The Reflection Formula for the Gamma and Digamma functions and the Infinite product of Sine

Today we are going to prove the famous Reflection formula of the Gamma function in two different ways. As a bonus, we show the proof of the infinite product of $\sin(x)$ and the reflection formula of the digamma function.

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

We will start by computing the following integral

$I(x)=\int_{0}^{\infty}\frac{u^{x-1}}{1+u}du=\Gamma(x)\Gamma(1-x)$

for 0<x<1 .

Reflection formula of The Gamma function first way

Let´s start

$I(x)=\int_{0}^{\infty}\frac{u^{x-1}}{1+u}du$

$\qquad =\int_{0}^{1}\frac{u^{x-1}}{1+u}du+\int_{1}^{\infty}\frac{u^{x-1}}{1+u}du$

Now let $u\longmapsto \frac{1}{u}$ in the second integral

$I(x)=\int_{0}^{1}\frac{u^{x-1}}{1+u}du+\int_{1}^{0}\frac{(\frac{1}{u})^{x-1}}{1+\frac{1}{u}}\frac{(-du)}{u^2}$

$=\int_{0}^{1}\frac{u^{x-1}}{1+u}du+\int_{0}^{1}\frac{u^{1-x}}{u(1+u)}du$

$=\int_{0}^{1}\frac{u^{x-1}}{1+u}du+\int_{0}^{1}\frac{u^{-x}}{(1+u)}du$

$=\int_{0}^{1}\frac{u^{x-1}+u^{-x}}{1+u}du$

$=\int_{0}^{1}(u^{x-1}+u^{-x})\sum_{k=0}^{\infty}(-1)^ku^kdu$

$=\sum_{k=0}^{\infty}(-1)^k\int_{0}^{1}u^{k+x-1}+u^{k-x}du$

$=\sum_{k=0}^{\infty} \frac{(-1)^k}{k+x}+\sum_{k=0}^{\infty}\frac{(-1)^k}{k-x+1}$

$=\frac{1}{x}+\sum_{k=1}^{\infty} \frac{(-1)^k}{k+x}+\sum_{k=0}^{\infty}\frac{(-1)^{k+1}(-1)^{-1}}{k+1-x}$

$=\frac{1}{x}+\sum_{k=1}^{\infty} \frac{(-1)^k}{k+x}-\sum_{k=1}^{\infty}\frac{(-1)^k}{k-x}$

$=\frac{1}{x}+\sum_{k=1}^{\infty} \frac{(-1)^k}{k+x}-\frac{(-1)^k}{k-x}$

$=\frac{1}{x}+\sum_{k=1}^{\infty} \frac{(-1)^{k}(k-x)-(-1)^{k}(k+x)}{k^2-x^2}$

$\int_{0}^{\infty}\frac{u^{x-1}}{1+u}du=\frac{1}{x}-2x\sum_{k=1}^{\infty} \frac{(-1)^{k}}{k^2-x^2}$

from our previous post we know that

$\boxed{\frac{\pi}{\sin(x\pi)}=\frac{1}{x}+2x\sum_{k=1}^{\infty} \frac{(-1)^{k}}{x^2-k^2}}$ (2)

Therefore

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

On the other hand, if we prove that

$\int_{0}^{\infty}\frac{u^{x-1}}{1+u}du=\Gamma(x)\Gamma(1-x)$

we conclude the proof.

Lets start by the noting the following fact

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

Than

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

$=\int_{0}^{\infty}\!\!\!\int_{0}^{\infty} u^{x-1}e^{-ut}e^{-t}du\,dt$

$=\int_{0}^{\infty}e^{-t}\!\!\!\int_{0}^{\infty} u^{x-1}e^{-ut}du\,dt$

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

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

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

From the standard integral representation of the Gamma function

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

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

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

Equating (3) and (4) we finally get the beautiful formula:

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

Infinite Product of $\sin(x)$

I´ll now show a second approach to prove this relation. But before we will show another beautiful formula, the infinite product for the $\sin(x \pi)$ !

We start with the partial fraction expansion of the cotangent proved in this post.

$\pi \cot(\pi x)-\frac{1}{x}=\sum_{k=0}^{\infty}\frac{2x}{x^2-k^2}$

Making a change of variable $x\longmapsto \frac{x}{\pi}$

$\pi \cot( x)-\frac{\pi}{x}=\frac{1}{\pi}\sum_{k=0}^{\infty}\frac{2x}{\frac{x^2}{\pi^2}-k^2}$

$=\frac{1}{\pi}\sum_{k=0}^{\infty}\frac{2x\pi^2}{{x^2}-\pi^2k^2}$

$=\pi\sum_{k=0}^{\infty}\frac{2x}{{x^2}-\pi^2k^2}$

$\cot( x)-\frac{1}{x}=\sum_{k=0}^{\infty}\frac{2x}{{x^2}-\pi^2k^2}$ (5)

Now we integrate (5) from to

$\int_{0}^{z}\Big(\cot( x)-\frac{1}{x} \Big)dx=\int_{0}^{z}\sum_{k=0}^{\infty}\frac{2x}{{x^2}-\pi^2k^2}dx$

For the first integral on the left side recall that $\cot(x)=\frac{\cos(x)}{\sin(x)}$ and make the substitution

$u=\sin(x) \text{,} \,\,\, du=\cos(x)dx \,\,\text{to get} \qquad \int\cot(x)dx=\int\frac{du}{u}=\log(\sin(x))$ .

The integral on the right hand side can be done either via inspection or substitution and is equal to $\log(x^2 - \pi^2k^2)$ (take the derivative to convince yourself).

$\log(sin(x))-\log(x)\Big|_{0}^{z}=\sum_{k=1}^{\infty}\{\log(x^2 - \pi^2k^2)\Big|_{0}^{z}\}$

$\log\Big(\frac{\sin(z)}{z} \Big)-\lim_{x \rightarrow 0}\log\Big(\frac{\sin(x)}{x} \Big)=\sum_{k=1}^{\infty}\Big\{\log(z^2 - \pi^2k^2)-\log(-\pi^2k^2) \Big\}$

By the continuity of $\log$ we can bring the $\lim_{x \rightarrow 0}$ inside

$\log\Big(\frac{\sin(z)}{z} \Big)-\log\Big(\lim_{x \rightarrow 0}\frac{\sin(x)}{x} \Big)=\sum_{k=1}^{\infty}\log(\frac{z^2 - \pi^2k^2}{ - \pi^2k^2})$

The $\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1$ , here is a proof. Another easy way to prove it is by expanding $\sin(x)$ in it´s Taylor series, divide by and then take the limit. Regardless the way $\log(1)=0$ . Also, by $\log$ properties we can bring the summation operator inside the log and it becomes a product operator, therefore:

$\log\Big(\frac{\sin(z)}{z} \Big)=\log(\prod_{k=1}^{\infty}1-\frac{z^2 }{ \pi^2k^2})$

Exponentiating both sides we finally get

$\boxed{\frac{\sin(z)}{z} =\prod_{k=1}^{\infty}\Big(1-\frac{z^2 }{ \pi^2k^2}\Big)}$ (6)

Reflection formula of The Gamma function second way

Now, lets proof the reflection formula based on the infinite product derived in the previous section. Recall the Gauss product for the Gamma function

$\Gamma(z)= \lim_{n \rightarrow \infty} \,\frac{n!\, n^z}{z (z+1)\cdots (z+n)}$

and the functional equation of the Gamma function

$\Gamma(x+1)=x\Gamma(x)$

Lets write $\Gamma(z)\Gamma(1-z)=-z\Gamma(z)\Gamma(-z)$ in terms of Gauss products

$\Gamma(z)\Gamma(1-z)=-z\lim_{n \rightarrow \infty} \,\frac{n!\, n^z}{z (z+1)\cdots (z+n)} \cdot\frac{n!\, n^{-z}}{-z (1-z)\cdots (n-z)}$

$=\frac{1}{z}\lim_{n \rightarrow \infty}\frac{(n!)^2\, n^{-z}n^z}{ \underbrace{(1+z)(1-z)}_{=(1^2-z^2)}\cdots \underbrace{(n+z)(n-z)}_{=(n^2-z^2)}}$

$=\frac{1}{z}\lim_{n \rightarrow \infty}\frac{(1\cdot 2 \cdot 3 \cdot n)^2\, }{ (1^2-z^2)(2^2-z^2)\cdots (n^2-z^2)}$

$=\frac{1}{z}\lim_{n \rightarrow \infty}\frac{1^1\cdot 2^2 \cdot 3^2 \cdot n^2\, }{ 1^2\cdot(1-\frac{z^2}{1^2})2^2\cdot(1-\frac{z^2}{2^2})\cdots n^2 \cdot(1-\frac{z^2}{n^2})}$

$=\frac{1}{z}\lim_{n \rightarrow \infty}\frac{1^1\cdot 2^2 \cdot 3^2 \cdot n^2\, }{1^1\cdot 2^2 \cdot 3^2 \cdot n^2 \cdot(1-\frac{z^2}{1^2})\cdot(1-\frac{z^2}{2^2})\cdots \cdot(1-\frac{z^2}{n^2})}$