒙] + 𝟏⋅∏𝐬𝐢𝐧 (
𝒌𝝅
𝟐𝒏 + 𝟏)
[𝒙]
𝒌=𝟏
𝒅𝒙𝒃
𝒂
≥𝟏
𝟐𝒂−𝟏
𝟐𝒃
1579.
𝒙𝒏 = (𝒏
𝟎)𝒑𝒏 + (
𝒏
𝟑)𝒑𝒏−𝟑 + (
𝒏
𝟔)𝒑𝒏−𝟔 +⋯ ;𝒚𝒏 = (
𝒏
𝟏)𝒑𝒏−𝟏 + (
𝒏
𝟒)𝒑𝒏−𝟒 + (
𝒏
𝟕)𝒑𝒏−𝟕
𝒛𝒏 = (𝒏
𝟐)𝒑𝒏−𝟐 + (
𝒏
𝟓)𝒑𝒏−𝟓 + (
𝒏
𝟖)𝒑𝒏−𝟖 +⋯ ;𝒏 ∈ ℕ, 𝒑 ≥ 𝟐. 𝐅𝐢𝐧𝐝:
𝛀(𝒑) = 𝐥𝐢𝐦𝒏→∞
√𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏𝒏
Proposed by Marian Ursărescu-Romania
Solution 1 by Adrian Popa-Romania
(𝒏
𝒌) = (
𝒏
𝒏 − 𝒌) ⇒ 𝒙𝒏 = (
𝒏
𝒏)𝒑𝒏 + (
𝒏
𝒏 − 𝟑)𝒑𝒏−𝟑 + (
𝒏
𝒏 − 𝟔)𝒑𝒏−𝟔 +⋯
𝒚𝒏 = (𝒏
𝒏 − 𝟏)𝒑𝒏−𝟏 + (
𝒏
𝒏 − 𝟒)𝒑𝒏−𝟒 + (
𝒏
𝒏 − 𝟕)𝒑𝒏−𝟕 +⋯
𝒛𝒏 = (𝒏
𝒏 − 𝟐)𝒑𝒏−𝟐 + (
𝒏
𝒏 − 𝟓)𝒑𝒏−𝟓 + (
𝒏
𝒏 − 𝟖)𝒑𝒏−𝟖 +⋯
Let 𝜺 be root by three order of unity, hence 𝜺𝟐 + 𝜺+ 𝟏 = 𝟎 and 𝜺𝟑 = 𝟏.
(𝟏 + 𝒑)𝒏 = (𝒏
𝟎) + (
𝒏
𝟏)𝒑 + (
𝒏
𝟐)𝒑𝟐 + (
𝒏
𝟑)𝒑𝟑 +⋯ ; (𝟏)
(𝟏 + 𝜺𝒑)𝒏 = (𝒏
𝟎) + (
𝒏
𝟏) 𝜺𝒑 + (
𝒏
𝟐) (𝜺𝒑)𝟐 + (
𝒏
𝟑) (𝜺𝒑)𝟑 +⋯ ; (𝟐)
(𝟏 + 𝜺𝟐𝒑)𝒏 = (𝒏
𝟎) + (
𝒏
𝟏) (𝜺𝟐𝒑) + (
𝒏
𝟐) (𝜺𝟐𝒑)𝟐 + (
𝒏
𝟑) (𝜺𝟐𝒑)𝟑 +⋯ ; (𝟑)
Adding (1),(2),(3) it follows that:
(𝟏 + 𝒑)𝒏 + (𝟏 + 𝜺𝒑)𝒏 + (𝟏 + 𝜺𝟐𝒑)𝒏 =(𝟏 + 𝒑)𝒏 + (𝟏 + 𝜺𝒑)𝒏 + (𝟏 + 𝜺𝟐𝒑)𝒏
𝟑= 𝒙𝒏
(𝟏 + 𝒑)𝒏 + 𝜺𝟐(𝟏 + 𝜺𝒑)𝒏 + 𝜺(𝟏 + 𝜺𝟐𝒑)𝒏 =(𝟏 + 𝒑)𝒏 + 𝜺𝟐(𝟏 + 𝜺𝒑)𝒏 + 𝜺(𝟏 + 𝜺𝟐𝒑)𝒏
𝟑= 𝒚𝒏
(𝟏 + 𝒑)𝒏 + 𝜺(𝟏 + 𝜺𝒑)𝒏 + 𝜺𝟐(𝟏 + 𝜺𝟐𝒑)𝒏 =(𝟏 + 𝒑)𝒏 + 𝜺(𝟏 + 𝜺𝒑)𝒏 + 𝜺𝟐(𝟏 + 𝜺𝟐𝒑)𝒏
𝟑= 𝒛𝒏
Let us denote: 𝒂 = (𝟏 + 𝒑)𝒏, 𝒃 = (𝟏 + 𝜺𝒑)𝒏, 𝒄 = (𝟏 + 𝜺𝟐𝒑)𝒏
𝒙𝒏𝒚𝒏 =(𝒂 + 𝒃 + 𝒄)(𝒂 + 𝜺𝟐𝒃 + 𝜺𝒄)
𝟗
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𝒙𝒏𝒛𝒏 =(𝒂+ 𝒃 + 𝒄)(𝒂 + 𝜺𝒃 + 𝜺𝟐𝒄)
𝟗
𝒚𝒏𝒛𝒏 =(𝒂 + 𝜺𝟐𝒃 + 𝜺𝒄)(𝒂 + 𝜺𝒃 + 𝜺𝟐𝒄)
𝟗
𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏 =𝟑𝒂𝟐 + 𝟑𝒃𝒄(𝜺 + 𝜺𝟐)
𝟗=𝒂𝟐 − 𝒃𝒄
𝟑
=(𝟏 + 𝒑)𝟐𝒏 − (𝟏 + 𝜺𝒑)𝒏(𝟏 + 𝜺𝟐𝒑)𝒏
𝟑
=(𝟏 + 𝒑)𝟐𝒏 − (𝟏 + 𝜺𝟐𝒑 + 𝜺𝒑 + 𝒑𝟐)𝒏
𝟑=(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏
𝟑
𝛀(𝒑) = 𝐥𝐢𝐦𝒏→∞
√𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏𝒏 = 𝐥𝐢𝐦
𝒏→∞√(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏
𝟑
𝒏
=𝑪−𝑫
= 𝐥𝐢𝐦𝒏→∞
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏= 𝐥𝐢𝐦𝒏→∞
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 [𝟏 −(𝒑𝟐 − 𝒑 + 𝟏)𝒏
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏]
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 [𝟏 −(𝒑𝟐 − 𝒑 + 𝟏)𝒏
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏]=
= (𝒑 + 𝟏)𝟐 𝐥𝐢𝐦𝒏→∞
𝟏 − 𝒂𝒏+𝟏
𝟏 − 𝒂𝒏= (𝒑 + 𝟏)𝟐; (𝒂 =
𝒑𝟐 − 𝒑 + 𝟏
𝒑𝟐 + 𝟐𝒑 + 𝟏< 1 ⇒ 𝒂𝒏 → 𝟎)
Solution 2 by Ravi Prakash-New Delhi-India
(𝒏
𝒌) = (
𝒏
𝒏 − 𝒌) ⇒ 𝒙𝒏 = (
𝒏
𝒏)𝒑𝒏 + (
𝒏
𝒏 − 𝟑)𝒑𝒏−𝟑 + (
𝒏
𝒏 − 𝟔)𝒑𝒏−𝟔 +⋯
𝒚𝒏 = (𝒏
𝒏 − 𝟏)𝒑𝒏−𝟏 + (
𝒏
𝒏 − 𝟒)𝒑𝒏−𝟒 + (
𝒏
𝒏 − 𝟕)𝒑𝒏−𝟕 +⋯
𝒛𝒏 = (𝒏
𝒏 − 𝟐)𝒑𝒏−𝟐 + (
𝒏
𝒏 − 𝟓)𝒑𝒏−𝟓 + (
𝒏
𝒏 − 𝟖)𝒑𝒏−𝟖 +⋯
Let 𝜺 be root by three order of unity, hence 𝜺𝟐 + 𝜺+ 𝟏 = 𝟎 and 𝜺𝟑 = 𝟏.
𝒙𝒏 + 𝒚𝒏 + 𝒛𝒏 = (𝒑 + 𝟏)𝒏
𝒙𝒏 + 𝜺𝒚𝒏 + 𝜺𝟐𝒛𝒏 = (𝒑 + 𝜺)
𝒏
𝒙𝒏 + 𝜺𝟐𝒚𝒏 + 𝜺𝒛𝒏 = (𝒑+ 𝜺
𝟐)𝒏
𝒙𝒏 =𝟏
𝟑[(𝒑 + 𝟏)𝒏 + (𝒑 + 𝜺)𝒏 + (𝒑 + 𝜺𝟐)𝒏]
𝒚𝒏 =𝟏
𝟑[(𝒑 + 𝟏)𝒏 + 𝜺𝟐(𝒑 + 𝜺)𝒏 + 𝜺(𝒑 + 𝜺𝟐)𝒏]
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𝒛𝒏 =𝟏
𝟑[(𝒑 + 𝟏)𝒏 + 𝜺𝟐(𝒑 + 𝜺)𝒏 + 𝜺(𝒑 + 𝜺𝟐)𝒏]
Let 𝒂 = (𝒑 + 𝟏)𝒏, 𝒃 = (𝒑 + 𝜺)𝒏, 𝒄 = (𝒑 + 𝜺𝟐)𝒏. Hence,
(𝒂 + 𝒃 + 𝒄)(𝒂 + 𝒃𝜺 + 𝒄𝜺𝟐) + (𝒂 + 𝒃 + 𝒄)(𝒂 + 𝒃𝜺𝟐 + 𝒄𝜺)
+ (𝒂 + 𝒃𝜺 + 𝒄𝜺𝟐)(𝒂 + 𝒃𝜺𝟐 + 𝒄𝜺)
= (𝒂 + 𝒃 + 𝒄)(𝟐𝒂 − 𝒃 − 𝒄) + 𝒂𝟐 + 𝒂𝒃𝜺 + 𝒂𝒄𝜺𝟐 + 𝒂𝒃𝜺𝟐 + 𝒃𝟐 + 𝒃𝒄𝜺 + +𝒂𝒄𝜺 + 𝒄𝟐 + 𝒃𝒄𝜺𝟐
= 𝟐𝒂𝟐 − 𝒂𝒃 − 𝒂𝒄 − 𝒃𝟐 − 𝒃𝒄 + 𝟐𝒂𝒃 + 𝟐𝒂𝒄 − 𝒃𝒄 − 𝒄𝟐 + 𝒂𝟐 − 𝒂𝒃 − 𝒂𝒄 + 𝒃𝟐 + 𝒄𝟐 − 𝒃𝒄
= 𝟑𝒂𝟐 − 𝟑𝒃𝒄 = 𝟑(𝒑 + 𝟏)𝒏 − 𝟑(𝒑𝟐 + 𝒑𝜺 + 𝒑𝜺𝟐 + 𝟏)𝒏 =
= 𝟑[(𝒑 + 𝟏)𝟐𝒏 − (𝒑𝟐 − 𝒑 + 𝟏)𝒏]. Thus,
𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏 =𝟏
𝟑[(𝒑 + 𝟏)𝒏 − (𝒑𝟏 − 𝒑 + 𝟏)𝒏]
Therefore,
𝛀(𝒑) = 𝐥𝐢𝐦𝒏→∞
√𝒙𝒏𝒚𝒏 + 𝒚𝒏𝒛𝒏 + 𝒛𝒏𝒙𝒏𝒏 =
𝟏
𝟑𝐥𝐢𝐦𝒏→∞
[(𝒑 + 𝟏)𝒏 − (𝒑𝟏 − 𝒑 + 𝟏)𝒏] =
𝐥𝐢𝐦𝒏→∞
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 [𝟏 −(𝒑𝟐 − 𝒑 + 𝟏)𝒏
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏]
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏 [𝟏 −(𝒑𝟐 − 𝒑 + 𝟏)𝒏
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏]= (𝒑 + 𝟏)𝟐;
(𝐰𝐡𝐞𝐫𝐞 𝒑𝟐 − 𝒑 + 𝟏
𝒑𝟐 + 𝟐𝒑 + 𝟏< 1 ⇒
(𝒑𝟐 − 𝒑 + 𝟏)𝒏
(𝒑𝟐 + 𝟐𝒑 + 𝟏)𝒏→ 𝟎)
1580. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
√∑𝒌𝟐 (𝒏
𝒌 − 𝟏) (𝒏
𝒌)
𝒏
𝒌=𝟏
𝒏
Proposed by Marian Ursărescu-Romania
Solution 1 by Ravi Prakash-New Delhi-India
𝒌𝟐 (𝒏
𝒌 − 𝟏) (𝒏
𝒌) = 𝒌𝟐 (
𝒏
𝒌)(𝒏
𝒌) = 𝒏𝟐 (
𝒏 − 𝟏
𝒌 − 𝟏)𝟐
⇒
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116 RMM-CALCULUS MARATHON 1501-1600
∑𝒌𝟐 (𝒏
𝒌 − 𝟏) (𝒏
𝒌)
𝒏
𝒌=𝟏
= 𝒏𝟐∑(𝒏− 𝟏
𝒌− 𝟏)𝟐𝒏
𝒌=𝟏
= 𝒏𝟐∑(𝒏− 𝟏
𝒌)𝟐𝒏−𝟏
𝒌=𝟎
= 𝒏𝟐 (𝟐𝒏 − 𝟐
𝒏 − 𝟏)
= 𝒏𝟐𝒂𝒏; (𝒂𝒏 = (𝟐𝒏 − 𝟐
𝒏 − 𝟏))
𝛀 = 𝐥𝐢𝐦𝒏→∞
√∑𝒌𝟐 (𝒏
𝒌 − 𝟏) (𝒏
𝒌)
𝒏
𝒌=𝟏
𝒏
= 𝐥𝐢𝐦𝒏→∞
√𝒏𝟐𝒂𝒏𝒏
=𝑪−𝑫
𝐥𝐢𝐦𝒏→∞
(𝟐𝒏)!
𝒏!𝒏!⋅(𝒏 − 𝟏)! (𝒏 − 𝟏)!
(𝟐𝒏 − 𝟐)!=
= 𝐥𝐢𝐦𝒏→∞
𝟐𝒏(𝟐𝒏 − 𝟏)
𝒏𝟐= 𝟒
Solution 2 by Adrian Popa-Romania
𝒌
𝒏(𝒏
𝒌) = (
𝒏 − 𝟏
𝒌 − 𝟏) ⇒ 𝒌𝟐 (
𝒏
𝒌 − 𝟏) (𝒏
𝒌) = 𝒌 (
𝒏
𝒌 − 𝟏) ⋅ 𝒏(
𝒏 − 𝟏
𝒌 − 𝟏) =
= (𝒌 − 𝟏 + 𝟏) (𝒏
𝒌 − 𝟏) ⋅ 𝒏 (
𝒏 − 𝟏
𝒌 − 𝟏) = (𝒌 − 𝟏) (
𝒏
𝒌 − 𝟏) ⋅ 𝒏 (
𝒏 − 𝟏
𝒌 − 𝟏) + 𝒏(
𝒏
𝒌 − 𝟏) ⋅ (
𝒏 − 𝟏
𝒌 − 𝟏)
=
= 𝒏(𝒏 − 𝟏
𝒌 − 𝟐) ⋅ 𝒏(
𝒏 − 𝟏
𝒌 − 𝟏) + 𝒏 (
𝒏
𝒌 − 𝟏) ⋅ (
𝒏 − 𝟏
𝒌 − 𝟏)
∑𝒌𝟐 (𝒏
𝒌 − 𝟏) (𝒏
𝒌)
𝒏
𝒌=𝟏
= 𝒏𝟐∑(𝒏− 𝟏
𝒌− 𝟐)
𝒏
𝒌=𝟏
(𝒏 − 𝟏
𝒌 − 𝟏) + 𝒏∑(
𝒏
𝒌− 𝟏)(𝒏 − 𝟏
𝒌 − 𝟏)
𝒏
𝒌=𝟏
𝐋𝐞𝐭: (𝟏 + 𝒙)𝒏−𝟏 = (𝒏 − 𝟏
𝟎) + 𝒙(
𝒏 − 𝟏
𝟏) + 𝒙𝟐 (
𝒏 − 𝟏
𝟐) +⋯+ 𝒙𝒏−𝟏 (
𝒏 − 𝟏
𝒏 − 𝟏)
(𝒙 + 𝟏)𝒏 = 𝒙𝒏−𝟏 (𝒏 − 𝟏
𝟎) + 𝒙𝒏−𝟐 (
𝒏 − 𝟏
𝟏) + ⋯+ (
𝒏 − 𝟏
𝒏 − 𝟏)
𝐓𝐡𝐞𝐧 𝑺𝟏 =∑(𝒏− 𝟏
𝒌 − 𝟐)
𝒏
𝒌=𝟏
(𝒏 − 𝟏
𝒌 − 𝟏) = (
𝟐𝒏 − 𝟐
𝒏 − 𝟐)
𝑺𝟐 = 𝒏∑(𝒏
𝒌− 𝟏)(𝒏 − 𝟏
𝒌 − 𝟏)
𝒏
𝒌=𝟏
=∑((𝒏 − 𝟏
𝒌 − 𝟏) + (
𝒏 − 𝟏
𝒌 − 𝟐)) (
𝒏 − 𝟏
𝒌 − 𝟏)
𝒏
𝒌=𝟏
=
=∑(𝒏− 𝟏
𝒌 − 𝟏)𝟐𝒏
𝒌=𝟏
+∑(𝒏− 𝟏
𝒌 − 𝟐) (𝒏 − 𝟏
𝒌 − 𝟏)
𝒏
𝒌=𝟏
= (𝟐𝒏 − 𝟐
𝒏 − 𝟏) + (
𝟐𝒏 − 𝟐
𝒏 − 𝟐) = (
𝟐𝒏 − 𝟏
𝒏 − 𝟏)
Hence,
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117 RMM-CALCULUS MARATHON 1501-1600
∑𝒌𝟐 (𝒏
𝒌 − 𝟏)(𝒏
𝒌)
𝒏
𝒌=𝟏
= 𝒏𝟐 (𝟐𝒏 − 𝟐
𝒏 − 𝟏) + 𝒏 (
𝟐𝒏 − 𝟏
𝒏 − 𝟏) = 𝒏𝟐 ⋅
(𝟐𝒏 − 𝟐)!
(𝒏 − 𝟐)!𝒏!+ 𝒏 ⋅
(𝟐𝒏 − 𝟏)!
(𝒏 − 𝟏)!𝒏!=
=(𝟐𝒏 − 𝟐)! (𝒏𝟑 + 𝒏𝟐 − 𝒏)
(𝒏 − 𝟏)! 𝒏!
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
√∑𝒌𝟐 (𝒏
𝒌 − 𝟏) (𝒏
𝒌)
𝒏
𝒌=𝟏
𝒏
= 𝐥𝐢𝐦𝒏→∞
√(𝟐𝒏 − 𝟐)! (𝒏𝟑 + 𝒏𝟐 − 𝒏)
(𝒏 − 𝟏)! 𝒏!
𝒏
=𝑪−𝑫
= 𝐥𝐢𝐦𝒏→∞
(𝟐𝒏)! [(𝒏 + 𝟏)𝟑 + (𝒏 + 𝟏)𝟐 − (𝒏 + 𝟏)]
𝒏! (𝒏 + 𝟏)!⋅
𝒏! (𝒏 − 𝟏)!
(𝟐𝒏 − 𝟐)! (𝒏𝟑 + 𝒏𝟐 − 𝒏)=
= 𝐥𝐢𝐦𝒏→∞
(𝟐𝒏)!𝒏! (𝒏 − 𝟏)!
𝒏! (𝒏 + 𝟏)! (𝟐𝒏 − 𝟐)!= 𝐥𝐢𝐦𝒏→∞
(𝟐𝒏 − 𝟏) ⋅ 𝟐𝒏
𝒏(𝒏 + 𝟏)= 𝟒
1581. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
∑
(
𝐬𝐢𝐧−𝟏 (
𝟐𝟖𝒆𝟏+𝐥𝐨𝐠𝒌
𝒌 + 𝒏𝒌)
𝐥𝐨𝐠 (𝟏 +𝒏
𝒌 + 𝒏𝒌√𝒏!𝒏 )
𝒏𝟐
)
𝒏
𝒌=𝟏
Proposed by Ruxandra Daniela Tonilă-Romania
Solution by Adrian Popa-Romania
𝛀 = 𝐥𝐢𝐦𝒏→∞
∑
(
𝐬𝐢𝐧−𝟏 (
𝟐𝟖𝒆𝟏+𝐥𝐨𝐠 𝒌
𝒌 + 𝒏𝒌)
𝐥𝐨𝐠 (𝟏 +𝒏
𝒌 + 𝒏𝒌√𝒏!𝒏 )
𝒏𝟐
)
𝒏
𝒌=𝟏
= 𝐥𝐢𝐦𝒏→∞
∑
(
𝐬𝐢𝐧−𝟏 (𝟐𝟖𝒆𝒌𝒌 + 𝒏𝒌
)
𝟐𝟖𝒆𝒌𝒌 + 𝒏𝒌
⋅𝟐𝟖𝒆𝒌𝒌 + 𝒏𝒌
𝒏𝟐 ⋅𝐥𝐨𝐠 (𝟏 +
𝒏
𝒌 + 𝒏𝒌√𝒏!𝒏 )
𝒏
𝒌 + 𝒏𝒌√𝒏!𝒏
⋅𝒏
𝒌 + 𝒏𝒌√𝒏!𝒏
)
𝒏
𝒌=𝟏
= 𝐥𝐢𝐦𝒏→∞
∑𝟐𝟖𝒆𝒌
𝒌 + 𝒏𝒌⋅(𝒌 + 𝒏𝒌)√𝒏!
𝒏
𝒏𝟑
𝒏
𝒌=𝟏
=
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118 RMM-CALCULUS MARATHON 1501-1600
= 𝐥𝐢𝐦𝒏→∞
𝟐𝟖𝒆√𝒏!𝒏
𝒏𝟑⋅∑𝒌
𝒏
𝒌=𝟏
= 𝐥𝐢𝐦𝒏→∞
𝟐𝟖𝒆√𝒏!𝒏
⋅ 𝒏(𝒏 + 𝟏)
𝟐𝒏𝟑= 𝟏𝟒𝒆 𝐥𝐢𝐦
𝒏→∞√𝒏! (𝒏 + 𝟏)𝒏
𝒏𝟐𝒏
𝒏
=𝑪−𝑫
= 𝟏𝟒𝒆 𝐥𝐢𝐦𝒏→∞
(𝒏 + 𝟏)! (𝒏 + 𝟐)𝒏+𝟏
(𝒏 + 𝟏)𝟐𝒏+𝟐⋅
𝒏𝟐𝒏
𝒏! (𝒏 + 𝟏)𝒏= 𝟏𝟒𝒆 𝐥𝐢𝐦
𝒏→∞
𝒏 + 𝟐
𝒏 + 𝟏⋅ (𝒏 + 𝟐
𝒏 + 𝟏)𝒏
⋅ (𝒏
𝒏 + 𝟏)𝟐𝒏
=
= 𝟏𝟒𝒆 𝐥𝐢𝐦𝒏→∞
(𝟏 +𝟏
𝒏 + 𝟏)𝒏
(𝟏 −𝟏
𝒏 + 𝟏)𝟐𝒏
= 𝟏𝟒𝒆 ⋅ 𝒆 ⋅ 𝒆−𝟐 = 𝟏𝟒
1582. If 𝒂, 𝒃, 𝒄 > 𝟏 then find:
𝛀 = 𝐥𝐢𝐦𝒙→𝟎(𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄
𝒃 ⋅ 𝒆−𝒙)))) (𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙))))
−𝟏
Proposed by Daniel Sitaru-Romania
Solution 1 by Adrian Popa-Romania
(𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙))))
′=
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙))
′
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠 𝒂=
=(𝐥𝐨𝐠𝒄(𝒄
𝒃 ⋅ 𝒆−𝒙))′
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒄(𝒄
𝒃 ⋅ 𝒆−𝒙) ⋅ 𝐥𝐨𝐠 𝒂 𝐥𝐨𝐠𝒃=
=−𝒆𝒃𝒆−𝒙
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙) ⋅ 𝒄𝒃𝒆−𝒙 𝐥𝐨𝐠𝒂 𝐥𝐨𝐠𝒃 𝐥𝐨𝐠 𝒄=
=−𝟏
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒆−𝒙) ⋅ 𝐥𝐨𝐠 𝒂 𝐥𝐨𝐠 𝒃 𝐥𝐨𝐠 𝒄
(𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙))))
′=
(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)))
′
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠 𝒄=
=(𝐥𝐨𝐠𝒂(𝒂
𝒃 ⋅ 𝒆−𝒙))′
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙) 𝐥𝐨𝐠 𝒃 𝐥𝐨𝐠 𝒄=
=−𝒂𝒃𝒆−𝒙
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)) 𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙) 𝒂𝒃𝒆−𝒙 𝐥𝐨𝐠𝒂 𝐥𝐨𝐠𝒃 𝐥𝐨𝐠 𝒄=
=−𝟏
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙)) 𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙) 𝐥𝐨𝐠 𝒂 𝐥𝐨𝐠 𝒃 𝐥𝐨𝐠 𝒄
𝛀 = 𝐥𝐢𝐦𝒙→𝟎(𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄
𝒃 ⋅ 𝒆−𝒙)))) (𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙))))
−𝟏=𝑳′𝑯
= 𝐥𝐢𝐦𝒙→𝟎
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒂(𝒂
𝒃𝒆−𝒙)
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃𝒆−𝒙)) ⋅ 𝐥𝐨𝐠𝒄(𝒄𝒃𝒆−𝒙)=𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂
𝒃)) ⋅ 𝐥𝐨𝐠𝒂(𝒂𝒃)
𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃)) ⋅ 𝐥𝐨𝐠𝒄(𝒄𝒃)= 𝟏
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119 RMM-CALCULUS MARATHON 1501-1600
Solution 2 by Florentin Vişescu-Romania
𝛀 = 𝐥𝐢𝐦𝒙→𝟎(𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄
𝒃 ⋅ 𝒆−𝒙)))) (𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙))))
−𝟏=
𝒚=𝒆−𝒙
= 𝐥𝐢𝐦𝒚→𝟏(𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄
𝒃 ⋅ 𝒚)))) (𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒚))))
−𝟏
𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒄(𝒄𝒃 ⋅ 𝒚))) = 𝐥𝐨𝐠𝒂(𝐥𝐨𝐠𝒃(𝒃 + 𝐥𝐨𝐠𝒄 𝒚)) =
= 𝐥𝐨𝐠𝒂 (𝐥𝐨𝐠𝒃 (𝒃 (𝟏 +𝐥𝐨𝐠𝒄 𝒚
𝒃))) = 𝐥𝐨𝐠𝒂 (𝟏 + 𝐥𝐨𝐠𝒃 (𝟏 +
𝐥𝐨𝐠𝒄 𝒚
𝒃))
Similarly,
𝐥𝐨𝐠𝒄(𝐥𝐨𝐠𝒃(𝐥𝐨𝐠𝒂(𝒂𝒃 ⋅ 𝒆−𝒙))) = 𝐥𝐨𝐠𝒄 (𝟏 + 𝐥𝐨𝐠𝒃 (𝟏 +
𝐥𝐨𝐠𝒂 𝒚
𝒃))
𝛀 = 𝐥𝐢𝐦𝒚→𝟏
𝐥𝐨𝐠𝒂 (𝟏 + 𝐥𝐨𝐠𝒃 (𝟏 +𝐥𝐨𝐠𝒄 𝒚𝒃 ))
𝐥𝐨𝐠𝒄 (𝟏 + 𝐥𝐨𝐠𝒃 (𝟏 +𝐥𝐨𝐠𝒂 𝒚𝒃 ))
= 𝐥𝐢𝐦𝒚→𝟏
𝐥𝐨𝐠 𝒄
𝐥𝐨𝐠 𝒂⋅𝐥𝐨𝐠𝒃 (𝟏 +
𝐥𝐨𝐠𝒄 𝒚𝒃 )
𝐥𝐨𝐠𝒃 (𝟏 +𝐥𝐨𝐠𝒂 𝒚𝒃 )
=
= 𝐥𝐢𝐦𝒚→𝟏
𝐥𝐨𝐠 𝒄
𝐥𝐨𝐠 𝒂⋅𝐥𝐨𝐠𝒄 𝒚
𝒃⋅𝒃
𝐥𝐨𝐠𝒂 𝒚= 𝐥𝐢𝐦𝒚→𝟏
𝐥𝐨𝐠 𝒄
𝐥𝐨𝐠𝒂⋅𝐥𝐨𝐠𝒚
𝐥𝐨𝐠 𝒄⋅𝐥𝐨𝐠 𝒂
𝐥𝐨𝐠 𝒚= 𝟏
1583. 𝒂𝟏 = 𝟒, 𝒂𝟐 = 𝟐, 𝒂𝒏 = 𝒂𝒏+𝟏
𝟑𝒏+𝟑
𝟕𝒏 ⋅ 𝒂𝒏+𝟐
𝟒𝒏+𝟖
𝟕𝒏 . Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
(∑(𝒂𝒌𝒂𝒌+𝟏
)𝒌
𝒏
𝒌=𝟏
)(𝟏
𝒏!∑𝒌 ⋅ 𝒌!
𝒏+𝟏
𝒌=𝟏
)
−𝟏
Proposed by Ruxandra Daniela Tonilă-Romania
Solution by George Florin Şerban-Romania
𝟏
𝒏!∑𝒌 ⋅ 𝒌!
𝒏+𝟏
𝒌=𝟏
=𝟏
𝒏!∑((𝒌 + 𝟏)! − 𝒌!)
𝒏+𝟏
𝒌=𝟏
=𝟏
𝒏!((𝒏 + 𝟐)! − 𝟏)
𝒂𝒏𝟕𝒏 = 𝒂𝒏+𝟏
𝟑(𝒏+𝟏) ⋅ 𝒂𝒏+𝟐𝟒(𝒏+𝟐). Let 𝒙𝒏 = 𝒂𝒏
𝒏 ⇒ 𝒙𝒏𝟕 = 𝒙𝒏+𝟏
𝟑 ⋅ 𝒙𝒏+𝟐𝟒 , 𝒙𝟏 = 𝒂𝟏 = 𝟒, 𝒙𝟐 = 𝒂𝟐
𝟐 = 𝟒.
𝒙𝟏𝟕 = 𝒙𝟐
𝟑 ⋅ 𝒙𝟑𝟒
𝒙𝟐𝟕 = 𝒙𝟑
𝟑 ⋅ 𝒙𝟒𝟒
…………
𝒙𝒏𝟕 = 𝒙𝒏+𝟏
𝟑 ⋅ 𝒙𝒏+𝟐𝟒
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120 RMM-CALCULUS MARATHON 1501-1600
𝑷𝒏 = 𝒙𝟏𝒙𝟐…𝒙𝒏 ⇒ 𝑷𝒏𝟕 =
𝑷𝒏𝟑
𝒙𝟏𝟑 ⋅ 𝒙𝒏+𝟏
𝟑 ⋅𝑷𝒏𝟒
𝒙𝟏𝟒𝒙𝟐𝟒 ⋅ 𝒙𝒏+𝟏
𝟒 𝒙𝒏+𝟐𝟒
We prove that 𝑷(𝒏): 𝒙𝒏 = 𝟒,∀𝒏 ≥ 𝟎 (by mathematical induction).
(I): 𝑷(𝟎): 𝟒𝟏𝟏 = 𝒙𝟏𝟕𝒙𝟐𝟒 = 𝟒𝟕 ⋅ 𝟒𝟒 = 𝟒𝟏𝟏 true.
(II): Suppose that: 𝑷(𝟎),𝑷(𝟏),… , 𝑷(𝒏 + 𝟏) are true, then
𝒙𝒏+𝟐𝟒 =
𝟒𝟏𝟏
𝒙𝒏+𝟏𝟕 =
𝟒𝟏𝟏
𝟒𝟕= 𝟒𝟒, because 𝒂𝒏, 𝒙𝒏 > 𝟎 ⇒ 𝒙𝒏+𝟐 = 𝟒 ⇒ 𝒂𝒏
𝒏 = 𝟒 ⇒ 𝒂𝒏 = √𝟒𝒏.
𝛀 = 𝐥𝐢𝐦𝒏→∞
(∑(𝒂𝒌𝒂𝒌+𝟏
)𝒌
𝒏
𝒌=𝟏
)(𝟏
𝒏!∑𝒌 ⋅ 𝒌!
𝒏+𝟏
𝒌=𝟏
)
−𝟏
= 𝐥𝐢𝐦𝒏→∞
(∑(√𝟒𝒌
√𝟒𝒌+𝟏 )
𝒌𝒏
𝒌=𝟏
) ⋅𝒏!
(𝒏 + 𝟐)! − 𝟏=
= 𝐥𝐢𝐦𝒏→∞
(𝟏
𝒏 + 𝟏(∑ √𝟒
𝒌+𝟏
𝒏
𝒌=𝟏
) ⋅𝒏! (𝒏 + 𝟏)
(𝒏 + 𝟐)! − 𝟏) =𝑪−𝑺
𝐥𝐢𝐦𝒏→∞
√𝟒𝒏+𝟐
𝒏 + 𝟐 −𝟏
(𝒏 + 𝟏)!
= 𝟎
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
(∑(𝒂𝒌𝒂𝒌+𝟏
)𝒌
𝒏
𝒌=𝟏
)(𝟏
𝒏!∑𝒌 ⋅ 𝒌!
𝒏+𝟏
𝒌=𝟏
)
−𝟏
= 𝟎
1584. 𝑺(𝒑) = {(𝒙, 𝒚)|𝒙𝟑 + 𝒚𝟑 ≤ 𝒑𝟑, 𝒙 ≥ 𝟎, 𝒚 ≥ 𝟎, 𝒑 ≥ 𝟎}
Find:
𝛀 = 𝐥𝐢𝐦𝒑→∞
𝑨𝒓𝒆𝒂(𝑺(𝒑))
𝒑𝟐
Proposed by Daniel Sitaru-Romania
Solution by Ty Halpen-Florida-USA
Rewrite the area bounded by 𝑺(𝒑) as 𝒚 ≤ (𝒑𝟑 − 𝒙𝟑)𝟏
𝟑 and integrate it from 𝟎 ≤ 𝒙 ≤ 𝒑:
𝑨𝒓𝒆𝒂(𝑺(𝒑)) = ∫ √𝒑𝟑 − 𝒙𝟑𝟑
𝒑
𝟎
𝒅𝒙 =
𝒕=𝒙𝟑
𝒑𝟑 𝒑𝟐
𝟑∫ 𝒕−
𝟐𝟑√𝟏 − 𝒕𝟑
𝟏
𝟎
𝒅𝒕 =
=𝒑𝟑
𝟔⋅𝚪 (𝟏𝟑)𝚪(
𝟒𝟑)
𝚪 (𝟓𝟑)
=𝒑𝟐
𝟔⋅𝚪𝟐 (
𝟏𝟑)
𝚪 (𝟐𝟑)
=𝒑𝟐
𝟔⋅
(√𝟑𝚪(𝟏𝟑))𝚪
𝟐 (𝟏𝟑)
𝟐𝝅=𝒑𝟐𝚪𝟑 (
𝟏𝟑)
𝟒𝝅√𝟑
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121 RMM-CALCULUS MARATHON 1501-1600
Therefore,
𝛀 = 𝐥𝐢𝐦𝒑→∞
𝑨𝒓𝒆𝒂(𝑺(𝒑))
𝒑𝟐=𝚪𝟑 (
𝟏𝟑)
𝟒𝝅√𝟑
1585. Prove that:
∑𝟏
𝒏+ 𝟏
∞
𝒏=𝟎
𝐜𝐨𝐬 (𝝅𝒏
𝟒) =
√𝟐
𝟏𝟔(𝟑𝝅 + 𝟒 𝐥𝐨𝐠(𝟏 + √𝟐) − 𝐥𝐨𝐠 𝟒)
Proposed by Asmat Qatea-Afghanistan
Solution by Mohammad Rostami-Afghanistan
First we prove that:
∵ ∑𝐬𝐢𝐧(𝒌𝜽)
𝒌
∞
𝒌=𝟏
=𝝅 − 𝜽
𝟐, (𝟎 < 𝜽 < 𝟐𝝅)
∑𝐬𝐢𝐧(𝒌𝜽)
𝒌
∞
𝒌=𝟏
= ∑
𝟏𝟐𝒊(𝒆𝒊𝒌𝜽 − 𝒆−𝒊𝒌𝜽)
𝒌
𝒏
𝒌=𝟏
=
=𝟏
𝟐𝒊(∑𝒆𝒊𝒌𝜽 ∫ 𝒆−𝒊𝒌
∞
𝟎
𝒅𝒙 −∑𝒆−𝒊𝒌𝜽∞
𝒌=𝟏
∫ 𝒆−𝒊𝒌𝒅𝒙∞
𝟎
∞
𝒌=𝟏
) =
=𝟏
𝟐𝒊[𝐥𝐨𝐠(𝟏 − 𝒆𝒊𝜽−𝒙) − 𝐥𝐨𝐠(𝟏 − 𝒆𝒊𝜽−𝒙)]
𝟎
∞=𝟏
𝟐𝒊[𝐥𝐨𝐠(𝟏 − 𝒆−𝒊𝜽) − 𝐥𝐨𝐠(𝟏 + 𝒆𝒊𝜽)] =
=𝟏
𝟐𝒊𝐥𝐨𝐠 (
𝟏 − 𝒆−𝒊𝜽
𝟏 − 𝒆𝒊𝜽) =
𝟏
𝟐𝒊𝐥𝐨𝐠 (
𝒆𝒊𝜽 − 𝟏
𝒆𝒊𝜽(𝟏 − 𝒆𝒊𝜽)) =
𝟏
𝟐𝒊𝐥𝐨𝐠(−𝒆−𝒊𝜽) =
=𝟏
𝟐𝒊(𝐥𝐨𝐠(−𝟏) + 𝐥𝐨𝐠(𝒆−𝒊𝜽)) =
{ 𝒆𝝅𝒊=−𝟏𝝅𝒊=𝐥𝐨𝐠(−𝟏) 𝟏
𝟐𝒊(𝝅𝒊 − 𝜽𝒊) =
𝒊(𝝅 − 𝜽)
𝟐𝒊=𝝅 − 𝜽
𝟐
Next, we prove that:
∵ ∑𝐜𝐨𝐬(𝒌𝜽)
𝒌
∞
𝒌=𝟏
= −𝟏
𝟐𝐥𝐨𝐠(𝟐 − 𝟐𝐜𝐨𝐬 𝜽) ; (𝜽 ∈ ℝ)
∑𝐜𝐨𝐬(𝒌𝜽)
𝒌
∞
𝒌=𝟏
=∑
𝟏𝟐(𝒆𝒊𝒌𝜽 + 𝒆−𝒊𝒌𝜽)
𝒌
𝒏
𝒌=𝟏
=
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122 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝟐(∑𝒆𝒊𝒌𝜽∫ 𝒆−𝒊𝒌
∞
𝟎
𝒅𝒙 +∑𝒆−𝒊𝒌𝜽∞
𝒌=𝟏
∫ 𝒆−𝒊𝒌𝒅𝒙∞
𝟎
∞
𝒌=𝟏
) =
=𝟏
𝟐(∫ ∑(𝒆𝒊𝜽−𝒙)
𝒌∞
𝒌=𝟏
𝒅𝒙∞
𝟎
+∫ ∑(𝒆−𝒊𝜽−𝒙)𝒌
𝒏
𝒌=𝟏
𝒅𝒙∞
𝟎
) =
=𝟏
𝟐(∫
𝒆𝒊𝜽−𝒙
𝟏 − 𝒆𝒊𝜽−𝒙
∞
𝟎
𝒅𝒙 +∫𝒆−𝒊𝜽−𝒙
𝟏 − 𝒆𝒊𝜽−𝒙𝒅𝒙
∞
𝟎
) =
=𝟏
𝟐[𝐥𝐨𝐠(𝟏 − 𝒆𝒊𝜽−𝒙) + 𝐥𝐨𝐠(𝟏 − 𝒆𝒊𝜽−𝒙)]
𝟎
∞=𝟏
𝟐[𝐥𝐨𝐠(𝟏 − 𝒆−𝒊𝜽) − 𝐥𝐨𝐠(𝟏 + 𝒆𝒊𝜽)] =
= −𝟏
𝟐𝐥𝐨𝐠(
(𝟏 − 𝒆𝒊𝜽)(𝒆𝒊𝜽 − 𝟏)
𝒆𝒊𝜽) = −
𝟏
𝟐𝐥𝐨𝐠 (−𝒆−𝒊𝜽(𝟏 − 𝟐𝒆𝒊𝜽 + 𝒆𝟐𝒊𝜽)) =
= −𝟏
𝟐𝐥𝐨𝐠(−𝒆−𝒊𝜽 + 𝟐 − 𝒆𝒊𝜽) = −
𝟏
𝟐𝐥𝐨𝐠 (𝟐 − (𝒆𝒊𝜽 + 𝒆−𝒊𝜽)) = −
𝟏
𝟐𝐥𝐨𝐠(𝟐 − 𝟐 𝐜𝐨𝐬𝜽)
Hence, we have:
∑𝟏
𝒏 + 𝟏
∞
𝒏=𝟎
𝐜𝐨𝐬 (𝝅𝒏
𝟒) = ∑
𝟏
𝒏𝐜𝐨𝐬 [
𝝅
𝟒(𝒏 − 𝟏)] =
∞
𝒏=𝟏
∑𝟏
𝒏𝐜𝐨𝐬 [(
𝝅
𝟒𝒏) −
𝝅
𝟒] =
∞
𝒏=𝟏
= ∑𝟏
𝒏[𝐜𝐨𝐬
𝝅
𝟒𝐜𝐨𝐬 (
𝝅
𝟒𝒏) + 𝐬𝐢𝐧
𝝅
𝟒𝐬𝐢𝐧 (
𝝅
𝟒𝒏)]
∞
𝒏=𝟏
=
=√𝟐
𝟐[∑
𝐜𝐨𝐬 (𝝅𝟒 𝒏)
𝒏
∞
𝒌=𝟏
+∑𝐬𝐢𝐧 (
𝝅𝟒 𝒏)
𝒏
∞
𝒏=𝟏
] =√𝟐
𝟐(−𝟏
𝟐𝐥𝐨𝐠(𝟐 − √𝟐) +
𝟑𝝅
𝟖) =
=√𝟐
𝟏𝟔[𝟑𝝅 − 𝟒 𝐥𝐨𝐠(𝟐 − √𝟐)] =
√𝟐
𝟏𝟔[𝟑𝝅 + 𝟒 𝐥𝐨𝐠(
𝟏(𝟐 + √𝟐)√𝟐
(𝟐 − √𝟐)(𝟐 + √𝟐)√𝟐) =
=√𝟐
𝟏𝟔[𝟑𝝅 + 𝟒 𝐥𝐨𝐠 (
𝟐√𝟐 + 𝟐
𝟐√𝟐)] =
√𝟐
𝟏𝟔[𝟑𝝅 + 𝟒 𝐥𝐨𝐠 (
𝟏 + √𝟐
𝟐)] =
=√𝟐
𝟏𝟔[𝟑𝝅 + 𝟒 𝐥𝐨𝐠(𝟏 + √𝟐) − 𝟒 𝐥𝐨𝐠 √𝟐] =
√𝟐
𝟏𝟔(𝟑𝝅 + 𝟒 𝐥𝐨𝐠(𝟏 + √𝟐) − 𝐥𝐨𝐠 𝟒)
⇒∑𝟏
𝒏 + 𝟏
∞
𝒏=𝟎
𝐜𝐨𝐬 (𝝅𝒏
𝟒) =
√𝟐
𝟏𝟔(𝟑𝝅 + 𝟒 𝐥𝐨𝐠(𝟏 + √𝟐) − 𝐥𝐨𝐠 𝟒)
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123 RMM-CALCULUS MARATHON 1501-1600
1586.
𝛀𝟏(𝒏) = ∫ 𝒙𝒏+𝟏√𝒙𝒏√𝒙𝒏−𝟏…√𝒙√𝟖𝒏
√𝟐𝒏
𝒅𝒙,𝛀𝟐(𝒏) = ∫ 𝒙√𝒙𝟐√𝒙𝟑…√𝒙𝒏+𝟏𝟐
𝟏
𝒅𝒙, 𝒏 ∈ ℕ∗ , 𝒏 ≥ 𝟐
Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟐𝟎𝟐𝟏𝛀𝟏(𝒏)𝛀𝟐(𝒏)
Proposed by Costel Florea-Romania
Solution by Kamel Gandouli Rezgui-Tunisia
𝒇(𝒙) = 𝒙𝒏+𝟏√𝒙𝒏√𝒙𝒏−𝟏…√𝒙 = 𝒙𝒏+𝟏 ⋅ 𝒙𝒏𝟐 ⋅ … ⋅ 𝒙
𝒏−𝒑
𝟐𝒑+𝟏 ⋅ … ⋅ 𝒙𝟏𝟐𝒏 = 𝒙𝒗𝒏
𝒗𝒏 = 𝒏+ 𝟏 +𝒏
𝟐+𝒏 − 𝟏
𝟐+𝒏 − 𝟐
𝟖+⋯+
𝟏
𝟐𝒏= 𝒏∑
𝟏
𝟐𝒌
𝒏
𝒌=𝟎
−∑𝒌− 𝟏
𝟐𝒌
𝒏
𝒌=𝟐
=
=∑𝟏
𝟐𝒌(𝒏 + 𝟏)
𝒏
𝒌=𝟐
+ 𝟏 +𝟏
𝟐−∑
𝒌
𝟐𝒌
𝒏
𝒌=𝟐
∑𝒌
𝟐𝒌
𝒏
𝒌=𝟐
=𝟑
𝟐− (𝒏 + 𝟐) (
𝟏
𝟐)𝒏
⇒ 𝒗𝒏 =𝟏
𝟒⋅𝟏 − (
𝟏𝟐)𝒏−𝟏
𝟏 −𝟏𝟐
⋅ (𝒏 + 𝟏) +𝟑
𝟐− (𝟑
𝟐− (𝒏 + 𝟐) (
𝟏
𝟐)𝒏
) =
= (𝟏
𝟐− (𝟏
𝟐)𝒏
)(𝒏 + 𝟏) + (𝒏 + 𝟐) (𝟏
𝟐)𝒏
= (𝟏
𝟐)𝒏
+𝟏
𝟐(𝒏 + 𝟏)
𝒈(𝒙) = 𝒙√𝒙𝟐√𝒙𝟑…√𝒙𝒏+𝟏 = 𝒙𝟏 ⋅ 𝒙𝟐𝟐 ⋅ 𝒙
𝟑𝟒 ⋅ 𝒙
𝟒
𝟐𝟑 ⋅ … ⋅ 𝒙𝒏+𝟏𝟐𝒏 = 𝒙𝒘𝒏 ,
𝒘𝒏 = 𝟏 +𝟐
𝟐+𝟑
𝟐𝟐+⋯+
𝒏 + 𝟏
𝟐𝒏=∑
𝒌+ 𝟏
𝟐𝒌
𝒏
𝒌=𝟎
== 𝟐 − (𝒏 + 𝟐) (𝟏
𝟐)𝒏
+ 𝟐 − (𝟏
𝟐)𝒏
= 𝟒 − (𝟏
𝟐)𝒏
(𝒏 + 𝟑) → 𝟒
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124 RMM-CALCULUS MARATHON 1501-1600
𝛀𝟏(𝒏) = ∫ 𝒙𝒏+𝟏√𝒙𝒏√𝒙𝒏−𝟏…√𝒙√𝟖𝒏
√𝟐𝒏
𝒅𝒙 = ∫ 𝒙𝒗(𝒏)√𝟖𝒏
√𝟐𝒏
𝒅𝒙 = [𝒙𝒗𝒏+𝟏
𝒗𝒏 + 𝟏]√𝟐𝒏
√𝟖𝒏
=
=𝟐𝟑(
𝒗𝒏+𝟏𝒏) − 𝟐
𝒗𝒏+𝟏𝒏
𝒗𝒏 + 𝟏
𝛀𝟐(𝒏) = ∫ 𝒙√𝒙𝟐√𝒙𝟑…√𝒙𝒏+𝟏𝟐
𝟏
𝒅𝒙 = ∫ 𝒙𝒘𝒏𝟐
𝟏
𝒅𝒙 = [𝒙𝒘𝒏+𝟏
𝒘𝒏 + 𝟏]𝟏
𝟐
=𝟐𝒘𝒏+𝟏 − 𝟏
𝒘𝒏 + 𝟏
Hence,
𝛀𝟏(𝒏)
𝛀𝟐(𝒏)=
𝟐𝟑(𝒗𝒏+𝟏𝒏) − 𝟐
𝒗𝒏+𝟏𝒏
𝒗𝒏 + 𝟏
𝟐𝒘𝒏+𝟏 − 𝟏𝒘𝒏 + 𝟏
=𝒘𝒏 + 𝟏
𝒗𝒏 + 𝟏⏟ →𝟎
⋅𝟐𝟑(
𝒗𝒏+𝟏𝒏) − 𝟐
𝒗𝒏+𝟏𝒏
𝟐𝟏+𝒘𝒏 − 𝟏⏟ 𝟐√𝟐/𝟑𝟏
> 0
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟐𝟎𝟐𝟏𝛀𝟏(𝒏)𝛀𝟐(𝒏) = 𝟏.
1587. Prove that:
∑(𝟏
(𝟔𝒌 − 𝟔)!+
𝟏
(𝟔𝒌 − 𝟓)!−
𝟏
(𝟔𝒌 − 𝟑)!−
𝟏
(𝟔𝒌 − 𝟐)!)
∞
𝒌=𝟏
= √𝟒𝒆
𝟑𝐬𝐢𝐧 (
𝟐𝝅 + 𝟑√𝟑
𝟔)
Proposed by Asmat Qatea-Afghanistan
Solution by Felix Marin-Romania
𝑯 −Hankel Contour.
𝜶 ∈ {𝟐, 𝟑, 𝟓, 𝟔},∑𝟏
(𝟔𝒌 − 𝜶)!
∞
𝒌=𝟏
=∑𝟏
𝚪(𝟔𝒌 − 𝜶 + 𝟏)!
∞
𝒌=𝟏
=∑∮𝒆𝒕
𝒕𝟔𝒌−𝜶+𝟏𝒅𝒕
𝟐𝝅𝒊𝑯
∞
𝒌=𝟏
=
= ∑∫𝒆𝒕
𝒕𝟔𝒌−𝜶+𝟏
𝟏++∞𝒊
𝟏+−∞𝒊
𝒅𝒕
𝟐𝝅𝒊
∞
𝒌=𝟏
= ∫ 𝒆𝒕𝒕𝜶−𝟏𝟏++∞𝒊
𝟏+−∞𝒊
∑(𝟏
𝒕𝟔)𝒌∞
𝒌=𝟏
𝒅𝒕
𝟐𝝅𝒊= ∫
𝒆𝒕𝒕𝜶−𝟏
𝒕𝟔 − 𝟏
𝒅𝒕
𝟐𝝅𝒊
𝟏++∞𝒊
𝟏+−∞𝒊
=
= ∑𝒆𝒑𝒏𝒑𝒏
𝜶−𝟏
𝟔𝒑𝒏𝟓|𝒑𝒏=𝒆
𝒏𝝅𝒊𝟑
𝟓
𝒏=𝟎
=𝟏
𝟔∑𝒆𝒑𝒏𝒑𝒏
𝜶
𝟓
𝒏=𝟎
Therefore,
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125 RMM-CALCULUS MARATHON 1501-1600
∑(𝟏
(𝟔𝒌 − 𝟔)!+
𝟏
(𝟔𝒌 − 𝟓)!−
𝟏
(𝟔𝒌 − 𝟑)!−
𝟏
(𝟔𝒌 − 𝟐)!)
∞
𝒌=𝟏
=𝟏
𝟔∑𝒆𝒑𝒏(𝒑𝒏
𝟔 + 𝒑𝒏𝟓 − 𝒑𝒏
𝟑 − 𝒑𝒏𝟐)
𝟓
𝒏=𝟎
=
= √𝒆 [𝐜𝐨𝐬 (√𝟑
𝟐) +
√𝟑
𝟑𝐬𝐢𝐧(
√𝟑
𝟐)] ≅ 𝟏. 𝟕𝟗𝟑𝟑
1588. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!∫
𝒙𝒏 𝐬𝐢𝐧 (𝒙 +𝝅𝟒)
𝒆𝒙
∞
𝟎
𝒅𝒙
Proposed by Daniel Sitaru-Romania
Solution by Mohammad Rostami-Afghanistan
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!∫
𝒙𝒏 𝐬𝐢𝐧 (𝒙 +𝝅𝟒)
𝒆𝒙
∞
𝟎
𝒅𝒙 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!∫
𝒙𝒏
𝒆𝒙⋅𝒆𝒊𝒙+
𝝅𝟒𝒊 − 𝒆−𝒊𝒙−
𝝅𝟒𝒊
𝟐𝒊𝒅𝒙
∞
𝟎
=
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!(𝒆𝝅𝟒𝒊
𝟐𝒊∫ 𝒙𝒏𝒆−(𝟏−𝒊)𝒙∞
𝟎
𝒅𝒙 −𝒆−𝒑𝒊𝟒𝒊
𝟐𝒊∫ 𝒙𝒏𝒆−(𝟏+𝒊)𝒙∞
𝟎
𝒅𝒙) =(𝟏±𝒙)𝒊=𝒖
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!(𝒆𝝅𝟒𝒊
𝟐𝒊∫
𝒖(𝒏+𝟏)−𝟏𝒆−𝒖
(𝟏 − 𝒊)𝒏+𝟏
∞
𝟎
𝒅𝒖 −𝒆−𝝅𝟒𝒊
𝟐𝒊∫
𝒖(𝒏+𝟏)−𝟏𝒆−𝒖
(𝟏 + 𝒊)𝒏+𝟏
∞
𝟎
𝒅𝒖) =
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝚪(𝒏 + 𝟏)⋅ 𝚪(𝒏 + 𝟏) [
𝒆𝝅𝟒𝒊
𝟐𝒊(𝟏 − 𝒊)𝒏+𝟏−
𝒆−𝝅𝟒𝒊
𝟐𝒊(𝒊 + 𝟏)𝒏+𝟏] =
= 𝐥𝐢𝐦𝒏→∞
[𝒆𝝅𝟒𝒊
𝟐𝒊 (√𝟐𝒆−𝝅𝟒𝒊)𝒏+𝟏 −
𝒆−𝝅𝟒𝒊
𝟐𝒊 (√𝟐𝒆𝝅𝟒𝒊)𝒏+𝟏] =
= 𝐥𝐢𝐦𝒏→∞
𝟏
(√𝟐)𝒏+𝟏(𝒆𝝅𝟐𝒊+𝝅𝟒𝒏𝒊
𝟐𝒊−𝒆−𝝅𝟐𝒊−𝝅𝟒𝒏𝒊
𝟐𝒊) =
= 𝐥𝐢𝐦𝒏→∞
𝟏
(√𝟐)𝒏+𝟏
[(𝐜𝐨𝐬
𝝅𝟐 + 𝒊 𝐬𝐢𝐧
𝝅𝟐) 𝒆
(𝝅𝟒𝒏)𝒊
𝟐𝒊−(𝐜𝐨𝐬 (−
𝝅𝟐) + 𝒊 𝒔𝒊𝒏 (–
𝝅𝟐))𝒆
(−𝝅𝟒𝒏)𝒊
𝟐𝒊] =
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126 RMM-CALCULUS MARATHON 1501-1600
= 𝐥𝐢𝐦𝒏→∞
𝟏
(√𝟐)𝒏+𝟏(𝒆(𝝅𝟒𝒏)𝒊 + 𝒆(−
𝝅𝟒𝒏)𝒊
𝟐) = 𝐥𝐢𝐦
𝒏→∞
𝟏
(√𝟐)𝒏+𝟏 𝐜𝐨𝐬 (
𝝅
𝟒𝒏) = 𝟎
Solution 2 by Syed Shahabudeen-India
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!∫
𝒙𝒏 𝐬𝐢𝐧 (𝒙 +𝝅𝟒)
𝒆𝒙
∞
𝟎
𝒅𝒙 =𝟏
√𝟐𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!∫
𝒙𝒏(𝐬𝐢𝐧 𝒙 + 𝐜𝐨𝐬 𝒙)
𝒆𝒙
∞
𝟎
𝒅𝒙 =
=𝟏
√𝟐𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!(𝑰𝒎∫ 𝒙𝒏𝒆−𝒙(𝟏−𝒊)
∞
𝟎
𝒅𝒙 + 𝑹𝒆∫ 𝒙𝒏𝒆−𝒙(𝟏−𝒊)∞
𝟎
𝒅𝒙) =
=𝟏
√𝟐𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!(𝑰𝒎𝑴(𝒆−𝒙(𝟏−𝒊)) + 𝑹𝒆𝑴(𝒆−𝒙(𝟏−𝒊))) ; (𝒂𝒑𝒑𝒍𝒚 𝑴𝒆𝒍𝒍𝒊𝒏 𝑻𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎)
=𝟏
√𝟐𝐥𝐢𝐦𝒏→∞
(𝑰𝒎(𝟏 − 𝒊)−(𝒏+𝟏) +𝑹𝒆(𝟏 − 𝒊)−(𝒏+𝟏)) =
= 𝐥𝐢𝐦𝒏→∞
𝟏
(√𝟐)𝒏+𝟐 (𝑰𝒎(𝒆
𝒊𝝅(𝒏+𝟏)𝟒 ) + 𝑹𝒆 (𝒆𝒊
𝝅(𝒏+𝟏)𝟒 )) =
= 𝐥𝐢𝐦𝒏→∞
𝟏
(√𝟐)𝒏+𝟐 (𝐬𝐢𝐧(
𝝅(𝒏 + 𝟏)
𝟒) + 𝐜𝐨𝐬 (
𝝅(𝒏 + 𝟏)
𝟒)) = 𝐥𝐢𝐦
𝒏→∞
𝐜𝐨𝐬 (𝝅𝒏𝟒 )
(√𝟐)𝒏+𝟏 = 𝟎, 𝐛𝐞𝐜𝐚𝐮𝐬𝐞
𝟎 ← −𝟏
(√𝟐)𝒏+𝟏 ≤
𝐜𝐨𝐬 (𝝅𝒏𝟒 )
(√𝟐)𝒏+𝟏 ≤
𝟏
(√𝟐)𝒏+𝟏 → 𝟎
Solution 3 by Muhammad Afzal-Pakistan
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!∫
𝒙𝒏 𝐬𝐢𝐧 (𝒙 +𝝅𝟒)
𝒆𝒙
∞
𝟎
𝒅𝒙 ; (𝟏)
𝝎 = ∫𝒙𝒏 𝐬𝐢𝐧 (𝒙 +
𝝅𝟒)
𝒆𝒙
∞
𝟎
𝒅𝒙 = 𝑰𝒎{∫ 𝒙𝒏𝒆𝒙(𝒊−𝟏)+
𝒊𝝅𝟒
∞
𝟎
𝒅𝒙} =
= 𝑰𝒎{𝒆𝒊𝝅𝟒∫ 𝒙𝒏𝒆𝒙(𝒊−𝟏)
∞
𝟎
𝒅𝒙} =𝒖=−𝒙(𝒊−𝟏)
𝑰𝒎{𝒆𝒊𝝅𝟒
(𝟏 − 𝒊)𝒏−𝟏∫ 𝒖𝒏𝒆−𝒖∞
𝟎
𝒅𝒖}
= 𝚪(𝒏 + 𝟏)𝑰𝒎{𝒆𝒊𝝅𝟒
𝟏
(𝟏 − 𝒊)𝒏−𝟏} = 𝒏! 𝑰𝒎{𝒆𝒊
𝝅𝟒(𝟏 + 𝒊)𝒏−𝟏
𝟐𝒏−𝟏} =
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127 RMM-CALCULUS MARATHON 1501-1600
=𝒏!
𝟐𝒏−𝟏𝑰𝒎{𝒆𝒊
𝝅𝟒𝟐𝒏−𝟏𝟐 𝒆𝒊
𝝅𝟒(𝒏−𝟏)} =
𝒏!
𝟐𝒏−𝟏𝟐
𝑰𝒎(𝒆𝒊𝒏𝝅𝟒)
𝛀 =𝒏!
𝟐𝒏−𝟏𝟐
𝐬𝐢𝐧 (𝒏𝝅
𝟒)(𝟏)⇒
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!⋅𝒏!
𝟐𝒏−𝟏𝟐
𝐬𝐢𝐧 (𝒏𝝅
𝟒) = √𝟐 𝐥𝐢𝐦
𝒏→∞
𝐬𝐢𝐧 (𝒏𝝅𝟒 )
√𝟐𝒏= √𝟐 𝐥𝐢𝐦
𝒏→∞
𝒏𝝅
𝟐⋅𝟏
√𝟐𝒏= 𝟎
Solution 4 by Ajenikoko Gbolahan-Nigeria
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!∫
𝒙𝒏 𝐬𝐢𝐧 (𝒙 +𝝅𝟒)
𝒆𝒙
∞
𝟎
𝒅𝒙 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!⋅𝟏
√𝟐∫
𝒙𝒏 𝐬𝐢𝐧 𝒙 + 𝒙𝒏 𝐜𝐨𝐬 𝒙
𝒆𝒙
∞
𝟎
𝒅𝒙 =
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!⋅𝟏
√𝟐∫ [𝑰𝒎(𝒙𝒏𝒆𝒊𝒙−𝒙) + 𝑹𝒆(𝒙𝒏𝒆𝒊𝒙−𝒙)]𝒅𝒙∞
𝟎
=
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!⋅𝟏
√𝟐∫ [𝑰𝒎(𝒙𝒏𝒆−(𝟏−𝒊)𝒙) + 𝑹𝒆(𝒙𝒏𝒆(𝟏−𝒊)𝒙)]∞
𝟎
𝒅𝒙 =
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!⋅𝟏
√𝟐[𝑰𝒎𝓛{𝒙𝒏}𝒔=𝟏−𝒊 + 𝑹𝒆𝓛{𝒙
𝒏}𝒔=𝟏−𝒊] =
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏!⋅𝟏
√𝟐(𝑰𝒎(
𝒏!
(𝟏 − 𝒊)𝒏+𝟏) + 𝑹𝒆 (
𝒏!
(𝟏 − 𝒊)𝒏+𝟏)) =
= 𝐥𝐢𝐦𝒏→∞
𝟏
√𝟐[(𝟏
√𝟐)𝒏+𝟏
(−𝐬𝐢𝐧(𝝅(−𝒏 − 𝟏)
𝟒) + 𝐜𝐨𝐬 (
𝝅(−𝒏 − 𝟏)
𝟒)) =
= 𝐥𝐢𝐦𝒏→∞
𝟏
√𝟐𝒏+𝟐(− 𝐬𝐢𝐧(
𝝅(𝒏 − 𝟏)
𝟒) + 𝐜𝐨𝐬 (
𝝅(−𝒏 − 𝟏)
𝟒)) = 𝟎
1589. Prove that:
∑𝚪(𝒏 +
𝟏𝟑)𝝍(𝒏 +
𝟏𝟑)
𝟑𝒏𝒏!
∞
𝒏=𝟎
= −√𝟑
𝟐
𝟑
𝚪 (𝟏
𝟑) {𝟏
𝟐𝐥𝐨𝐠 (
𝟏𝟎𝟖
𝟗) + 𝜸 +
𝝅
𝟐√𝟑}
Proposed by Ajetunmobi Abdulqoyyum-Nigeria
Solution by Dawid Bialek-Poland
𝛀 = ∑𝚪(𝒏 +
𝟏𝟑)𝝍(𝒏 +
𝟏𝟑)
𝟑𝒏𝒏!
∞
𝒏=𝟎
= ∑𝚪′ (𝒏 +
𝟏𝟑)
𝟑𝒏𝒏!
∞
𝒏=𝟎
= ∑𝟏
𝟑𝒏𝒏!⋅𝝏
𝝏𝒏(∫ 𝒆−𝒕 ⋅ 𝒕𝒏−
𝟐𝟑
∞
𝟎
𝒅𝒕)
∞
𝒏=𝟎
=
= ∑𝟏
𝟑𝒏𝒏!⋅ ∫ 𝒆−𝒕 ⋅ 𝒕𝒏−
𝟐𝟑 ⋅ 𝐥𝐨𝐠 𝒕
∞
𝟎
𝒅𝒕
∞
𝒏=𝟎
= ∫ 𝒆−𝒕 ⋅ 𝐥𝐨𝐠 𝒕 ⋅ 𝒕−𝟐𝟑∑
(𝒕𝟑)𝒏
𝒏!
∞
𝒏=𝟎
𝒅𝒕∞
𝟎
=
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128 RMM-CALCULUS MARATHON 1501-1600
= ∫ 𝒆−𝒕 ⋅ 𝐥𝐨𝐠 𝒕 ⋅ 𝒕−𝟐𝟑 ⋅ 𝒆
𝒕𝟑
∞
𝟎
𝒅𝒕 = ∫ 𝒕−𝟐𝟑 ⋅ 𝐥𝐨𝐠 𝒕 ⋅ 𝒆−
𝟐𝟑𝒕
∞
𝟎
𝒅𝒕 =
=𝝏
𝝏𝒔∫ 𝒕𝒔 ⋅ 𝒆−
𝟐𝟑𝒕𝒅𝒕|
𝒔=−𝟐𝟑
∞
𝟎
=𝝏
𝝏𝒔(𝚪(𝒔 + 𝟏)
(𝟐𝟑)𝒔+𝟏 )
𝒔=−𝟐𝟑
=
= ((𝟐
𝟑)−𝒔−𝟏
⋅ 𝚪(𝒔 + 𝟏) ⋅ (𝝍𝟎(𝒔 + 𝟏) − 𝐥𝐨𝐠 (𝟐
𝟑))𝒔=−
𝟐𝟑
=
= (𝟐
𝟑)−𝟏𝟑⋅ 𝚪 (
𝟏
𝟑) (𝝍𝟎 (
𝟏
𝟑) + 𝐥𝐨𝐠 (
𝟐
𝟑))
𝝍𝟎 (𝟏
𝟑) = −
𝝅
𝟐√𝟑− 𝜸 −
𝟑
𝟐𝐥𝐨𝐠 𝟑
𝛀 = −√𝟑
𝟐
𝟑
𝚪(𝟏
𝟑) {𝟑
𝟐𝐥𝐨𝐠 𝟑 − 𝐥𝐨𝐠 (
𝟑
𝟐) + 𝜸 +
𝝅
𝟐√𝟑} =
= −√𝟑
𝟐
𝟑
𝚪(𝟏
𝟑) {𝟑
𝟐𝐥𝐨𝐠𝟑 − 𝐥𝐨𝐠𝟑 + 𝐥𝐨𝐠 𝟐 + 𝜸 +
𝝅
𝟐√𝟑} =
= −√𝟑
𝟐
𝟑
𝚪(𝟏
𝟑) {𝟑
𝟐𝐥𝐨𝐠𝟑 − 𝐥𝐨𝐠𝟑 +
𝟏
𝟐𝐥𝐨𝐠 𝟒 + 𝜸 +
𝝅
𝟐√𝟑} =
= −√𝟑
𝟐
𝟑
𝚪(𝟏
𝟑) {𝜸 +
𝝅
𝟐√𝟑+𝟏
𝟐𝐥𝐨𝐠 𝟏𝟐}
1590. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)
𝒌
𝒊=𝟏
]
−𝟏𝒏
𝒌=𝟏
Proposed by Vasile Mircea Popa-Romania
Solution 1 by Kamel Gandouli Rezgui-Tunisia
𝒔𝒌 =∑(𝒊 + 𝟏)(𝒌 + 𝟐)
𝒌
𝒊=𝟏
−∑𝒊(𝒊 + 𝟏)
𝒌
𝒊=𝟏
= (𝒌 + 𝟐)∑𝒊
𝒌
𝒊=𝟏
+ 𝒌(𝒌 + 𝟐) −∑𝒊𝟐𝒌
𝒊=𝟏
−∑𝒊
𝒌
𝒊=𝟏
=
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129 RMM-CALCULUS MARATHON 1501-1600
=𝒌(𝒌 + 𝟏)𝟐
𝟐+ 𝒌(𝒌 + 𝟐) −∑𝒊𝟐
𝒌
𝒊=𝟏
=𝒌𝟑 + 𝟗𝒌𝟐 + 𝟏𝟒𝒌
𝟔⇒
⇒𝟏
𝒔𝒌=
𝟔
𝒌𝟑 + 𝟗𝒌𝟐 + 𝟏𝟒𝒌=
𝟔
𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)=
𝟔
𝟑𝟓(𝒌 + 𝟕)+𝟑
𝟕𝒌−
𝟑
𝟓(𝒌 + 𝟐)
⇒∑𝟏
𝒔𝒌
𝒏
𝒌=𝟏
=∑(𝟔
𝟑𝟓(𝒌 + 𝟕)+𝟑
𝟕𝒌−
𝟑
𝟓(𝒌 + 𝟐))
𝒏
𝒌=𝟏
=
=∑𝟔
𝟑𝟓(𝒌 + 𝟕)
𝒏
𝒌=𝟏
+∑𝟑
𝟕𝒌
𝒏
𝒌=𝟖
−∑𝟑
𝟓(𝒌 + 𝟐)
𝒏
𝒌=𝟔
+∑𝟑
𝟕𝒌
𝟕
𝒌=𝟏
−∑𝟑
𝟓(𝒌 + 𝟐)
𝟓
𝒌=𝟏
𝟔
𝟑𝟔+𝟑
𝟕−𝟑
𝟓= 𝟎 ⇒ ∑
𝟔
𝟑𝟓(𝒌 + 𝟕)
𝒏
𝒌=𝟏
+∑𝟑
𝟕𝒌
𝒏
𝒌=𝟖
−∑𝟑
𝟓(𝒌 + 𝟐)
𝒏
𝒌=𝟔
→ −
𝛀 = 𝐥𝐢𝐦𝒏→∞
∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)
𝒌
𝒊=𝟏
]
−𝟏𝒏
𝒌=𝟏
= ∑𝟑
𝟕𝒌
𝟕
𝒌=𝟏
−∑𝟑
𝟓(𝒌 + 𝟐)
𝟓
𝒌=𝟏
=
= ∑𝟑
𝟕𝒌
𝟕
𝒌=𝟏
−∑𝟑
𝟓𝒌
𝟕
𝒌=𝟑
=𝟑
𝟕𝑯𝟕 −
𝟑
𝟓𝑯𝟕 +
𝟑
𝟓+𝟑
𝟏𝟎= −
𝟔
𝟑𝟓𝑯𝟕 +
𝟗
𝟏𝟎
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)
𝒌
𝒊=𝟏
]
−𝟏𝒏
𝒌=𝟏
= −𝟔
𝟑𝟓𝑯𝟕 +
𝟗
𝟏𝟎
Solution 2 by Ravi Prakash-New Delhi-India
∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)
𝒌
𝒊=𝟏
=∑[𝒊(𝒌 + 𝟏) − 𝒊𝟐 + (𝒌 + 𝟐)]
𝒌
𝒊=𝟏
=
=(𝒌 + 𝟏)(𝒌 + 𝟏)𝒌
𝟐−𝟏
𝟔𝒌(𝒌 + 𝟏)(𝟐𝒌 + 𝟏) + (𝒌 + 𝟐)𝒌 =
=𝒌
𝟔(𝒌𝟐 + 𝟗𝒌+ 𝟏𝟒) =
𝟏
𝟔𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)
∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)
𝒌
𝒊=𝟏
]
−𝟏𝒏
𝒌=𝟏
=∑𝟔
𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)
𝒏
𝒌=𝟏
= ∑[𝟑
𝟕𝒌−
𝟑
𝟓(𝒌 + 𝟐)+
𝟔
𝟑𝟓(𝒌 + 𝟕)]
𝒏
𝒌=𝟏
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130 RMM-CALCULUS MARATHON 1501-1600
=𝟏𝟓
𝟑𝟓∑(
𝟏
𝒌−
𝟏
𝒌 + 𝟐)
𝒏
𝒌=𝟏
−𝟔
𝟑𝟓∑(
𝟏
𝒌 + 𝟐−
𝟏
𝒌 + 𝟕)
𝒏
𝒌=𝟏
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)
𝒌
𝒊=𝟏
]
−𝟏𝒏
𝒌=𝟏
=𝟏𝟓
𝟑𝟓(𝟏 +
𝟏
𝟐) −
𝟔
𝟑𝟓(𝟏
𝟑+𝟏
𝟒+𝟏
𝟓+𝟏
𝟔+𝟏
𝟕) =
𝟓𝟓𝟖
𝟏𝟐𝟐𝟓.
Solution 3 by Amrit Awasthi-Punjab-India
𝑺𝒌 =∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)
𝒌
𝒊=𝟏
= (𝒌 + 𝟏)∑𝒊
𝒌
𝒊=𝟏
−∑𝒊𝟐𝒌
𝒊=𝟏
+ (𝒌 + 𝟐)∑𝟏
𝒌
𝒊=𝟏
=
=𝒌(𝒌 + 𝟏)𝟐
𝟐−𝒌(𝒌 + 𝟏)(𝟐𝒌 + !)
𝟔+ 𝒌(𝒌 + 𝟐) =
𝟏
𝟔𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)
𝑺′𝒏 =∑𝟏
𝑺𝒌
𝒏
𝒌=𝟏
=∑𝟔
𝒌(𝒌 + 𝟐)(𝒌 + 𝟕)
𝒏
𝒌=𝟏
=∑𝟑
𝟕𝒌
𝒏
𝒌=𝟏
−∑𝟑
𝟓(𝒌 + 𝟐)
𝒏
𝒌=𝟏
+∑𝟔
𝟑𝟓(𝒌 + 𝟕)
𝒏
𝒌=𝟏
=
=𝟑
𝟕𝑯𝒏 −
𝟑
𝟓(𝑯𝒏 − 𝑯𝟐) +
𝟔
𝟑𝟓(𝑯𝒏 − 𝑯𝟕)
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
∑[∑(𝒊 + 𝟏)(𝒌 − 𝒊 + 𝟐)
𝒌
𝒊=𝟏
]
−𝟏𝒏
𝒌=𝟏
=𝟗
𝟏𝟎−𝟏𝟎𝟖𝟏
𝟐𝟒𝟓𝟎=𝟓𝟓𝟖
𝟏𝟐𝟐𝟓
1591. Prove that:
𝐥𝐢𝐦𝒏→∞
(√𝒏𝟐 − 𝒏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏))
𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏)
= 𝒆−𝟏𝟖
Proposed by Abdul Mukhtar-Nigeria
Solution 1 by Asmat Qatea-Afghanistan
𝛀 = 𝐥𝐢𝐦𝒏→∞
(√𝒏𝟐 − 𝒏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏))
𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏)
= 𝐥𝐢𝐦𝒏→∞
((𝟏 + 𝒖 − 𝟏)𝟏𝒖−𝟏)
(𝒖−𝟏)(𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏))
=
= 𝒆(𝒖−𝟏)(𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏)) = 𝒆𝑺
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131 RMM-CALCULUS MARATHON 1501-1600
𝑺 = 𝐥𝐢𝐦𝒏→∞
(√𝒏𝟐 − 𝒏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏) − 𝟏) (𝒏𝟐 + 𝐬𝐢𝐧(𝟑𝒏)) =
= 𝐥𝐢𝐦𝒏→∞
(𝒏𝟐√𝟏 −𝟏
𝒏+𝒏𝟐
𝟒𝐬𝐢𝐧 (
𝟐
𝒏) − 𝒏𝟐) + 𝐬𝐢𝐧(𝟑𝒏)(√𝟏 −
𝟏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏) − 𝟏)
⏟ 𝟎
=
= 𝐥𝐢𝐦𝒏→∞
(𝒏𝟐√𝟏 −𝟏
𝒏+𝒏
𝟐− 𝒏𝟐) =
√𝟏−𝟏𝒏=𝒖
𝐥𝐢𝐦𝒖→𝟏
𝟐 − 𝟐𝒖
𝟒(𝟏 − 𝒖𝟐)(−𝟐𝒖)= −
𝟏
𝟖
Therefore,
𝐥𝐢𝐦𝒏→∞
(√𝒏𝟐 − 𝒏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏))
𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏)
= 𝒆−𝟏𝟖
Solution 2 by Kamel Gandouli Rezgui-Tunisia
(√𝒏𝟐 − 𝒏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏))
𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏)
= (√𝟏 −𝟏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏))
𝒏𝟐+𝐬𝐢𝐧𝟑𝒏
=
= 𝒆𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏) 𝐥𝐨𝐠(√𝟏−
𝟏𝒏+𝟏𝟒𝐬𝐢𝐧(
𝟐𝒏))
𝐥𝐨𝐠 (√𝟏 −𝟏𝒏 +
𝟏𝟒𝐬𝐢𝐧 (
𝟐𝒏))
√𝟏 −𝟏𝒏 +
𝟏𝟒𝐬𝐢𝐧 (
𝟐𝒏) − 𝟏
→ 𝟏; (𝒏 → ∞)
𝐥𝐢𝐦𝒏→∞
(𝒏𝟐 + 𝐬𝐢𝐧(𝟑𝒏))(√𝟏 −𝟏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏) − 𝟏) =
= 𝐥𝐢𝐦𝒏→∞
𝒏𝟐 + 𝐬𝐢𝐧(𝟑𝒏)
𝒏𝟐(√𝟏 −
𝟏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏) − 𝟏) ⋅ 𝒏𝟐
Now, we want to find:
𝝎 = 𝐥𝐢𝐦𝒏→∞
(√𝟏 −𝟏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏) − 𝟏) ⋅ 𝒏𝟐
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132 RMM-CALCULUS MARATHON 1501-1600
√𝟏 − 𝒏 ≅ 𝟏 −𝒏
𝟐−𝒏𝟐
𝟖−𝒏𝟑
𝟏𝟔
𝐬𝐢𝐧 (𝟐
𝒏) ≅(𝟐𝒏 −
𝟒𝒏𝟑
𝟑)
𝟏
𝒏𝟐(𝟏 −
𝒏
𝟐−𝒏𝟐
𝟖−𝒏𝟑
𝟏𝟔+𝟏
𝟒(𝟐𝒏 −
𝟒𝒏𝟑
𝟑) − 𝟏) =; (𝒏 → 𝟎)
𝟏
𝒏𝟐(−𝒏𝟐
𝟖−𝒏𝟑
𝟏𝟔+ (−
𝒏𝟑
𝟑)) ⋅ 𝒏 = −
𝟏
𝟖−𝒏
𝟏𝟔−𝒏
𝟑
Hence,
𝝎 = 𝐥𝐢𝐦𝒏→∞
(√𝟏 −𝟏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏) − 𝟏) ⋅ 𝒏𝟐 = −
𝟏
𝟖
Therefore,
𝐥𝐢𝐦𝒏→∞
(√𝒏𝟐 − 𝒏
𝒏+𝟏
𝟒𝐬𝐢𝐧 (
𝟐
𝒏))
𝒏𝟐+𝐬𝐢𝐧(𝟑𝒏)
= 𝒆−𝟏𝟖
1592. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
√𝐜𝐨𝐬𝟐𝒏𝝅
𝟕− 𝟐𝟏−𝟐𝒏 ∑ (
𝒏
𝒊) (𝒏
𝒋) 𝐜𝐨𝐬
𝟐(𝒋 − 𝒊)𝝅
𝟕𝟎≤𝒊<𝑗≤𝑛
𝒏
Proposed by Daniel Sitaru-Romania
Solution 1 by Surjeet Singhania-India
𝐃𝐞𝐧𝐨𝐭𝐞:𝑿𝒏 = ∑ (𝒏
𝒊) (𝒏
𝒋) 𝐜𝐨𝐬
𝟐𝝅(𝒋 − 𝒊)
𝟕𝟎≤𝒊<𝑗≤𝑛
Observe that 𝟎 ≤ 𝒊 < 𝑗 ≤ 𝑛 it meand 𝒊 can take value from 𝟎 to 𝒏 − 𝟏 and 𝒋 take value
from 𝒊 + 𝟏 to 𝒏,
𝑿𝒏 = ∑ (𝒏
𝒊) (𝒏
𝒋) 𝐜𝐨𝐬
𝟐(𝒋 − 𝒊)𝝅
𝟕𝟎≤𝒊<𝑗≤𝑛
= ∑ ∑ (𝒏
𝒊) (𝒏
𝒋) 𝐜𝐨𝐬
𝟐(𝒋 − 𝒊)𝝅
𝟕
𝒏
𝒋=𝒊+𝟏
=
𝒏−𝟏
𝒊=𝟎
= ∑(𝒏
𝒋)
𝒏−𝟏
𝒊=𝟎
∑ (𝒏
𝒋) 𝐜𝐨𝐬
𝟐𝝅(𝒋 − 𝒊)
𝟕
𝒏
𝒋=𝒊+𝟏
=𝒋−𝒊=𝒌
∑∑(𝒏
𝒊) (
𝒏
𝒌 + 𝒊) 𝐜𝐨𝐬
𝟐𝒌𝝅
𝟕
𝒏
𝒌=𝟏
𝒏−𝟏
𝒊=𝟎
=
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133 RMM-CALCULUS MARATHON 1501-1600
= ∑∑(𝒏
𝒊)(
𝒌
𝒌 + 𝒊) 𝐜𝐨𝐬
𝟐𝒌𝝅
𝟕
𝒏−𝟏
𝒊=𝟏
𝒏
𝒌=𝟏
=𝟏
𝟐𝝅𝒊∑∮
(𝟏 + 𝒛)𝒏
𝒛𝒌+𝟏∑(𝒏𝒊)𝟏
𝒛𝒊𝒅𝒛
𝒏−𝟏
𝒊=𝟎
𝐜𝐨𝐬 (𝟐𝒌𝝅
𝟕)
𝒏
𝒌=𝟏
=
=𝟏
𝟐𝝅𝒊∑𝐜𝐨𝐬 (
𝟐𝒌𝝅
𝟕)∮(
(𝟏 + 𝒛)𝟐𝒏
𝒛𝒏+𝒌+𝟏−(𝟏 + 𝒛)𝒏
𝒛𝒌+𝟏+𝒏)𝒅𝒛
𝒏
𝒌=𝟏
=∑(𝟐𝒏
𝒏 + 𝒌) 𝐜𝐨𝐬 (
𝟐𝒌𝝅
𝟕)
𝒏
𝒌=𝟏
=
=∑(𝟐𝒏
𝒏 − 𝒌)
𝒏
𝒌=𝟏
𝐜𝐨𝐬 (𝟐𝒌𝝅
𝟕) = ∑(
𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕
𝒏−𝟏
𝒌=𝟏⏟ 𝑨
𝑨𝒍𝒔𝒐,𝑿𝒏 =∑(𝟐𝒏
𝒏 + 𝒌)
𝒏
𝒌=𝟏
𝐜𝐨𝐬𝟐𝒌𝝅
𝟕= ∑ (
𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕
𝟐𝒏
𝒌=𝒏+𝟏
⇒ 𝑿𝒏 =𝟏
𝟐∑(
𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕
𝟐𝒏
𝒌=𝟎
−𝟏
𝟐(𝟐𝒏
𝒏)
𝑿𝒏 =𝟏
𝟐𝕽∑(
𝟐𝒏
𝒌) 𝐞𝐱𝐩 (
𝟐𝝅(𝒏 − 𝒌)
𝟕)
𝟐𝒏
𝒌=𝟎
−𝟏
𝟐(𝟐𝒏
𝒏) =
=𝟏
𝟐𝕽(𝐞𝐱𝐩 (
𝟐𝝅𝒊𝒏
𝟕)∑(
𝟐𝒏
𝒌)
𝟐𝒏
𝒌=𝟎
𝐞𝐱𝐩 (−𝟐𝝅𝒊𝒌
𝟕))−
𝟏
𝟐(𝟐𝒏
𝒏) =
=𝟏
𝟐𝕽(𝐞𝐱𝐩 (
𝟐𝝅𝒊
𝟕) (𝟏 + 𝐞𝐱𝐩 (−
𝟐𝝅
𝟕))𝟐𝒏
) −𝟏
𝟐(𝟐𝒏
𝒏)
𝑿𝒏 = ∑ (𝒏
𝒊) (𝒏
𝒋) 𝐜𝐨𝐬
𝟐𝝅(𝒋 − 𝒊)
𝟕𝟎≤𝒊<𝑗≤𝑛
= 𝟐𝟐𝒏−𝟏 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅
𝟕) −
𝟏
𝟐(𝟐𝒏
𝒏)
𝐥𝐢𝐦𝒏→∞
√𝐜𝐨𝐬𝟐𝒏 (𝝅
𝟕) − 𝟐𝟏−𝟐𝒏𝑿𝒏
𝒏= 𝐥𝐢𝐦𝒏→∞
√𝟒−𝒏 (𝟐𝒏
𝒏)
𝒏
= 𝐥𝐢𝐦𝒏→∞
√𝟏
√𝒏𝝅
𝒏
= 𝟏
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
√𝐜𝐨𝐬𝟐𝒏𝝅
𝟕− 𝟐𝟏−𝟐𝒏 ∑ (
𝒏
𝒊) (𝒏
𝒋) 𝐜𝐨𝐬
𝟐(𝒋 − 𝒊)𝝅
𝟕𝟎≤𝒊<𝑗≤𝑛
𝒏= 𝟏.
Solution 2 by Ravi Prakash-New Delhi-India
𝑺 = ∑ (𝒏
𝒊) (𝒏
𝒋) 𝐜𝐨𝐬
𝟐(𝒋 − 𝒊)𝝅
𝟕𝟎≤𝒊<𝑗≤𝑛
=
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134 RMM-CALCULUS MARATHON 1501-1600
= ∑ (𝒏
𝒊) (𝒏
𝒋) 𝐜𝐨𝐬
𝟐(𝒋 − 𝒊)𝝅
𝟕𝟎≤𝒊<𝑗≤𝑛
+ ∑ (𝒏
𝒊)(𝒏
𝒋) 𝐜𝐨𝐬
𝟐(𝒊 − 𝒋)𝝅
𝟕𝟎≤𝒊<𝑗≤𝑛
=
= 𝑹𝒆 [∑(𝒏
𝒊) (𝒏
𝒋) 𝒆
(𝒊−𝒋)𝝅𝟕
𝒊≠𝒋
] ⇒∑(𝒏
𝒋)𝟐
𝒏
𝒋=𝟎
+ 𝑺 =
= 𝑹𝒆 {[(𝒏
𝟎)+ (
𝒏
𝟏)𝒆
𝒊𝝅𝟕 + (
𝒏
𝟐)𝒆
𝟐𝝅𝒊𝟕 +⋯+ (
𝒏
𝒏)𝒆
𝒏𝝅𝒊𝟕 ]
⋅ [(𝒏
𝟎) + (
𝒏
𝟏)𝒆−
𝒊𝝅𝟕 + (
𝒏
𝟐)𝒆−
𝟐𝝅𝒊𝟕 +⋯+ (
𝒏
𝒏)𝒆−
𝒏𝝅𝒊𝟕 ]}
= 𝑹𝒆 [(𝟏 + 𝒆𝒊𝝅𝟕 )
𝒏
(𝟏 + 𝒆−𝒊𝝅𝟕 )
𝒏
] = 𝑹𝒆 [(𝟏 + 𝟐𝐜𝐨𝐬𝝅𝒊
𝟕+ 𝟏)
𝒏
] =
= 𝑹𝒆[𝟐𝒏 (𝟏 + 𝐜𝐨𝐬𝝅
𝟕)𝒏
] = 𝟐𝒏 (𝟏 + 𝐜𝐨𝐬𝝅
𝟕)𝒏
= 𝟐𝒏 (𝟐 𝐜𝐨𝐬𝟐𝟐𝝅
𝟕)𝒏
= 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅
𝟕)
⇒ 𝑺 = 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅
𝟕)−∑(
𝒏
𝒋)𝟐
𝒏
𝒋=𝟎
= 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅
𝟕) − (
𝟐𝒏
𝒏)
⇒ 𝟐−𝟐𝒏 = 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅
𝟕) −
𝟏
𝟐𝟐𝒏(𝟐𝒏
𝒏)
⇒ 𝐜𝐨𝐬𝟐𝒏 (𝟐𝝅
𝟕) − 𝟐𝟏−𝟐𝒏 ∑ (
𝒏
𝒊)(𝒏
𝒋) 𝐜𝐨𝐬
(𝒋 − 𝒊)𝝅
𝟕𝟎≤𝒊<𝑗≤𝑛
=𝟏
𝟐𝟐𝒏(𝟐𝒏
𝒏)
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
√𝐜𝐨𝐬𝟐𝒏𝝅
𝟕− 𝟐𝟏−𝟐𝒏 ∑ (
𝒏
𝒊)(𝒏
𝒋) 𝐜𝐨𝐬
𝟐(𝒋 − 𝒊)𝝅
𝟕𝟎≤𝒊<𝑗≤𝑛
𝒏= 𝐥𝐢𝐦𝒏→∞
√𝟏
𝟐𝟐𝒏(𝟐𝒏
𝒏)
𝒏
=
=𝑪−𝑫 𝟏
𝟒𝐥𝐢𝐦𝒏→∞
(𝟐𝒏 + 𝟐)(𝟐𝒏 + 𝟏)
(𝒏 + 𝟏)(𝒏 + 𝟏)= 𝟏
1593. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌) (𝐭𝐚𝐧−𝟏 (
𝟏
√𝒌))𝟐
𝒏
𝒌=𝟏
Proposed by Florică Anastase-Romania
Solution 1 by Asmat Qatea-Afghanistan
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐∑𝐥𝐨𝐠(𝟏 +
𝟏
𝒌)(𝐭𝐚𝐧−𝟏 (
𝟏
√𝒌))𝟐
𝒏
𝒌=𝟏
=𝑪−𝑺
= 𝐥𝐢𝐦𝒏→∞
∑ 𝐥𝐨𝐠 (𝟏 +𝟏𝒌) (𝐭𝐚𝐧
−𝟏 (𝟏
√𝒌))𝟐
𝒏+𝟏𝒌=𝟏 –∑ 𝐥𝐨𝐠 (𝟏 +
𝟏𝒌)(𝐭𝐚𝐧
−𝟏 (𝟏
√𝒌))𝟐
𝒏𝒌=𝟏
(𝒏 + 𝟏)𝟐 − 𝒏𝟐=
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135 RMM-CALCULUS MARATHON 1501-1600
= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠 (𝟏 +𝟏
𝒏 + 𝟏)(𝐭𝐚𝐧−𝟏 (
𝟏
√𝒏 + 𝟏))𝟐
𝟐𝒏 + 𝟏= 𝟎
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌) (𝐭𝐚𝐧−𝟏 (
𝟏
√𝒌))𝟐
𝒏
𝒌=𝟏
= 𝟎.
Solution 2 Ruxandra Daniela Tonilă-Romania
We have: |𝐭𝐚𝐧−𝟏 𝒙| ≤𝝅
𝟐, ∀𝒙 ∈ ℝ ⇔ (𝐭𝐚𝐧−𝟏 𝒙)𝟐 ≤
𝝅𝟐
𝟐, ∀𝒙 ∈ ℝ and 𝛀 ≥ 𝟎. Thus,
𝟎 ≤ 𝛀 ≤𝝅𝟐
𝟒𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌)
𝒏
𝒌=𝟏
𝟎 ≤ 𝛀 ≤ 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐𝐥𝐨𝐠 (∏
𝒌+ 𝟏
𝒌
𝒏
𝒌=𝟏
) ⇔ 𝟎 ≤ 𝛀 ≤ 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐𝐥𝐨𝐠 (
(𝒏 + 𝟏)!
𝒏!) ⇔
𝟎 ≤ 𝛀 ≤𝝅𝟐
𝟒𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠(𝒏 + 𝟏)
𝒏𝟐= 𝟎
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌) (𝐭𝐚𝐧−𝟏 (
𝟏
√𝒌))𝟐
𝒏
𝒌=𝟏
= 𝟎.
Solution 3 by Ravi Prakash-New Delhi-India
For 𝒙 > 𝟎, we have: 𝟎 < 𝐭𝐚𝐧−𝟏 𝒙 < 𝒙 and 𝟎 < 𝐥𝐨𝐠(𝟏 + 𝒙) < 𝒙. Thus,
𝟎 < 𝐭𝐚𝐧−𝟏 (𝟏
√𝒌) <
𝟏
√𝒌, ∀𝒌 > 𝟎 and 𝟎 < 𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌) <
𝟏
𝒌, ∀𝒌 > 𝟎.
𝟎 < 𝐥𝐨𝐠 (𝟏 +𝟏
𝒌) (𝐭𝐚𝐧−𝟏 (
𝟏
√𝒌))𝟐
≤𝟏
𝒌𝟐≤ 𝟏,∀𝒌 ≥ 𝟏
Hence,
𝟎 ≤∑𝐥𝐨𝐠 (𝟏 +𝟏
𝒌) (𝐭𝐚𝐧−𝟏 (
𝟏
√𝒌))𝟐
𝒏
𝒌=𝟏
≤ 𝒏, ∀𝒏 ∈ ℕ,𝒏 ≥ 𝟏
𝟎 ≤𝟏
𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌)(𝐭𝐚𝐧−𝟏 (
𝟏
√𝒌))𝟐
𝒏
𝒌=𝟏
≤𝟏
𝒏,∀𝒏 ∈ ℕ,𝒏 ≥ 𝟏
Therefore,
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136 RMM-CALCULUS MARATHON 1501-1600
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐∑𝐥𝐨𝐠 (𝟏 +
𝟏
𝒌) (𝐭𝐚𝐧−𝟏 (
𝟏
√𝒌))𝟐
𝒏
𝒌=𝟏
= 𝟎.
1594. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
√𝐬𝐢𝐧𝟐𝒏𝝅
𝟕− 𝟐𝟏−𝟐𝒏∑(−𝟏)𝒏−𝒌 (
𝟐𝒏
𝒏) 𝐜𝐨𝐬
(𝟐𝒏 − 𝟐𝒌)𝝅
𝟕
𝒏−𝟏
𝒌=𝟎
𝒏
Proposed by Daniel Sitaru-Romania
Solution by Kamel Gandouli Rezgui-Tunisia
𝒌𝒏 = ∑(−𝟏)𝒏−𝒌 (𝟐𝒏
𝒏) 𝐜𝐨𝐬
(𝟐𝒏 − 𝟐𝒌)𝝅
𝟕
𝒏−𝟏
𝒌=𝟎
=∑(−𝟏)𝒌 (𝟐𝒏
𝒏 − 𝒌) 𝐜𝐨𝐬
𝟐𝒌𝝅
𝟕
𝒏
𝒌=𝟏
=
∑(−𝟏)𝒌 (𝟐𝒏
𝒏 + 𝒌) 𝐜𝐨𝐬
𝟐𝒌𝝅
𝟕
𝒏
𝒌=𝟏
= ∑ (−𝟏)𝒌−𝒏 (𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕
𝟐𝒏
𝒌=𝒏+𝟏
=
= ∑(−𝟏)𝒏−𝒌 (𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕=
𝟐𝒏
𝒌=𝟎
= ∑(−𝟏)𝒏−𝒌 (𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕
𝒏−𝟏
𝒌=𝟎
+ ∑ (−𝟏)𝒏−𝒌 (𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕
𝟐𝒏
𝒌=𝒏+𝟏
+ (𝟐𝒏
𝒏)
= 𝒌𝒏 + 𝒌𝒏 + (𝟐𝒏
𝒏)
⇒ 𝒌𝒏 =𝟏
𝟐(∑(−𝟏)𝒏−𝒌 (
𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕
𝟐𝒏
𝒌=𝟎
− (𝟐𝒏
𝒏))
∑(−𝟏)𝒏−𝒌 (𝟐𝒏
𝒌) 𝐜𝐨𝐬
𝟐𝝅(𝒏 − 𝒌)
𝟕
𝟐𝒏
𝒌=𝟎
= 𝑹𝒆(∑(−𝟏)𝒏−𝒌 (𝟐𝒏
𝒌) 𝒆
𝒊𝟐𝝅(𝒏−𝒌)𝟕
𝟐𝒏
𝒏=𝟎
) =
= 𝑹𝒆(∑(−𝟏)𝒏−𝒌𝒆𝟐𝒊𝝅𝒌𝟕 (
𝟐𝒏
𝒌) 𝒆𝒊⋅(−
𝟐𝝅𝒌𝟕)
𝟐𝒏
𝒌=𝟎
) = (−𝟏)𝒏𝑹(∑𝒆𝟐𝒊𝝅𝒏𝟕
𝟐𝒏
𝒌=𝟎
(𝟐𝒏
𝒌) 𝒆𝒊⋅
𝟓𝝅𝒌𝟕 ) =
= (−𝟏)𝒏𝑹𝒆(𝒆𝟐𝒊𝝅𝒏𝟕 (𝟏 + 𝒆𝒊(
𝟓𝒌𝝅𝟕))𝟐𝒏
) = (−𝟏)𝒏𝑹𝒆(𝒆𝟐𝒊𝝅𝒏𝟕 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏
𝟓𝝅
𝟕𝒆𝟓𝒊𝝅𝟏𝟒 )
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137 RMM-CALCULUS MARATHON 1501-1600
= (−𝟏)𝒏𝑹𝒆 (𝒆𝟐𝒊𝝅𝒏𝟕 𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏
𝟓𝝅
𝟏𝟒𝒆𝟓𝒏𝒊𝝅𝟕 ) = (−𝟏)𝟏+𝒏𝟐𝟐𝒏 𝐜𝐨𝐬𝟐𝒏
𝟓𝝅
𝟕
𝒌𝒏 = (−𝟏)𝟏+𝒏𝟐𝟐𝒏−𝟏 𝐜𝐨𝐬𝟐𝒏
𝟓𝝅
𝟕−𝟏
𝟐(𝟐𝒏
𝒏)
𝟐𝟏−𝟐𝒏𝒌𝒏 = (−𝟏)𝟏+𝒏 𝐜𝐨𝐬𝟐𝒏
𝟓𝝅
𝟕−𝟏
𝟐𝟐𝒏(𝟐𝒏
𝒏)
⇒ 𝐬𝐢𝐧𝟐𝒏𝝅
𝟕− 𝟐𝟏−𝟐𝒏𝒌𝒏 = 𝐬𝐢𝐧
𝟐𝒏𝝅
𝟕+ (−𝟏)𝒏 𝐜𝐨𝐬𝟐𝒏
𝝅
𝟕⏟ →𝟎
+ 𝟐−𝟐𝒏 (𝟐𝒏
𝒏)
𝛀 = 𝐥𝐢𝐦𝒏→∞
√(𝟐𝒏)!
𝟐𝟐𝒏(𝒏!)𝟐𝒏
= 𝐞𝐱𝐩 (𝐥𝐢𝐦𝒏→∞
𝟏
𝒏(𝐥𝐨𝐠(𝟐𝒏)! − 𝐥𝐨𝐠(𝟐𝟐𝒏) − 𝟐 𝐥𝐨𝐠(𝒏!))) =
= 𝐞𝐱𝐩 (𝐥𝐢𝐦𝒏→∞
𝟏
𝒏(∑ 𝐥𝐨𝐠 𝒌 −∑𝐥𝐨𝐠𝒌
𝒏
𝒌=𝟏
𝟐𝒏
𝒌=𝟏
)) = 𝐞𝐱𝐩(𝐥𝐢𝐦𝒏→∞
𝟏
𝒏∑𝐥𝐨𝐠 (𝟏 +
𝒌
𝒏)
𝒏
𝒌=𝟏
) =
= 𝐞𝐱𝐩 (∫ 𝐥𝐨𝐠 𝒙𝒅𝒙𝟐
𝟏
) =𝟏
𝒆
1595. −𝟏 < 𝒂 ≤ 𝒃 < 𝟏, 𝒏 ∈ ℕ∗, 𝑷𝒏 −Legendre’s polynomials. Find:
𝛀(𝒂, 𝒃) = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏∫
𝑷𝒏′ (𝒙)
𝑷𝒏−𝟏(𝒙) − 𝒙𝑷𝒏(𝒙)𝒅𝒙
𝒃
𝒂
Proposed by Daniel Sitaru-Romania
Solution by Amrit Awasthi-Punjab-India
It is known that:
𝒅
𝒅𝒙𝑷𝒏(𝒙) =
𝒏
𝒙𝟐 − 𝟏(𝑷𝒏−𝟏(𝒙) − 𝒙𝑷𝒏(𝒙))
Rearrange and integrating:
∫𝑷𝒏′ (𝒙)
𝑷𝒏−𝟏(𝒙) − 𝒙𝑷𝒏(𝒙)𝒅𝒙
𝒃
𝒂
= ∫𝒏
𝟏 − 𝒙𝟐𝒅𝒙
𝒃
𝒂
=𝒏
𝟐𝒍𝒐𝒈(
(𝟏 + 𝒃)(𝟏 − 𝒂)
(𝟏 + 𝒂)(𝟏 − 𝒃))
𝛀(𝒂, 𝒃) = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏∫
𝑷𝒏′ (𝒙)
𝑷𝒏−𝟏(𝒙) − 𝒙𝑷𝒏(𝒙)𝒅𝒙
𝒃
𝒂
=𝟏
𝟐𝒍𝒐𝒈 (
(𝟏 + 𝒃)(𝟏 − 𝒂)
(𝟏 + 𝒂)(𝟏 − 𝒃))
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138 RMM-CALCULUS MARATHON 1501-1600
Note by editors (Daniel Sitaru, Florică Anastase)
∑𝑷𝒏(𝒙)𝒕𝒏
∞
𝒏=𝟎
= (𝟏 − 𝟐𝒕𝒙 + 𝒕𝟐)−𝟏𝟐 = 𝑲(𝒙, 𝒕)
−𝟏
𝟐⋅ (−𝟐𝒙 + 𝟐𝒕)(𝟏 − 𝟐𝒕𝒙 + 𝒕𝟐)−
𝟑𝟐 =
𝝏𝑲
𝝏𝒕
(𝒙 − 𝒕)(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)−𝟏 ⋅ 𝑲(𝒙, 𝒕) =𝝏𝑲
𝝏𝒕
(𝒙 − 𝒕)𝑲(𝒙, 𝒕) = (𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)𝝏𝑲
𝝏𝒕
(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)𝝏𝑲
𝝏𝒕+ (𝒕 − 𝒙)𝑲(𝒙, 𝒕) = 𝟎
(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)∑𝒏𝑷𝒏(𝒙)𝒕𝒏−𝟏
∞
𝒏=𝟎
+ (𝒕 − 𝒙)∑𝑷𝒏(𝒙)𝒕𝒏
∞
𝒏=𝟎
= 𝟎
Coefficient of 𝒕𝒏 is:
∑𝒏𝑷𝒏(𝒙)𝒕𝒏−𝟏
∞
𝒏=𝟎
− 𝟐𝒙𝒏∑𝑷𝒏(𝒙)𝒕𝒏
∞
𝒏=𝟎
+ 𝒏𝒙𝟐∑𝑷𝒏(𝒙)𝒕𝒏−𝟏
∞
𝒏=𝟎
+∑𝑷𝒏(𝒙)𝒕𝒏+𝟏
∞
𝒏=𝟎
− 𝒙∑𝑷𝒏(𝒙)𝒕𝒏
∞
𝒏=𝟎
= 𝟎
−𝟐𝒙𝒏𝑷𝒏(𝒙) − 𝒙𝑷𝒏(𝒙) + 𝒏𝑷𝒏−𝟏(𝒙) + (𝒏 + 𝟏)𝑷𝒏+𝟏(𝒙) = 𝟎
(𝒏 + 𝟏)𝑷𝒏+𝟏(𝒙) − (𝟐𝒏 + 𝟏)𝒙𝑷𝒏(𝒙) + 𝒏𝑷𝒏−𝟏(𝒙) = 𝟎; (𝟏)
(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐)𝝏𝑲
𝝏𝒙− 𝒕𝑲(𝒙, 𝒕) =
= (𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐) ⋅ (−𝟏
𝟐) ⋅ (−𝟐𝒕) ⋅ (𝟏 − 𝟐𝒕𝒙 + 𝒕𝟐)−
𝟑𝟐 − 𝒕𝑲(𝒙, 𝒕) =
= 𝒕 ⋅ 𝑲(𝒙, 𝒕) − 𝒕 ⋅ 𝑲(𝒙, 𝒕) = 𝟎
(𝟏 − 𝟐𝒕𝒙 + 𝒙𝟐) ⋅ ∑𝑷𝒏′ (𝒙)𝒕𝒏
∞
𝒏=𝟎
− 𝒕 ⋅ ∑𝑷𝒏(𝒙)𝒕𝒏
∞
𝒏=𝟎
= 𝟎
Coefficient of 𝒕𝒏+𝟏 is:
𝑷𝒏+𝟏′ (𝒙) − 𝟐𝒙𝑷𝒏
′ (𝒙) + 𝑷𝒏−𝟏′ (𝒙) − 𝑷𝒏(𝒙) = 𝟎; (𝟐)
Derivative of (1):
(𝒏 + 𝟏)𝑷𝒏+𝟏′ (𝒙) − (𝟐𝒏 + 𝟏)𝑷𝒏(𝒙) − (𝟐𝒏 + 𝟏)𝒙𝑷𝒏
′ (𝒙) + 𝒏𝑷𝒏−𝟏′ (𝒙) = 𝟎; (𝟑)
By (2): 𝑷𝒏−𝟏′ (𝒙) = 𝑷𝒏(𝒙) + 𝟐𝒙𝑷𝒏
′ (𝒙) − 𝑷𝒏+𝟏′ (𝒙).
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139 RMM-CALCULUS MARATHON 1501-1600
Replace in (3):
(𝒏 + 𝟏)𝑷𝒏+𝟏′ (𝒙) − (𝟐𝒏 + 𝟏)𝑷𝒏(𝒙) − (𝟐𝒏 + 𝟏)𝒙𝑷𝒏
′ (𝒙) + 𝒏𝑷𝒏(𝒙) + 𝟐𝒏𝒙𝑷𝒏′ (𝒙) − 𝒏𝑷𝒏+𝟏
′ = 𝟎
𝑷𝒏+𝟏′ (𝒙) − 𝒙𝑷𝒏
′ (𝒙) = (𝒏 + 𝟏)𝑷𝒏(𝒙)
By (2): 𝑷𝒏+𝟏′ (𝒙) = 𝑷𝒏(𝒙) + 𝟐𝒙𝑷𝒏
′ (𝒙) − 𝑷𝒏−𝟏′ (𝒙)
Replace in (3):
(𝒏 + 𝟏)(𝑷𝒏(𝒙) + 𝟐𝒙𝑷𝒏′ (𝒙) − 𝑷𝒏−𝟏
′ (𝒙)) − (𝟐𝒏 + 𝟏)𝑷𝒏(𝒙) − (𝟐𝒏 + 𝟏)𝒙𝑷𝒏′ (𝒙) + 𝒏𝑷𝒏−𝟏
′ (𝒙) = 𝟎
𝒙𝑷𝒏′ (𝒙) − 𝑷𝒏−𝟏
′ (𝒙) = 𝒏𝑷𝒏(𝒙), 𝑷𝒏−𝟏′ (𝒙) = 𝒙𝑷𝒏
′ (𝒙) + 𝒏𝑷𝒏(𝒙)
𝑷𝒏′ (𝒙) − 𝒙(𝒙𝑷𝒏
′ (𝒙) − 𝒏𝑷𝒏(𝒙)) = 𝒏𝑷𝒏−𝟏(𝒙)
(𝟏 − 𝒙𝟐)𝑷𝒏′ (𝒙) = 𝒏(𝑷𝒏−𝟏(𝒙) − 𝒙𝑷𝒏(𝒙))
1596. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝟐∑
𝟏
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟏
)
𝒏
Proposed by Daniel Sitaru-Romania
Solution 1 by Asmat Qatea-Afghanistan
𝟐∑𝟏
𝟐𝒌+ 𝟓
𝒏
𝒌=𝟏
=𝟐
𝟕+𝟐
𝟗+𝟐
𝟏𝟏+𝟐
𝟏𝟑+𝟐
𝟏𝟓+𝟐
𝟏𝟕+∑
𝟐
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟕
𝟐∑𝟏
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟏
=𝟖𝟑𝟖𝟏𝟗𝟐
𝟕𝟔𝟓𝟕𝟔𝟓⏟ >𝟏
+∑𝟐
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟕
; ∀𝒏 ≥ 𝟔
Let put 𝑹 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝟏)𝒏; (𝛀 > 𝑹)
𝑹 = 𝐥𝐢𝐦𝒏→∞
𝟐𝒏
𝒏𝟐=𝑳′𝑯𝐥𝐢𝐦𝒏→∞
𝟐𝒏 𝐥𝐨𝐠𝟐
𝟐𝒏= 𝐥𝐢𝐦𝒏→∞
𝟐𝒏 𝐥𝐨𝐠𝟐 𝟐
𝟐= +∞
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝟐∑
𝟏
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟏
)
𝒏
= ∞
Solution 2 by Amrit Awasthi-Punjab-India
Using the definition of digamma function, we have:
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140 RMM-CALCULUS MARATHON 1501-1600
𝝍(𝒏 + 𝟏 + 𝒛) = 𝝍(𝒛 + 𝟏) +∑𝟏
𝒌 + 𝒛
𝒏
𝒌=𝟏
;
𝝍(𝒛) = 𝐥𝐨𝐠 𝒛 −𝟏
𝟐𝒛−
𝟏
𝟏𝟐𝒛𝟐+
𝟏
𝟏𝟐𝟎𝒛𝟒−
𝟏
𝟐𝟓𝟐𝒛𝟔+⋯
For 𝒛 =𝟓
𝟐⇒
𝟐∑𝟏
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟏
= 𝝍(𝒏 +𝟕
𝟐) −𝝍(
𝟕
𝟐)
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝟐∑
𝟏
𝟐𝒌+ 𝟓
𝒏
𝒌=𝟏
)
𝒏
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝝍(𝒏 +
𝟕
𝟐) − 𝝍(
𝟕
𝟐))
𝒏
=𝑳′𝑯
= 𝐥𝐢𝐦𝒏→∞
(𝐥𝐨𝐠 (𝒏 +𝟕𝟐) −
𝟏
𝟐(𝒏 +𝟕𝟐)−
𝟏
𝟏𝟐(𝒏 +𝟕𝟐)𝟐 +
𝟏
𝟏𝟐𝟎(𝒏 +𝟕𝟐)𝟒 −
𝟏
𝟐𝟓𝟐(𝒏 +𝟕𝟐)𝟔+. . )
𝒏−𝟏
𝟐
= 𝐥𝐢𝐦𝒏→∞
(𝐥𝐨𝐠 (𝒏 +𝟕𝟐))
𝒏−𝟏
𝟐= ∞
Solution 3 by Syed Shahabudeen-Kerala-India
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝟐∑
𝟏
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟏
)
𝒏
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝝍(𝒏 +
𝟕
𝟐) − 𝝍(
𝟕
𝟐))
𝒏
∵ 𝝍(𝒙) > 𝐥𝐨𝐠 (𝒙 +𝟏
𝟐) −
𝟏
𝒙⇒
𝛀 > 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝐥𝐨𝐠(𝒏 + 𝟒) −
𝟐
𝟐𝒏 + 𝟕+ 𝝍(
𝟕
𝟐))
𝒏
; (𝒍𝒆𝒕 𝟏 + 𝝍(𝟕
𝟐) = 𝒄)
> 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝐥𝐨𝐠(𝒏 + 𝟒) −
𝟐
𝟐𝒏 + 𝟕+ 𝒄)
𝒏
; (∵ 𝐥𝐨𝐠(𝟏 + 𝒙) >𝒙
𝟏 + 𝒙)
> 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝒏 + 𝟑
𝒏 + 𝟒−
𝟐
𝟐𝒏 + 𝟕+ 𝒄)
𝒏
> 𝐥𝐢𝐦𝒏→∞
(
𝒏 + 𝟑𝒏 + 𝟒 −
𝟐𝟐𝒏 + 𝟕
𝒏𝟐𝒏
+𝒄
𝒏𝟐𝒏
)
𝒏
>
> 𝐥𝐢𝐦𝒏→∞
(𝟏 + 𝒄)𝒏
Therefore,
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141 RMM-CALCULUS MARATHON 1501-1600
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝟐∑
𝟏
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟏
)
𝒏
= ∞
Solution 4 by Remus Florin Stanca-Romania
Let 𝒂𝒏 =𝟏
𝒏𝟐(𝟏 + 𝟐∑
𝟏
𝟐𝒌+𝟓
𝒏𝒌=𝟏 )
𝒏
.
𝐥𝐢𝐦𝒏→∞
𝒂𝒏+𝟏𝒂𝒏
= 𝐥𝐢𝐦𝒏→∞
𝟏(𝒏 + 𝟏)𝟐
(𝟏 + 𝟐∑𝟏
𝟐𝒌 + 𝟓𝒏+𝟏𝒌=𝟏 )
𝒏
𝟏𝒏𝟐(𝟏 + 𝟐∑
𝟏𝟐𝒌 + 𝟓
𝒏𝒌=𝟏 )
𝒏 =
= 𝐥𝐢𝐦𝒏→∞
(𝟏 + 𝟐∑𝟏
𝟐𝒌 + 𝟓
𝒏+𝟏
𝒌=𝟏
)(𝟏 + 𝟐∑
𝟏𝟐𝒌 + 𝟓
𝒏+𝟏𝒌=𝟏
𝟏 + 𝟐∑𝟏
𝟐𝒌 + 𝟓𝒏𝒌=𝟏
)
𝒏
𝟐𝒌 + 𝟓 ≤ 𝟐𝒌 + 𝟔 ⇒𝟐
𝟐𝒌 + 𝟓>
𝟏
𝟐𝒌 + 𝟑⇒ ∑
𝟏
𝟐𝒌 + 𝟓
𝒏+𝟏
𝒌=𝟏
≥ ∑𝟏
𝒌+ 𝟑
𝒏+𝟏
𝒌=𝟏
; (𝐥𝐢𝐦𝒌→∞
𝟏
𝒌 + 𝟑= ∞)
⇒ 𝐥𝐢𝐦𝒏→∞
∑𝟐
𝟐𝒌 + 𝟓
𝒏+𝟏
𝒌=𝟏
= ∞ ⇒ 𝟏 + 𝟐∑𝟐
𝟐𝒌 + 𝟓
𝒏+𝟏
𝒌=𝟏
= ∞; (𝟏)
𝒍𝒆𝒕 𝒖(𝒏) =𝟏 + 𝟐∑
𝟏𝟐𝒌 + 𝟓
𝒏+𝟏𝒌=𝟏
𝟏 + 𝟐∑𝟏
𝟐𝒌 + 𝟓𝒏𝒌=𝟏
𝐥𝐢𝐦𝒏→∞
(𝟏 + 𝟐∑
𝟏𝟐𝒌 + 𝟓
𝒏+𝟏𝒌=𝟏
𝟏 + 𝟐∑𝟏
𝟐𝒌 + 𝟓𝒏𝒌=𝟏
)
𝒏
= 𝐥𝐢𝐦𝒏→∞
(𝟏 + (𝒖(𝒏) − 𝟏))𝟏
𝒖(𝒏)−𝟏⋅𝒏(𝒖(𝒏)−𝟏)
=
= 𝐞𝐱𝐩 {𝐥𝐢𝐦𝒏→∞
𝟐𝒏
𝟐𝒏 + 𝟕⋅
𝟏
𝟏 + 𝟐∑𝟏
𝟐𝒌 + 𝟓𝒏𝒌=𝟏
} = 𝒆𝟎 = 𝟏; (𝟐)
From (1),(2) it follows that:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏𝟐(𝟏 + 𝟐∑
𝟏
𝟐𝒌 + 𝟓
𝒏
𝒌=𝟏
)
𝒏
= ∞
1597. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑
𝚪′(𝒌)
𝚪(𝒌)
𝒏
𝒌=𝟐
)
Proposed by Daniel Sitaru-Romania
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Solution 1 by Remus Florin Stanca-Romania
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑
𝚪′(𝒌)
𝚪(𝒌)
𝒏
𝒌=𝟐
) = 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠(𝒏!) − ∑ 𝝍(𝒌)𝒏𝒌=𝟐
𝒏(𝑯𝒏 − 𝟏)=𝑪−𝑺
= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠(𝒏 + 𝟏) − 𝝍(𝒏 + 𝟏)
(𝒏 + 𝟏)(𝑯𝒏 +𝟏
𝒏 + 𝟏) − 𝒏𝑯𝒏 + 𝒏=
= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠(𝒏 + 𝟏) − 𝝍(𝒏 + 𝟏)
𝒏𝑯𝒏 +𝑯𝒏 + 𝟏 − 𝒏− 𝟏 − 𝒏𝑯𝒏 + 𝒏= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠(𝒏 + 𝟏) − 𝝍(𝒏 + 𝟏)
𝑯𝒏=𝑪−𝑺
= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠 (𝒏 + 𝟐𝒏 + 𝟏) − 𝝍
(𝒏 + 𝟐) + 𝝍(𝒏 + 𝟏)
𝟏𝒏 + 𝟏
= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠 (𝒏 + 𝟐𝒏 + 𝟏) −
𝟏𝒏 + 𝟏
𝟏𝒏 + 𝟏
=
= 𝐥𝐢𝐦𝒏→∞
(𝒏 + 𝟏) 𝐥𝐨𝐠 (𝒏 + 𝟐
𝒏 + 𝟏) − 𝟏 = 𝐥𝐢𝐦
𝒏→∞𝐥𝐨𝐠 (𝟏 +
𝟏
𝒏 + 𝟏)𝒏+𝟏
− 𝟏 = 𝐥𝐨𝐠 𝒆 − 𝟏 = 𝟎
Solution 2 by Syed Shahabudeen-Kerala-India
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑
𝚪′(𝒌)
𝚪(𝒌)
𝒏
𝒌=𝟐
) =
= 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝚪(𝒏 + 𝟏)) − 𝝍(𝒏)(𝒏 − 𝟏)) =
=𝑪−𝑺
𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠 (𝚪(𝒏 + 𝟐)𝚪(𝒏 + 𝟏)
) − 𝝍(𝒏 + 𝟏)(𝒏) + 𝝍(𝒏)(𝒏 − 𝟏)
(𝒏 + 𝟏)(𝑯𝒏+𝟏 − 𝟏) − (𝒏)(𝑯𝒏 − 𝟏)=
= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠(𝒏 + 𝟏) − 𝝍(𝒏) − 𝟏
𝑯𝒏; (𝝍(𝒙)~ 𝐥𝐨𝐠 𝒙)
= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠 (𝒏 + 𝟏𝒏 ) − 𝟏
𝑯𝒏= 𝐥𝐢𝐦𝒏→∞
−𝟏
𝑯𝒏= 𝟎
Solution 3 by Kamel Gandouli Rezgui-Tunisia
𝐥𝐨𝐠 𝒏! −∑𝚪′(𝒌)
𝚪(𝒌)
𝒏
𝒌=𝟐
=∑𝐥𝐨𝐠 𝒌 −∑𝝍(𝒏)
𝒏
𝒌=𝟐
𝒏
𝒌=𝟏
= ∑𝐥𝐨𝐠𝒌
𝒏
𝒌=𝟏
−∑(𝑯𝒏−𝟏 − 𝜸)
𝒏
𝒌=𝟐
⇒
𝟏
𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑
𝚪′(𝒌)
𝚪(𝒌)
𝒏
𝒌=𝟐
) =𝟏
𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠 𝒏! − (𝒏 − 𝟏)(𝑯𝒏−𝟏 − 𝜸))
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143 RMM-CALCULUS MARATHON 1501-1600
=𝟏
𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠 𝒏! − (𝒏 − 𝟏) 𝐥𝐨𝐠(𝒏 − 𝟏)); (∵ 𝑯𝒏−𝟏 − 𝜸 ≅ 𝐥𝐨𝐠(𝒏 − 𝟏))
𝐥𝐨𝐠 𝒏! = 𝐥𝐨𝐠(𝒏 + 𝟏) ≅ (𝒏 +𝟏
𝟐) 𝐥𝐨𝐠(𝒏 + 𝟏) − 𝒏 − 𝟏 +
𝟏
𝟐𝐥𝐨𝐠𝟐𝝅
(𝐥𝐨𝐠𝒏! − (𝒏 − 𝟏) 𝐥𝐨𝐠(𝒏 − 𝟏)) ≅ (𝒏 +𝟏
𝟐) 𝐥𝐨𝐠(𝒏 + 𝟏) − 𝒏 − 𝟏 +
𝟏
𝟐𝐥𝐨𝐠 𝟐𝝅 −
−(𝒏 − 𝟏) 𝐥𝐨𝐠(𝒏 − 𝟏)
(𝒏+𝟏
𝟐) 𝐥𝐨𝐠(𝒏+𝟏)
𝒏(𝑯𝒏−𝟏)≅ 𝟏 and
(𝒏−𝟏) 𝐥𝐨𝐠(𝒏−𝟏)
𝒏(𝑯𝒏−𝟏)≅ 𝟏
𝒏 − 𝟏 +𝟏𝟐𝐥𝐨𝐠 𝟐𝝅
𝒏(𝑯𝒏 − 𝟏)→ 𝟎
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
𝟏
𝒏(𝑯𝒏 − 𝟏)(𝐥𝐨𝐠(𝒏!) −∑
𝚪′(𝒌)
𝚪(𝒌)
𝒏
𝒌=𝟐
)
1598.
Find a closed form:
𝛀 = ∑𝟏
𝒏∫ (𝟏 + 𝒙𝟐)−𝒏 ⋅ 𝐭𝐚𝐧−𝟏 𝒙𝟏
𝟎
𝒅𝒙
∞
𝒌=𝟏
Proposed by Ajetunmobi Abdulqoyyum-Nigeria
Solution 1 by Remus Florin Stanca-Romania
𝛀 =∑𝟏
𝒏∫ (𝟏 + 𝒙𝟐)−𝒏 ⋅ 𝐭𝐚𝐧−𝟏 𝒙𝟏
𝟎
𝒅𝒙
∞
𝒌=𝟏
= ∫ 𝐭𝐚𝐧−𝟏 𝒙 ⋅∑(
𝟏𝟏+ 𝒙𝟐
)𝒏
𝒏
∞
𝒏=𝟏
𝒅𝒙𝟏
𝟎
∑𝒙𝒏
𝒏
∞
𝒏=𝟏
= ∫(∑𝒙𝒏−𝟏∞
𝒏=𝟏
)𝒅𝒙 = ∫𝟏
𝟏 − 𝒙𝒅𝒙 , 𝒇𝒐𝒓 |𝒙| < 𝟏 ⇒
∑𝒙𝒏
𝒏
∞
𝒏=𝟏
= − 𝐥𝐨𝐠 |𝒙 − 𝟏| ⇒ ∑(
𝟏𝒙𝟐 + 𝟏
)𝒏
𝒏
∞
𝒏=𝟏
= − 𝐥𝐨𝐠 |𝒙𝟐
𝒙𝟐 + 𝟏| = 𝐥𝐨𝐠 (
𝒙𝟐 + 𝟏
𝒙𝟐)
𝛀 = ∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏)𝟏
𝟎
𝒅𝒙 − 𝟐∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
; (𝟏)
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∫𝐥𝐨𝐠(𝒙𝟐 + 𝟏) ⋅ 𝒙′𝒅𝒙 = 𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) −∫𝟐𝒙𝟐 + 𝟐 − 𝟐
𝒙𝟐 + 𝟏𝒅𝒙
= 𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) − 𝟐𝒙 + 𝟐 𝐭𝐚𝐧−𝟏 𝒙
∫ 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) 𝐭𝐚𝐧−𝟏 𝒙𝟏
𝟎
𝒅𝒙 = (𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) − 𝟐𝒙 + 𝟐 𝐭𝐚𝐧−𝟏 𝒙)|𝟎
𝟏−
−∫ (𝒙 𝐥𝐨𝐠(𝒙𝟐 + 𝟏)
𝒙𝟐 + 𝟏−
𝟐𝒙
𝒙𝟐 + 𝟏+𝟐 𝐭𝐚𝐧−𝟏 𝒙
𝒙𝟐 + 𝟏)𝒅𝒙
𝟏
𝟎
=
= (𝐥𝐨𝐠𝟐 − 𝟐 +𝝅
𝟐) ⋅𝝅
𝟒− (𝐥𝐨𝐠𝟐(𝒙𝟐 + 𝟏) − 𝐥𝐨𝐠(𝒙𝟐 + 𝟏) + (𝐭𝐚𝐧−𝟏 𝒙)𝟐)|
𝟎
𝟏=
=𝝅
𝟒𝐥𝐨𝐠 𝟐 −
𝝅
𝟐+𝝅𝟐
𝟖−𝟏
𝟒𝐥𝐨𝐠𝟐 𝟐 −
𝝅𝟐
𝟏𝟔; (𝟐)
∫ 𝐥𝐨𝐠𝒙 𝐭𝐚𝐧−𝟏 𝒙𝟏
𝟎
𝒅𝒙 = (𝒙 𝐥𝐨𝐠𝒙 − 𝒙) 𝐭𝐚𝐧−𝟏 𝒙|𝟎
𝟏−∫ (
𝒙 𝐥𝐨𝐠𝒙
𝒙𝟐 + 𝟏−
𝒙
𝒙𝟐 + 𝟏)𝒅𝒙
𝟏
𝟎
=
= −𝝅
𝟒−∫
𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟐 + 𝟏
𝟏
𝟎
𝒅𝒙 +𝟏
𝟐𝐥𝐨𝐠(𝒙𝟐 + 𝟏)|
𝟎
𝟏
= −𝝅
𝟒+𝟏
𝟐𝐥𝐨𝐠 𝟐 − ∫
𝒙 𝐥𝐨𝐠𝒙
𝒙𝟐 + 𝟏𝒅𝒙
𝟏
𝟎
𝟏
𝟐∫𝟐𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟐 + 𝟏
𝟏
𝟎
𝒅𝒙 =𝟏
𝟐(−∫
𝐥𝐨𝐠(𝒙𝟐 + 𝟏)
𝒙𝒅𝒙
𝟏
𝟎
)
∫𝐥𝐨𝐠(𝒙𝟐 + 𝟏)
𝒙𝟐 + 𝟏𝒅𝒙
𝟏
𝟎
=𝒙𝟐=𝒕
∫𝐥𝐨𝐠(𝒕 + 𝟏)
√𝒕⋅𝟏
𝟐√𝒕
𝟏
𝟎
𝒅𝒕 =𝟏
𝟐∫𝐥𝐨𝐠(𝟏 + 𝒕)
𝒕
𝟏
𝟎
𝒅𝒕 =
=𝟏
𝟐∑(−𝟏)𝒏−𝟏∞
𝒏=𝟏
∫𝒕𝒏−𝟏
𝒏𝒅𝒕
𝟏
𝟎
=𝟏
𝟐∑(−𝟏)𝒏−𝟏
𝒏𝟐
∞
𝒏=𝟏
=𝝅𝟐
𝟐𝟒⇒
∫𝒙 𝐥𝐨𝐠 𝒙
𝒙𝟐 + 𝟏
𝟏
𝟎
𝒅𝒙 = −𝝅𝟐
𝟒𝟖⇒ ∫ 𝐥𝐨𝐠 𝒙 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙
𝟏
𝟎
= −𝝅
𝟒+𝟏
𝟐𝐥𝐨𝐠 𝟐 +
𝝅𝟐
𝟒𝟖; (𝟑)
From (1),(2),(3) it follows that:
𝛀 =𝟏
𝟒𝟖(𝝅𝟐 + 𝟏𝟐𝝅 𝐥𝐨𝐠 𝟐 − 𝟏𝟐 𝐥𝐨𝐠𝟐 𝟐)
Solution 2 by Amrit Awasthi-India
𝛀 = ∫ ∑((𝟏 + 𝒙𝟐)−𝟏)𝒏
𝒏
∞
𝒏=𝟏
𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝟏
𝟎
= ∫ − 𝐥𝐨𝐠 (𝟏 −𝟏
𝟏 + 𝒙𝟐) 𝐭𝐚𝐧−𝟏 𝒙
𝟏
𝟎
𝒅𝒙 =
= ∫ 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) 𝐭𝐚𝐧−𝟏 𝒙 𝒅𝒙𝟏
𝟎
− 𝟐∫ 𝐥𝐨𝐠 𝒙 𝐭𝐚𝐧−𝟏 𝒙𝟏
𝟎
𝒅𝒙 = 𝑰𝟏 − 𝟐𝑰𝟐,
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𝑰𝟏 = ∫ 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝟏
𝟎
= 𝐥𝐨𝐠(𝟏 + 𝒙𝟐)∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙 −∫𝟐𝒙
𝟏 + 𝒙𝟐⋅ ∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝒅𝒙 =
= 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) [𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 −𝟏
𝟐𝐥𝐨𝐠(𝟏 + 𝒙𝟐)] − 𝟐∫
𝒙𝟐 𝐭𝐚𝐧−𝟏 𝒙
𝟏 + 𝒙𝟐𝒅𝒙 +∫
𝒙 𝐥𝐨𝐠(𝟏 + 𝒙𝟐)
𝟏 + 𝒙𝟐𝒅𝒙
=
= (𝐥𝐨𝐠(𝟏 + 𝒙𝟐) − 𝟐) [𝒙 ⋅ 𝐭𝐚𝐧−𝟏 𝒙 −𝟏
𝟐𝐥𝐨𝐠(𝟏 + 𝒙𝟐)] + (𝐭𝐚𝐧−𝟏 𝒙)𝟐 +
𝟏
𝟒𝐥𝐨𝐠𝟐(𝟏 + 𝒙𝟐)
Putting limits, we get:
𝑰𝟏 =𝟏
𝟒𝐥𝐨𝐠𝟐 𝟐 +
𝝅𝟐
𝟏𝟔+ (𝐥𝐨𝐠𝟐 − 𝟐) (
𝝅
𝟒−𝟏
𝟐𝐥𝐨𝐠𝟐)
Now,
𝑰𝟐 = ∑(−𝟏)𝒏+𝟏
𝟐𝒏 − 𝟏
∞
𝒏=𝟏
∫ 𝒙𝟐𝒏−𝟏 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
=𝐥𝐨𝐠 𝒙=−𝒖
−∑(−𝟏)𝒏+𝟏
𝟐𝒏 − 𝟏∫ 𝒖𝒆−𝟐𝒖𝒏∞
𝟎
𝒅𝒖
∞
𝒏=𝟏
=𝒕=𝟐𝒏
= −𝟏
𝟒∑
(−𝟏)𝒏+𝟏
𝒏𝟐(𝟐𝒏 − 𝟏)
∞
𝒏=𝟏
∫ 𝒕𝒆−𝒕∞
𝟎
𝒅𝒕 = −𝟏
𝟒∑
(−𝟏)𝒏+𝟏
𝒏𝟐(𝟐𝒏 − 𝟏)
∞
𝒏=𝟏
Using partial fraction decomposition, we get:
𝑰𝟐 =𝟏
𝟒∑(−𝟏)𝒏+𝟏
𝒏𝟐
∞
𝒏=𝟏
+𝟏
𝟐∑(−𝟏)𝒏+𝟏
𝒏
∞
𝒏=𝟏
−∑(−𝟏)𝒏+𝟏
𝟐𝒏 − 𝟏
∞
𝒏=𝟏
=𝟏
𝟒⋅𝝅𝟐
𝟏𝟐+𝟏
𝟐⋅ 𝐥𝐨𝐠 𝟐 −
𝝅
𝟒
𝛀 = 𝑰𝟏 − 𝟐𝑰𝟐 =𝟏
𝟒𝐥𝐨𝐠𝟐 𝟐 +
𝝅𝟐
𝟏𝟔+ (𝐥𝐨𝐠𝟐 − 𝟐) (
𝝅
𝟒−𝟏
𝟐𝐥𝐨𝐠𝟐) − 𝟐 (
𝝅𝟐
𝟒𝟖+𝟏
𝟐⋅ 𝐥𝐨𝐠 𝟐 −
𝝅
𝟒) =
=𝝅
𝟒𝐥𝐨𝐠𝟐 −
𝟏
𝟒𝐥𝐨𝐠𝟐 𝟐 +
𝝅𝟐
𝟒𝟖
Solution 3 by Katrick Chandra Betal-India
𝛀 = ∑𝟏
𝒏∫
𝐭𝐚𝐧−𝟏 𝒙
(𝟏 + 𝒙𝟐)𝒏𝒅𝒙
𝟏
𝟎
∞
𝒏=𝟏
= −∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠 (𝟏 −𝟏
𝟏 + 𝒙𝟐)𝒅𝒙
𝟏
𝟎
=
= −∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠 (𝒙𝟐
𝟏 + 𝒙𝟐)
𝟏
𝟎
𝒅𝒙 = ∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠(𝟏 + 𝒙𝟐) 𝒅𝒙𝟏
𝟎
− 𝟐∫ 𝐭𝐚𝐧−𝟏 𝒙 𝐥𝐨𝐠 𝒙𝒅𝒙𝟏
𝟎
=
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= [𝐥𝐨𝐠(𝟏 + 𝒙𝟐) {𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)
𝟐}]𝟎
𝟏
− 𝟐∫ (𝒙𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)
𝟐)𝒙𝒅𝒙
𝟏 + 𝒙𝟐
𝟏
𝟎
−
−𝟐 [𝐥𝐨𝐠 𝒙 {𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)
𝟐}]𝟎
𝟏
+ 𝟐∫ (𝒙 𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)
𝟐)𝒅𝒙
𝒙
𝟏
𝟎
=
= 𝐥𝐨𝐠 𝟐 (𝝅
𝟒−𝐥𝐨𝐠𝟐
𝟐) − ∫
𝟐𝒙𝟐 𝐭𝐚𝐧−𝟏 𝒙 − 𝒙 𝐥𝐨𝐠(𝟏 + 𝒙𝟐)
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
+∫ (𝟐 𝐭𝐚𝐧−𝟏 𝒙 −𝐥𝐨𝐠(𝟏 + 𝒙𝟐)
𝒙)𝒅𝒙
𝟏
𝟎
=
=𝝅
𝟒𝐥𝐨𝐠 𝟐 −
𝐥𝐨𝐠𝟐 𝟐
𝟐− 𝟐∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙
𝟏
𝟎
+ 𝟐∫𝐭𝐚𝐧−𝟏 𝒙
𝟏 + 𝒙𝟐𝒅𝒙
𝟏
𝟎
+𝟏
𝟐∫𝐥𝐨𝐠(𝟏 + 𝒙)
𝟏 + 𝒙𝒅𝒙
𝟏
𝟎
+
+𝟐∫ 𝐭𝐚𝐧−𝟏 𝒙𝒅𝒙𝟏
𝟎
−𝟏
𝟐∫𝐥𝐨𝐠(𝟏 + 𝒙)
𝒙𝒅𝒙
𝟏
𝟎
=
=𝝅
𝟒𝐥𝐨𝐠𝟐 −
𝐥𝐨𝐠𝟐 𝟐
𝟐+ [(𝐭𝐚𝐧−𝟏 𝒙)𝟐]𝟎
𝟏 +𝟏
𝟐[𝐥𝐨𝐠𝟐(𝟏 + 𝒙)
𝟐]𝟎
𝟏
−𝜻(𝟐)
𝟒=
=𝝅
𝟒𝐥𝐨𝐠𝟐 −
𝐥𝐨𝐠𝟐 𝟐
𝟐+𝝅𝟐
𝟏𝟔+𝐥𝐨𝐠𝟐 𝟐
𝟒−𝝅𝟐
𝟐𝟒=𝝅
𝟒𝐥𝐨𝐠𝟐 −
𝐥𝐨𝐠𝟐 𝟐
𝟒+𝝅𝟐
𝟐𝟒
1599. Find:
𝛀 = 𝐥𝐢𝐦𝒏→∞
(𝒏 − 𝟏)!∑𝟏
(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌
𝒏
𝒌=𝟎
Proposed by Daniel Sitaru-Romania
Solution 1 by Kamel Gandouli Habib Rezgui-Tunisia
𝝎𝒏 = ∑𝟏
(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌
𝒏
𝒌=𝟎
; 𝒗𝒏(𝒌) =𝟏
(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌
𝒗𝟐𝒏(𝒌) =𝟏
(𝒌 + 𝟏)𝒌(𝟐𝒏 + 𝟏 − 𝒌)𝟐𝒏−𝒌; 𝒗𝒏(𝟐𝒏) =
𝟏
(𝟐𝒏 + 𝟏)𝟐𝒏 𝒇𝒐𝒓 𝒌 = 𝟎 𝒂𝒏𝒅
𝒏 = 𝐦𝐢𝐧 𝒗𝟐𝒏(𝒌) ⇒ 𝐦𝐚𝐱𝒗𝟐𝒏(𝒌) =𝟏
(𝒏 + 𝟏)𝒏(𝒏 + 𝟏)𝒏=
𝟏
(𝒏 + 𝟏)𝟐𝒏 𝒇𝒐𝒓 𝒌 = 𝒏.
𝟏
(𝟐𝒏 + 𝟏)𝟐𝒏≤ 𝒗𝟐𝒏(𝒌) ≤
𝟏
(𝒏 + 𝟏)𝟐𝒏, ∀𝒌 ∈ ℕ
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147 RMM-CALCULUS MARATHON 1501-1600
𝟐𝒏 + 𝟏
(𝟐𝒏 + 𝟏)𝟐𝒏≤∑𝒗𝟐𝒏(𝒌)
𝟐𝒏
𝒌=𝟎
≤𝟐𝒏 + 𝟏
(𝒏 + 𝟏)𝟐𝒏, ∀𝒌 ∈ ℕ
(𝟐𝒏 + 𝟏)(𝟐𝒏 − 𝟏)!
(𝟐𝒏 + 𝟏)𝟐𝒏≤ (𝟐𝒏 − 𝟏)!∑𝒗𝒏(𝟐𝒏)
𝟐𝒏
𝒌=𝟎
≤(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)
(𝒏 + 𝟏)𝟐𝒏
(𝟐𝒏 − 𝟏)!
(𝟐𝒏 + 𝟏)𝟐𝒏−𝟏≤ (𝟐𝒏 − 𝟏)!∑𝒗𝒏(𝟐𝒏)
𝟐𝒏
𝒌=𝟎
≤(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)
(𝒏 + 𝟏)𝟐𝒏
(𝟐𝒏 − 𝟏)!
(𝟐𝒏 + 𝟏)𝟐𝒏−𝟏≤ 𝝎𝟐𝒏 ≤
(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)
(𝒏 + 𝟏)𝟐𝒏
∵ 𝒏! = √𝟐𝒏𝝅 (𝒏
𝒆)𝒏
(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)
(𝒏 + 𝟏)𝟐𝒏≅√𝟐𝝅(𝟐𝒏 − 𝟏) (
𝟐𝒏 − 𝟏𝟐 )
𝒏
(𝒏 + 𝟏)𝟐𝒏(𝟐𝒏 + 𝟏) =
=√𝟐𝝅(𝟐𝒏 − 𝟏)(𝟐𝒏− 𝟏)𝒏
(𝒏 + 𝟏)𝟐𝒏𝒆𝒏(𝟐𝒏 + 𝟏)
(𝟐𝒏 − 𝟏)𝒏
(𝒏 + 𝟏)𝟐𝒏= (
(𝟐𝒏− 𝟏)
𝒏𝟐 + 𝟐𝒏 + 𝟏)
𝒏
→ 𝟎
√𝟐𝝅(𝟐𝒏 − 𝟏)(𝟐𝒏 + 𝟏)
𝒆𝒏≅ √𝟐𝝅(𝟐𝒏 − 𝟏)(𝟐𝒏 + 𝟏)𝒆−𝒏 ≤ (𝟐𝒏 + 𝟏)𝟐𝒆−𝒏 → 𝟎
Similarly for 𝟐𝒏 + 𝟏
𝒗𝟐𝒏+𝟏(𝒌) ≤𝟏
(𝟐𝒏 − 𝟏)𝟐𝒏(𝟐𝒏)𝟐𝒏,
(𝟐𝒏)! ∑ 𝒗𝟐𝒏+𝟏(𝒌)
𝟐𝒏+𝟏
𝒌=𝟎
≤(𝟐𝒏 + 𝟐)𝟐𝒏!
(𝟐𝒏 − 𝟏)𝟐𝒏(𝟐𝒏)𝟐𝒏→ 𝟎 ⇒ 𝝎𝟐𝒏+𝟏 → 𝟎
Therefore,
𝛀 = 𝐥𝐢𝐦𝒏→∞
(𝒏 − 𝟏)!∑𝟏
(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌
𝒏
𝒌=𝟎
= 𝟎
Solution 2 by Ravi Prakash-New Delhi-India
Let 𝒎 ∈ ℕ− {𝟎}, 𝒇(𝒙) = (𝒙 + 𝟏)𝒙(𝒎 + 𝟏 − 𝒙)𝒎−𝒙, 𝟎 ≤ 𝒙 ≤ [𝒎
𝟐]
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148 RMM-CALCULUS MARATHON 1501-1600
𝐥𝐨𝐠 𝒇(𝒙) = 𝒙 𝐥𝐨𝐠(𝒙 + 𝟏) + (𝒎− 𝒙) 𝐥𝐨𝐠(𝒎+ 𝟏 − 𝒙)
𝒇′(𝒙)
𝒇(𝒙)= 𝐥𝐨𝐠(𝒙 + 𝟏) − 𝐥𝐨𝐠(𝒎+ 𝟏 − 𝒙) −
𝒎 − 𝒙
𝒎 + 𝟏 − 𝒙=
= 𝐥𝐨𝐠 (𝒙 + 𝟏
𝒎+ 𝟏 − 𝒙) −
𝒎 − 𝒙
𝒎 + 𝟏 − 𝒙< 𝟎,∀𝟎 < 𝒙 < [
𝒎
𝟐]
Thus, 𝒇 decreases on [𝟎, [𝒎
𝟐]] ⇒ 𝒇(𝒙) ≥ 𝒇([
𝒎
𝟐]) , ∀𝒙 ∈ [𝟎, [
𝒎
𝟐]].
Hence,
𝟏
(𝒌 + 𝟏)𝒌(𝟐𝒏 + 𝟏 − 𝒌)𝟐𝒏−𝒌≤
𝟏
(𝒏 + 𝟏)𝒏(𝒏 + 𝟏)𝒏=
𝟏
(𝒏 + 𝟏)𝟐𝒏
and
𝟏
(𝒌 + 𝟏)𝒌(𝟐𝒏 + 𝟐 − 𝒌)𝟐𝒏+𝟏−𝒌≤
𝟏
(𝒏 + 𝟏)𝟐𝒏+𝟏
(𝟐𝒏 − 𝟏)!∑𝟏
(𝒌 + 𝟏)𝒌(𝟐𝒏 − 𝒌 + 𝟏)𝟐𝒏−𝒌
𝟐𝒏
𝒌=𝟎
<(𝟐𝒏 − 𝟏)! (𝟐𝒏 + 𝟏)
(𝒏 + 𝟏)𝟐𝒏=
(𝟐𝒏 + 𝟏)!
𝟐𝒏(𝒏 + 𝟏)𝟐𝒏
Let 𝒃𝒏 =(𝟐𝒏+𝟏)!
𝟐𝒏(𝒏+𝟏)𝟐𝒏 and
(𝟐𝒏 + 𝟏 − 𝟏)! ∑𝟏
(𝒌 + 𝟏)𝒌(𝟐𝒏 + 𝟏 − 𝒌 + 𝟏)𝟐𝒏+𝟏−𝒌
𝟐𝒏+𝟏
𝒌=𝟎
<(𝟐𝒏)! (𝟐𝒏 + 𝟐)
(𝒏 + 𝟏)𝟐𝒏+𝟏
=(𝟐𝒏 + 𝟐)!
(𝟐𝒏 + 𝟏)(𝒏 + 𝟏)𝟐𝒏+𝟏
Let 𝒄𝒏 =(𝟐𝒏+𝟐)!
(𝟐𝒏+𝟏)(𝒏+𝟏)𝟐𝒏+𝟏. We prove that: 𝒃𝒏 , 𝒄𝒏 → ∞ for 𝒏 → ∞.
𝒃𝒏𝒃𝒏+𝟏
= (𝟏+𝟏
𝒏) ⋅
(𝒏 + 𝟐)𝟐
(𝟐𝒏 + 𝟐)(𝟐𝒏+ 𝟑)[(𝟏 +
𝟏
𝒏 + 𝟏)𝒏+𝟏
]
𝟐
→𝒆𝟐
𝟒
As 𝒆𝟐
𝟒> 𝟏, 𝒃𝒏 → 𝟎 as 𝒏 → ∞. Similarly, 𝒄𝒏 → 𝟎 as 𝒏 → ∞.
Now,
𝟎 <∑𝟏
(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌
𝒏
𝒌=𝟎
< 𝒃𝒏, 𝒄𝒏
As 𝒃𝒏 → 𝟎, 𝒄𝒏 → 𝟎 as 𝒏 → ∞. Therefore,
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149 RMM-CALCULUS MARATHON 1501-1600
𝛀 = 𝐥𝐢𝐦𝒏→∞
(𝒏 − 𝟏)!∑𝟏
(𝒌 + 𝟏)𝒌(𝒏 − 𝒌 + 𝟏)𝒏−𝒌
𝒏
𝒌=𝟎
= 𝟎
1600.
𝛀𝒏(𝒙) = ∫𝒅𝒙
𝒙(𝟏 + 𝒙𝒏), 𝒏 ∈ ℕ∗, 𝛀𝒏(𝟏) = 𝐥𝐨𝐠 𝟐
Find:
𝛀(𝒙) = 𝐥𝐢𝐦𝒏→∞
(𝒏𝛀𝒏(𝒙)) , 𝒙 > 𝟎
Proposed by Daniel Sitaru-Romania
Solution 1 by Ravi Prakash-New Delhi-India
𝛀𝒏(𝒙) = ∫𝒅𝒙
𝒙(𝟏 + 𝒙𝒏)= ∫
𝒙𝒏−𝟏
𝒙𝒏(𝟏 + 𝒙𝒏)𝒅𝒙 = ∫(
𝟏
𝒙𝒏−
𝟏
𝟏 + 𝒙𝒏)𝒙𝒏−𝟏 𝒅𝒙 =
=𝟏
𝒏∫(𝟏
𝒕−
𝟏
𝒕 + 𝟏)𝒅𝒕 =
𝟏
𝒏𝐥𝐨𝐠 (
𝒕
𝒕 + 𝟏) + 𝑪 =
𝟏
𝒏𝐥𝐨𝐠 (
𝒙𝒏
𝟏 + 𝒙𝒏) + 𝑪
𝛀𝒏(𝟏) =𝟏
𝒏𝐥𝐨𝐠 (
𝟏
𝟐) = −
𝟏
𝒏𝐥𝐨𝐠𝟐 + 𝑪 ⇒ 𝑪 =
𝒏 + 𝟏
𝒏𝐥𝐨𝐠 𝟐
Thus,
𝒏𝛀𝒏(𝒙) = (𝒏 + 𝟏) 𝐥𝐨𝐠 𝟐 + 𝐥𝐨𝐠 (𝒙𝒏
𝟏 + 𝒙𝒏) = 𝐥𝐨𝐠𝟐 + 𝐥𝐨𝐠 (
𝟐𝒏𝒙𝒏
𝟏 + 𝒙𝒏)
If 𝟎 < 𝒙 <𝟏
𝟐, 𝟎 < 𝟐𝒙 < 𝟏 ⇒ (𝟐𝒙)𝒏 → 𝟎
𝐥𝐢𝐦𝒏→∞
𝒏𝛀𝒏(𝒙) = −∞ 𝒊𝒇 𝟎 < 𝒙 <𝟏
𝟐.
For 𝒙 =𝟏
𝟐 we have: 𝒏𝛀𝒏(𝒙) = 𝐥𝐨𝐠 𝟐 + 𝐥𝐨𝐠 (
𝟏
𝟏+(𝟏
𝟐)𝒏) → 𝐥𝐨𝐠 𝟐 as 𝒏 → ∞.
For 𝟏
𝟐< 𝒙 < 𝟏, (𝟐𝒙)𝒏 → ∞,𝒙𝒏 → ∞ ad 𝒏𝛀𝒏(𝒙) → ∞ as 𝒏 → ∞.
For 𝒙 ≥ 𝟏,𝒏𝛀𝒏(𝒙) = 𝐥𝐨𝐠 𝟐 − 𝐥𝐨𝐠 (𝟏
𝟐𝒏+
𝟏
𝟐𝒏𝒙𝒏) → ∞ as 𝒏 → ∞.
Solution 3 by Kamel Gandouli Rezgui-Tunisia
For 𝐱 > 𝟏:
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150 RMM-CALCULUS MARATHON 1501-1600
𝛀𝒏(𝒙) = ∫𝟏
𝒙(𝒙𝒏 + 𝟏)𝒅𝒙 = 𝐥𝐨𝐠 𝒙 −
𝐥𝐨𝐠(𝒙𝒏 + 𝟏)
𝒏
Hence,
𝛀(𝒙) = 𝐥𝐢𝐦𝒏→∞
(𝒏𝛀𝒏(𝒙)) = 𝐥𝐢𝐦𝒏→∞
𝒏(𝐥𝐨𝐠𝒙 −𝐥𝐨𝐠(𝒙𝒏 + 𝟏)
𝒏) = 𝐥𝐢𝐦
𝒏→∞[𝒏 𝐥𝐨𝐠𝒙 − 𝐥𝐨𝐠(𝒙𝒏 + 𝟏)] =
= 𝐥𝐢𝐦𝒏→∞
𝐥𝐨𝐠 (𝒙𝒏
𝒙𝒏 + 𝟏) = 𝐥𝐢𝐦
𝒏→∞
𝟏
𝟏𝒙𝒏 + 𝟏
= 𝟎
Solution 3 by Satyam Roy-India
For 𝐱 > 𝟏:
𝛀𝒏(𝒙) = ∫𝒅𝒙
𝒙(𝟏 + 𝒙𝒏)= ∫
𝒙−𝒏−𝟏
𝟏𝒙𝒏 + 𝟏
𝒅𝒙 =
𝟏𝒙𝒏+𝟏=𝒖
−𝟏
𝒏∫𝟏
𝒖𝒅𝒖 = −
𝟏
𝒏𝐥𝐨𝐠|𝒖| + 𝑪 =
= −𝟏
𝒏𝐥𝐨𝐠 |
𝟏
𝒙𝒏+ 𝟏| + 𝑪
Hence,
𝛀(𝒙) = 𝐥𝐢𝐦𝒏→∞
(𝒏𝛀𝒏(𝒙)) = 𝐥𝐢𝐦𝒏→∞
(𝒏 ⋅−𝟏
𝒏𝐥𝐨𝐠 |
𝟏
𝒙𝒏+ 𝟏|) = − 𝐥𝐢𝐦
𝒏→∞𝐥𝐨𝐠 |
𝟏
𝒙𝒏+ 𝟏| = 𝟎
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151 RMM-CALCULUS MARATHON 1501-1600
It’s nice to be important but more important it’s to be nice.
At this paper works a TEAM.
This is RMM TEAM.
To be continued!
Daniel Sitaru