improper integrals ii. improper integrals ii by mika seppälä improper integrals an integral is...
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Improper Integrals II
Improper Integrals II
Improper Integrals II by Mika Seppälä
Improper Integrals
An integral is improper if either:
•the interval of integration is infinitely long or
•if the function f is either discontinuous or undefined at some points of the interval of integration (or both).
DefinitionDefinition
f x( )dx
a
b
∫
Improper Integrals II by Mika Seppälä
Improper Integrals
DefinitionDefinition Assume that the function f is continuous on the interval [a, b).
If the limit exists and is finite, the improper integral converges, and
limβ→ b−
f x( )dxa
β
∫
f x( )dxa
b
∫
If the integral does not converge, then it diverges.
f x( )dxa
b
∫
Improper Integrals II by Mika Seppälä
Improper Integrals
ExampleExample
dx
xp1
∞
∫ =limb→ ∞
dx
xp1
b
∫
=
1p −1
.
Assuming that 1 - p < 0. If 1 - p ≥ 0, the integral diverges.
Assume that p ≠ 1.
=lim
b→ ∞x−p dx
1
b
∫
Improper Integrals II by Mika Seppälä
Improper Integrals
ExampleExample
dx
xp0
1
∫ = lima→ 0 +
dx
xpa
1
∫
= lim
a→ 0+
11 −p
−a1−p
1 −p
⎛
⎝⎜
⎞
⎠⎟
=
11 −p
.
Assuming that 1 - p > 0, i.e. p < 1.
Assume that p > 0, and p ≠ 1. If p ≤ 0, the integral is not improper.
= lim
a→ 0 +x−p dx
a
1
∫
Improper Integrals II by Mika Seppälä
Improper Integrals
ExampleExample
eax dx1
∞
∫ =limb→ ∞
eax dx1
b
∫
=lim
b→ ∞
eax
a
⎡
⎣⎢
⎤
⎦⎥1
b
=lim
b→ ∞
eab
a−
ea
a
⎛
⎝⎜
⎞
⎠⎟
=−
ea
a.
Assuming that a < 0. If a ≥ 0, the integral diverges.
Assume that a ≠ 0.
Improper Integrals II by Mika Seppälä
Basic Improper Integrals
11
eax dx
1
∞
∫
dx
xp1
∞
∫ converges if p > 1.
22
33
dx
xp0
1
∫ converges if p < 1.
converges if a < 0.
Improper Integrals II by Mika Seppälä
Basic Improper Integrals
44 xp dx
1
∞
∫ converges if p < -1.
55 xp dx
0
1
∫ converges if p > -1.
For all other values of the parameter the above integrals 1 - 5 diverge.
Improper Integrals II by Mika Seppälä
Convergence of Improper Integrals
Often it is not possible to compute the limit defining a given improper integral directly.
In order to find out whether such an integral converges or not one can try to
compare the integral to a known integral of which we know that it either converges or
diverges.
Improper Integrals II by Mika Seppälä
Convergence of Improper Integrals
Consider the improper integral
3
(1 + 2 sin2 (4πx))(1 + x2 )dx
0
∞
∫ .
3
(1 + 2 sin2 (4πx))(1 + x2 )
The graph of the function
is shown in the figure.
Improper Integrals II by Mika Seppälä
Idea of the Comparison Theorem
The improper integral converges if the areaunder the red curve is finite.
Improper Integrals II by Mika Seppälä
Idea of the Comparison Theorem
We show this by finding a simple function, whose graph is the blue curve such that the area of the domain under the blue curve is finite.
The improper integral converges if the areaunder the red curve is finite.
Improper Integrals II by Mika Seppälä
Idea of the Comparison Theorem
0 <
3(1 + 2 sin2 (4πx))(1 + x2 )
≤3
1 + x2Observe that,
for all x:
3
1 + x2
Here the blue curve is
the graph of and
the red curve is that
of
3
(1 + 2 sin2 (4πx))(1 + x2 ).
Improper Integrals II by Mika Seppälä
Comparison Theorem
3
1 + x2dx
0
∞
∫ =limb→ ∞
3dx
1 + x20
∞
∫Since
=lim
b→ ∞3 arctan x⎡
⎣⎤⎦0
b
=lim
b→ ∞3 arctanb −3 arctan0( )
=3π2
,
and since
we conclude that converges.
0 <
3(1 + 2 sin2 (4πx))(1 + x2 )
≤3
1 + x2,
3
(1 + 2 sin2 (4πx))(1 + x2 )dx
0
∞
∫
Improper Integrals II by Mika Seppälä
Comparison Theorem
Theorem ATheorem A Let -∞ ≤ a < b ≤ ∞ . Assume
that 0 ≤ f(x) ≤ g(x) for all x, a < x < b.
If the integral converges, then
also converges, and
g x( )dxa
b
∫
f x( )dx
a
b
∫
0 ≤ f x( )dx
a
b
∫ ≤ g x( )dxa
b
∫ .
Improper Integrals II by Mika Seppälä
Comparison Theorem
Theorem BTheorem B Let ∞ ≤ a < b ≤ ∞ . Assume
that 0 ≤ f(x) ≤ g(x) for all x, a < x < b.
If the integral diverges, then also
diverges. g x( )dx
a
b
∫
f x( )dxa
b
∫
Improper Integrals II by Mika Seppälä
Comparison Theorem
ExampleExample
Does the integral converge?
dx
sin x0
1
∫
We know that 0 < sin x < x, for 0 < x ≤ 1.
SolutionSolution
0 <
1x
<1
sin x
Hence for 0 < x ≤ 1.
Improper Integrals II by Mika Seppälä
Comparison Theorem
ExampleExample
Does the integral converge?
0 <
1x
<1
sin x
dx
sin x0
1
∫
Solution (cont’d)Solution (cont’d)
Since for 0 < x ≤ 1, and since
the integral diverges, also the integral
diverges.
dx
x0
1
∫
dx
sin x0
1
∫
Improper Integrals II by Mika Seppälä
Basic Improper Integrals
11
eax dx
1
∞
∫
dx
xp1
∞
∫ converges if p > 1.
22
33
dx
xp0
1
∫ converges if p < 1.
converges if a < 0.
Improper Integrals II by Mika Seppälä
Basic Improper Integrals
44 xp dx
1
∞
∫ converges if p < -1.
55 xp dx
0
1
∫ converges if p > -1.
For all other values of the parameter the above integrals 1 - 5 diverge. Use the Comparison Theorem to study the convergence (or divergence) of improper integrals by comparing them to an integral of one of the types 1 - 5.