Gamma Function

Last Updated : 16 Jun, 2020

Gamma function is one commonly used extension of the factorial function to complex numbers. The gamma function is defined for all complex numbers except the non-positive integers. Gamma function denoted by $\Gamma\left (p \right)$ is defined as: $\Gamma\left(p \right) = \int_{0}^{\infty}e^{-t} t^{p-1} dt$ where p>0. Gamma function is also known as Euler’s integral of second kind. Integrating Gamma function by parts we get, $\Gamma\left (p+1 \right) = \int_{0}^{\infty}e^{-t} t^{p} dt$ $=-e^{-t} t^p \Biggr |_{0}^{\infty}+p\int_{0}^{\infty}e^{-t} t^{p-1} dt$ $=0+p\Gamma\left (p \right)$ Thus $\Gamma\left (p+1 \right) = p\Gamma\left (p \right)$ Some standard results:

$\Gamma\left (1/2 \right) = \sqrt \pi$ We know that $\Gamma\left(1/2 \right) = \int_{0}^{\infty}t^{-\frac{1}{2}}e^{-t} dt$ Put t=u^2 Thus $\Gamma\left(1/2 \right) = 2\int_{0}^{\infty}e^{{-u^2}}du$ $\Gamma\left(1/2 \right) .\Gamma\left(p \right) = (2\int_{0}^{\infty}e^{{-u^2}}du)(2\int_{0}^{\infty}e^{{-u^2}}du)$ $=4\int_{0}^{\infty} \int_{0}^{\infty}e^{-{u^2 + v^2}} du dv$ Now changing to polar coordinates by using u = r cosθ and v = r sinθ Thus ${\Gamma\left(1/2 \right)}^2 = 4\int_{\theta=0}^{\pi/2}\int_{r=0}^{\infty}e^{-{r^2}} dr d\theta$ $=4\int_{0}^{\pi/2} -\frac{1}{2}e^{-r^2}\Biggr|_{r=0}^{\infty}$ $=2\int_{0}^{\pi/2}d\theta = 2.\theta \Biggr|_{0}^{\pi/2}=\pi$ Hence $\Gamma\left (1/2 \right) = \sqrt \pi$

$\Gamma\left(n+1 \right) = (m+1)^{n+1}(-1)^n \int_{0}^{1}x^m (ln x)^n dx$ Where n is a positive integer and m>-1 Put x=e^-y such that dx=-e^-ydy=-x dy $\int_{0}^{1}x^m(ln x)^n dx= \int_{0}^{\infty}e^{-my} . (-y)^n e^{-y} dy$ $(-1)^n \int_{0}^{\infty} y^n . e^{-(m+1)y} dy$ Put (m+1)y = u $=(-1)^n \int_{0}^{\infty}\frac{u^n}{(m+1)^n}.e^{-u} .\frac{du}{m+1}$ $=\frac{(-1)^n}{(m+1)^n+1}\int_{0}^{\infty}e^{-u} .u^n du = \frac{(-1)^n}{(m+1)^{n+1}}.\Gamma\left(n+1\right)$

Example-1: Compute $\Gamma\left(4.5\right).$ Explanation : Using $\Gamma\left(p+1\right)=p\Gamma\left(p\right)$ $\Gamma\left(4.5\right)=\Gamma\left(3.5+1 \right)=3.5\Gamma\left(3.5\right)$ $=(3.5)(2.5)\Gamma\left(2.5\right)$ $=(3.5)(2.5)(1.5)\Gamma\left(1.5\right)$ $=(3.5)(2.5)(1.5)(0.5)\Gamma\left(0.5\right)$ We know $\Gamma\left(0.5\right)=\sqrt\pi$ Thus $\Gamma\left(4.5\right)=6.5625\sqrt\pi$
Example-2: Evaluate $I=\int_{0}^{\infty}x^4 e^-{x^4} dx$ Explanation : Put x⁴ = t, 4x³dx = dt, dx = ¼ t^-3/4 dt $I=\int_{0}^{\infty}t.e^{-t} \frac{t^{-3/4}}{4}dt$ $= \frac{1}{4}\int_{0}^{\infty}e^{-t} t^{3/4} dt$ $= \frac{1}{4}\Gamma\left(1+\frac{1}{4}\right)$ $= \frac{1}{4}\Gamma\left(\frac{5}{4}\right)$

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Gamma Function

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