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Gamma Function

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  • Difficulty Level : Medium
  • Last Updated : 16 Jun, 2020
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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:

  1. \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

  2. \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)

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)
We know \Gamma\left(0.5\right)=\sqrt\pi
Thus \Gamma\left(4.5\right)=6.5625\sqrt\pi

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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