GAMMA REFLECTION FORMULA VIA CONTOUR INTEGRATION

$\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin(a\pi)}$

We have already proved this famous result in a previous post via real methods. Today I want to proof it via contour integration and in the end of the post, use it to compute a famous infinite series.

We start by considering the following integral

$\int_{0}^{\infty} \frac{x^{a-1}}{1+x}dx=\frac{\pi}{\sin a \pi}$

0<a<1 . Lets evaluate the complex integral

$I=\oint_Cf(z)dz$

where

$f(z)=\frac{z^{a-1}}{1+z}$

It´s clear that f(z) has a pole @ $-1=e^{i \pi}$ and a branch @ z=0 . We choose to be the keyhole contour below.

We can Write

$\oint_Cf(z)dz=\int_{\epsilon}^{R}\frac{x^{a-1}}{1+x}dx+\int_{\Gamma}\frac{z^{a-1}}{1+z}dz+\int_{R}^{\epsilon}\frac{(xe^{2 i \pi})^{a-1}}{1+xe^{2 i \pi}}dx+\int_{\gamma}\frac{z^{a-1}}{1+z}dz=2 \pi i Res\left(f(z) \right)\Big|_{z=-1}$ (1)

Let´s first calculate the residues

$2 \pi i Res\left(f(z) \right)\Big|_{z=-1}=2 \pi i \lim_{z \rightarrow -1}(1+z)\frac{z^{a-1}}{(1+z) }=2 \pi i (e^{i \pi})^{a-1}$ (2)

Now let´s focus in each integral. Lets start by the integral around the big Arc and show that it vanishes as $R \longrightarrow \infty$

$I_{\Gamma}= \int_{\Gamma}\frac{z^{a-1}}{1+z}dz$

$|I_{\Gamma}| \leq \int_{0}^{2 \pi}\frac{|z^{a-1}|}{|1+z|}|dz|$

letting $z=Re^{i \theta} \, \Rightarrow dz=iRe^{i \theta} d \theta$ . We have:

|z|=|R|

$|dz|=R d \theta \, \text{and }\,|z^{a-1}|=R^{a-1}$

$z=(z+1)-1 \Rightarrow$

$|z| \leq|z+1|-|1| \Rightarrow$

$|z+1| \geq R-1$

Then

$|I_{\Gamma}| \leq R\int_{0}^{2 \pi}\frac{R^{a-1}}{R-1}|d \theta|=\frac{R^{a}}{R-1}\int_{0}^{2 \pi} d \theta=\frac{2 \pi R^{a}}{R-1}$

$=2 \pi \frac{R^{a+1}}{1-\frac{1}{R}} \, \longrightarrow 0 \, \text{as}\, R \longrightarrow \infty$

Since 0<a<1

Similarly, letting $z= \epsilon e^{i \phi}$ over $\gamma$ we get

$I_{\gamma}=\int_{\gamma}\frac{z^{a-1}}{1+z}dz$

$|I_{\gamma}| \leq\epsilon\int_{0}^{2 \pi}\frac{\epsilon^{a-1}}{1-\epsilon}d\phi$

$|I_{\gamma}| \leq 2 \pi \lim_{\epsilon \rightarrow 0} \frac{\epsilon^{a}}{1-\epsilon} \, \longrightarrow 0$

Therefore, from (1) and (2) we have

$\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx-\int_{0}^{\infty}\frac{(xe^{2 i \pi})^{a-1}}{1+x}dx=2 \pi i (e^{i \pi})^{a-1}$

$\left(1-e^{2 i \pi(a-1)} \right)\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx=2 \pi i (e^{i \pi})^{a-1}$

$e^{i \pi (a-1)}\left(e^{ -i \pi(a-1)}-e^{ i \pi(a-1)} \right)\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx=2 \pi i e^{i \pi}^{(a-1)}$

$-\frac{\left(e^{ i \pi(a-1)}-e^{- i \pi(a-1)} \right)}{2i}\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx= \pi$

$-\sin(\pi(a-1))\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx= \pi$

$\sin(\pi a)\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx= \pi$

$\boxed{\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx= \frac{\pi}{\sin(\pi a)} }$ (3)

Lets start by the noting the following fact

$\frac{1}{1+x}=\int_{0}^{\infty}e^{-(1+x)t}dt$

Than

$\int_{0}^{\infty}\frac{x^{x-1}}{1+x}du=\int_{0}^{\infty} x^{a-1}\int_{0}^{\infty}e^{-(1+x)t}dt\,dx$

$=\int_{0}^{\infty}\!\!\!\int_{0}^{\infty} x^{a-1}e^{-xt}e^{-t}dx\,dt$

$=\int_{0}^{\infty}e^{-t}\!\!\!\int_{0}^{\infty} x^{a-1}e^{-xt}dx\,dt$

now let $xt\longmapsto w$ in the inner integral

$=\int_{0}^{\infty}e^{-t} \bigg[\int_{0}^{\infty} (\frac{w}{t})^{a-1}e^{-w}\frac{dw}{t} \bigg]\,dt$

$=\int_{0}^{\infty}e^{-t} \bigg[\frac{1}{t^{a}}\int_{0}^{\infty} {w}^{a-1}e^{-w}dw \bigg]\,dt$

From the standard integral representation of the Gamma function

$\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx=\int_{0}^{\infty}e^{-t} \bigg[\frac{\Gamma(a)}{t^{a}} \bigg]\,dt$

$=\Gamma(a)\int_{0}^{\infty}e^{-t} t^{-a}dt$

$\boxed{\int_{0}^{\infty}\frac{x^{a-1}}{1+x}dx=\Gamma(a)\Gamma(1-a)}$ (4)

Equating (3) and (4) obtain the reflection formula for the Gamma function:

$\boxed{\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin(a\pi)}}$ (5)

if we enforce $x \longrightarrow \frac{1}{2}-x$

$\Gamma(\tfrac{1}{2}-x)\Gamma(1-(\tfrac{1}{2}-x))=\Gamma(\tfrac{1}{2}-x)\Gamma(\tfrac{1}{2}+x)=\frac{\pi}{\sin\pi(\tfrac{1}{2}-x)}=\frac{\pi}{\cos\pi x}$