beta gamma functins material

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Beta and Gamma Functions If (i) the interval [a, b] is finite (ii) the function f(x) is bounded in [a, b], that is, f(x) does not become infinite at any point in the interval, and (iii) d dx [ ϕ ( x ) ] = f( x) , then a b f ( x ) dx = (b) - (a) is called a proper integral. In condition (i) is not satisfied (that is, a, b or both are infinite), the integral a b f ( x ) dx is called an improper integral of first kind. Definition: If m and n are positive, then 0 1 x m1 ( 1x) n1 dx is called a Beta function and denoted by B(m, n) or (m, n). Note: If m and n are both greater than or equal to 1, then the above integral is a proper integral. On the other hand if either m or n is less than 1, then the integral becomes improper but may be convergent. Properties of Beta Functions: 1. Symmetry: B(m, n) = B(n, m) Verification: By definition, B(m, n) = 0 1 x m1 ( 1x) n1 dx = 0 1 ( 1x) m1 ¿¿ (since 0 a f ( x ) dx= ¿ 0 a f ( ax ) dx ¿ = 0 1 ( 1x) m1 x n1 dx

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Beta and Gamma Functions

If (i) the interval [a, b] is finite (ii) the function f(x) is bounded in [a, b], that is, f(x) does not become infinite at any point in the interval, and(iii) , then = (b) - (a) is called a proper integral.In condition (i) is not satisfied (that is, a, b or both are infinite), the integral is called an improper integral of first kind.Definition: If m and n are positive, then is called a Beta function and denoted by B(m, n) or (m, n).Note: If m and n are both greater than or equal to 1, then the above integral is a proper integral. On the other hand if either m or n is less than 1, then the integral becomes improper but may be convergent. Properties of Beta Functions: 1. Symmetry: B(m, n) = B(n, m)Verification: By definition, B(m, n) = = (since = = = B(n, m).

2. If p > -1, q > -1 then = BVerification: = Put sin2 = x. Then cos2 d = 1- x. When = 0, we have x = 0 and = , we have x = 1. Therefore = = B

Note: By taking = m and = n, we get p = 2m-1 and q = 2n-1. Therefore B(m, n) = .3. B(m, n + 1) + B(m + 1, n) = B(m, n).

4. = Verification: Left to the students practiceDefinition: If n is positive then is called a Gamma function denoted by (n).Properties of Gamma function: Property 1: (1) = 1Verification: By definition (n) = Therefore (1) = = = = = -(0-1) = 1.

Property 2: Reduction formula: (n + 1) = n (n). Verification: By definition, (n + 1) = . Applying integration by parts, (n + 1) = - = = -(0-0) + n(n) = n (n).

Property 3: If n is a positive integer then (n) = (n-1)Verification: (n + 1) = n (n). Changing n to n-1, we get (n) = (n-1)(n-1). Similarly (n -1) = (n -2)(n-2). By substitution (n) = (n-1) (n-2)(n-2). Similarly, (n - 2) = (n -3)(n-3), and again by substitution, (n) = (n-1)(n-2)(n-3)(n-3). This process can be continued successively, and it ends with (1), since n is a positive integer. Therefore (n) = (n-1) (n-2)(n-3) 1(1). But (1) = 1. Hence (n) = (n-1) (n-2)(n-3) 1.Property 4: Relation between beta and gamma function: .Verification: By definition, , using t as the variable. Put t = x2. Then dt = 2x dx. When t = 0, we have x = 0 and when t = , we have x = .Therefore (n) = =. Similarly (m) = .Multiplying, (m) (n) = = The range of this integral is the entire first quadrant of the XOY plane. Converting to polar coordinates, by writing x = r cos , y = r sin so that dx dy = r d dr. The limits being from 0 to and r from to .Therefore (m) (n) = = = 4 = B(m, n) (m + n).

Property 5: ; .Verification: Use , by taking , we get . But (1) = 1. Therefore = = 2Therefore () = .Now (n) = (n-1)(n-1) (n-1) = Put , we get = -2.Property 6: Duplication formula: (m) = Verification: Therefore = ______(equation 1)But = = . Put 2 = . Then d = . When = 0, we have = 0 and when = we get = . Therefore B(m, m) = = = = = _______(equation 2).Hence from (1) and (2) we get Cancelling (m) both sides and write () = we get (m) = Property 7: (n) = Hint: Use (n) = . Put e-x = y so that ex = y or x = log . Example: Evaluate Solution: By definition = B(m, n). Take m = p + 1 and n = q + 1. We get = B(p+1, q+1). Therefore = B(5, 4) = = = Example: Evaluate Solution: Use B = B = = Example: Evaluate Solution: Put x = a + (b - a)t. Then dx = (b - a)dt, x a = (b - a)t, b x = (b - a)(1- t). When x = a, we have t = 0 and when x = b we have t = 1. Therefore = (b-a) p+1, q+1).Example: Show that = .Solution: Use B taking p = and q = 0, we get B = B _______(equation 1) Taking p = -1/2, q = 0, we get B = B ________(equation 2).Multiplying (1) and (2) = BB

= .Example: Show that = .Solution: By definition of gamma function (n) = . Put t = x2. Then dt = 2x dx. When t = 0, we get x = 0; when t = we get x = .Therefore (n) = = . Taking n = 1/2, we get () = 2 = 2 = 2. Therefore = Example: Evaluate dxSolution: Put x2 = a2t. Then dx = . When x = 0, we have t = 0; when x = a, we have t = 1.Therefore dx = = = = .Example: Show that = Solution: d = = = = = .Using Duplication formula with m = , we get = .Therefore = .

Example: Express in terms of gamma function. Solution: Put x2 = sin . Then Therefore = = = = .

Exercises:1. Compute (3.5), (4.5), () ()2. Express in terms of gamma function.3. Prove that 4. Evaluate .5. Prove that .