Beta and Gamma Functions
Dr. Kamlesh Jangid
Department of HEAS (Mathematics)
Rajasthan Technical University, Kota-324010, India
E-mail: [email protected]
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 1 / 20 Outline
Outline
1 Improper Integrals The Gamma Function The Beta Function Properties
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 2 / 20 Improper Integrals
Improper Integrals
Improper integrals are definite integrals that cover an unbounded Z ∞ 1 area. For example, 2 dx 1 x An improper integral is a definite integral that has either one or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration.
Hence improper integrals are of two type.
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 3 / 20 Improper Integrals
1 Type-1: (Infinite limits of integration): In this kind of integrals one Z ∞ 1 or both limits of integration are infinity. For example, 2 dx is 1 x an improper integral. Z b 1 It can be viewed as the limit lim 2 dx. b→∞ 1 x 2 Type-2: (Discontinuous integrand or integrands with vertical asymptotes): In this type of improper integrals the endpoints are finite, but the integrand function is unbounded at one (or two) of the Z 1 1 endpoints. For example, √ dx. 0 x Z 1 1 It can be viewed as the limit lim √ dx. a→0+ a x
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 4 / 20 Improper Integrals
Remark Not all improper integrals have a finite value, but some of them definitely do.
When the limit exists we say the integral is convergent, and when it doesn’t we say it’s divergent.
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 5 / 20 Improper Integrals The Gamma Function
The Gamma Function
The Gamma function is important as it is an extension to the
factorial function f (n) = n! for all n ∈ N. The Gamma function is defined as the single variable function
Z ∞ Γ(x) = e−t tx−1 dt, x > 0. (1) 0 By using integration by parts we find that
Z ∞ Z ∞ Γ(x + 1) = e−t tx dt = x e−t tx−1 dt = xΓ(x). 0 0 + and Γ(x + 1) = x! , if x ∈ Z .
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 6 / 20 Improper Integrals The Gamma Function
From Γ(x + 1) = xΓ(x), it is clear that if Γ(x) is known throughout a unit interval say: 1 ≤ x ≤ 2, then the value of Γ(x) throughout the next unit interval 2 < x ≤ 3 are found and so on. In this way, the values of Γ(x) for all positive values of x > 1 may be found. Γ(x + 1) From Γ(x) = , it is clear that if Γ(x) is known throughout x a unit interval say: 1 < x ≤ 2, then the value of Γ(x) throughout the previous unit interval 0 < x ≤ 1 are found and so on.
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 7 / 20 Improper Integrals The Gamma Function
Γ(x + 1) Note: From Γ(x + 1) = xΓ(x) and Γ(x) = , it is clear that Γ(x) x is exists for all values of x except when x = 0 or a negative integer.
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 8 / 20 Improper Integrals The Beta Function
The Beta Function
The Beta Function is defined as the two variable function
Z 1 B(x, y) = tx−1 (1 − t)y−1 dt, for x, y > 0. (2) 0
Z π/2 B(x, y) = 2 (sin θ)2x−1(cos θ)2y−1dθ, put t = sin2 θ. 0 Z ∞ ux−1 1 B(x, y) = x+y du, put t = . 0 (1 + u) 1 + u From the definition it is easily seen that B(x, y) = B(y, x).
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 9 / 20 Improper Integrals The Beta Function
Example Z 1 dx Evaluate √ . 4 0 1 − x
Solution Substitute x2 = sin θ, then we obtain
Z 1 dx 1 Z π/2 √ = (sin θ)−1/2dθ 4 0 1 − x 2 0
using the defintion of beta function (polar form), we get √ π Γ(1/4) = 4 Γ(3/4)
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 10 / 20 Improper Integrals The Beta Function
Example Z ∞ xc Evaluate x dx. 0 c
Solution
Z ∞ c Z ∞ x −x log c c x dx = e x dx 0 c 0 Z ∞ t c dt = e−t put x log c = t 0 log c log c Z ∞ 1 −t c = c+1 e t dt (log c) 0 1 = Γ(c + 1) (by the Gamma function definition) (log c)c+1
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 11 / 20 Improper Integrals The Beta Function
Example Z ∞ Evaluate e−axxm−1 sin bx dx. 0
Solution R ∞ −x m−1 By substituting x = ay in Γ(m) = 0 e x dx, we obtain Γ(m) R ∞ e−ayym−1dy = . Thus 0 am Z ∞ Z ∞ −ax m−1 −ax m−1 ibx ibx e x sin bx dx = IP e x e dx (∵ sin bx = IP(e )) 0 0 Z ∞ −(a−ib)x m−1 Γ(m) = IP e x dx = IP m 0 (a − ib) using a = r cos θ, b = r sin θ, r2 = a2 + b2 Γ(m) = sin mθ. (a2 + b2)m/2
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 12 / 20 Improper Integrals Properties
Properties
Theorem (Relation between beta and gamma functions) The connection between the beta function and the gamma function is given by Γ(x)Γ(y) B(x, y) = . Γ(x + y)
In order to prove, we use the definition (1) to obtain
Z ∞ Z ∞ Γ(x)Γ(y) = tx−1e−tdt sy−1e−sds 0 0 Z ∞ Z ∞ = e−(t+s)tx−1sy−1dt ds. 0 0
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 13 / 20 Improper Integrals Properties
Now we apply change of variables t = u v and s = u(1 − v) to this double integral. Note that t + s = u and that 0 < t < ∞ and 0 < s < ∞ imply that 0 < u < ∞ and 0 < v < 1. The Jacobian of this transformation is
∂(t, s) v u = = −u. ∂(u, v) 1 − v −u
∂(t, s) Therefore, dt ds = du dv = u du dv. ∂(u, v)
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 14 / 20 Improper Integrals Properties
Hence, we have
Z 1 Z ∞ Γ(x)Γ(y) = e−u (u v)x−1 (u(1 − v))y−1 u du dv v=0 u=0 Z ∞ Z 1 = e−u ux+y−1 du vx−1(1 − v)y−1dv u=0 v=0 = Γ(x + y).B(x, y)
Dividing both sides by Γ(x + y) gives the desired result.
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 15 / 20 Improper Integrals Properties
Let us see the application of the previous Theorem. Example √ Prove that Γ(1/2) = π.
Solution Γ(x)Γ(y) Since B(x, y) = , if x = y = 1/2, then we have Γ(x + y)
2 B (1/2, 1/2) = {Γ(1/2)} , ∵ Γ(1) = 1.
Now, using the definition of beta function (polar form), we have
√ B (1/2, 1/2) = π =⇒ Γ(1/2) = π
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 16 / 20 Improper Integrals Properties
Example Z ∞ xn−1 π Given that dx = , then show that 0 1 + x sin(nπ) π Γ(n)Γ(1 − n) = . sin(nπ)
Solution Γ(m)Γ(n) Z ∞ xn−1 We know that B(m, n) = = m+n dx. Γ(m + n) 0 (1 + x)
Taking m + n = 1 or m = 1 − n and using given result, we obtain the required result.
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 17 / 20 Improper Integrals Properties
Example Prove that B(m, n) = B(m + 1, n) + B(m, n + 1).
Solution By the definition of the beta function, we have
Z π/2 B(m + 1, n) = 2 (sin θ)2m+1(cos θ)2n−1dθ 0 Z π/2 = 2 (sin θ)2m−1(cos θ)2n−1(sin θ)2dθ 0 Z π/2 Z π/2 = 2 (sin θ)2m−1(cos θ)2n−1dθ − 2 (sin θ)2m−1(cos θ)2n+1dθ 0 0 = B(m, n) − B(m, n + 1).
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 18 / 20 Improper Integrals Properties
Example (duplication formula) √ Prove that Γ(n)Γ(n + 1/2) = 21−2n π Γ(2n).
Solution By the definition of beta function, we have
Z π/2 B(n, n) = 2 (sin θ)2n−1(cos θ)2n−1dθ 0 Z π/2 = 2.21−2n (sin 2θ)2n−1dθ (put 2θ = φ) 0 Z π Z π/2 = 21−2n (sin φ)2n−1dφ = 21−2n.2 (sin φ)2n−1dφ 0 0 Γ(n)Γ(n) Γ(n)Γ(1/2) = 21−2n.B(n, 1/2) = 21−2n Γ(2n) Γ(n + 1/2)
after simplification, we obtain the required result.
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 19 / 20 Improper Integrals Properties
For the video lecture use the following link https: //youtube.com/channel/UCk9ICMqdkO0GREITx-2UaEw THANK YOU
Dr. Kamlesh Jangid (RTU Kota) Improper Integrals 20 / 20