The Beta Function

Today´s post I want to talk about the Beta function and in the end of the post show it´s usefulness to compute a family of hard and important integrals.

Lets start observing the following fact: Consider the integral

$I=\int_{0}^{\infty}e^{-tu}u^{x-1}du \qquad \qquad \qquad (1)$

Now, let´s substitute $tu=w \, \Rightarrow \, du=\frac{dw}{t}$ ,the limits of integration remain the same.

$I=\int_{0}^{\infty}e^{-w}\Big(\frac{w}{t}\Big)^{x-1}\frac{dw}{t}$

$=\frac{1}{t^x}\underbrace{\int_{0}^{\infty}e^{-w}w^{x-1}dw}_{=\Gamma(x)}$

and we conclude that

$\int_{0}^{\infty}e^{-tu}u^{x-1}du = \frac{\Gamma(x)}{t^x}\qquad \qquad \qquad (2)$

$\boxed{\frac{1}{t^x}=\frac{1}{\Gamma(x)}\int_{0}^{\infty}e^{-tu}u^{x-1}du} \qquad \qquad \qquad (3)$

Now, lets use (3) to evaluate the integral

$I= \int_{0}^{\infty}\frac{t^{x-1}}{\big( 1+t \big)^{x+y}}dt\qquad \qquad \qquad (4)$

According to (3) we can write

$\frac{1}{t^{x+y}}=\frac{1}{\Gamma(x+y)}\int_{0}^{\infty}e^{-tu}u^{x+y-1}du \qquad \qquad \qquad (5)$

Substituting (5) in (4) we get

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

swapping the order of integration

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

The inner integral inside the curly brackets is in the same form as (2) above, therefore

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

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

$= \frac{\Gamma(x)}{\Gamma(x+y)}\underbrace{\int_{0}^{\infty} e^{-u}u^{y-1} \,du}_{\Gamma(y)}$

And finally!

$\boxed{\int_{0}^{\infty}\frac{t^{x-1}}{\big( 1+t \big)^{x+y}}dt=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}}$ (6)

(6) is known as the Euler´s Beta Function and is usually written as

$B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

An important fact to observe is that the integral evaluated here

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

is a special case of (6) where x+y=1

Consider now the following substitution in (6)

$u=\frac{t}{1+t}=1-\frac{1}{1+t}$

$1-u=\frac{1}{1+t}$

$1+t=\frac{1}{1-u}$

$dt=\frac{du}{(1-u)^2}$

The limits change to

$\text{when} \,\,\,\, t=0 \Rightarrow u=0$

$\text{when} \,\,\,\, t=\infty \Rightarrow u=1$

therefore, we get

$\int_{0}^{\infty}\frac{t^{x-1}}{\big( 1+t \big)^{x+y}}dt=\int_{0}^{1}\bigg( 1-\frac{1}{1-u}\bigg)^{x-1}\Big( 1-u\Big)^{x+y}\frac{du}{(1-u)^2}$

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

$=\int_{0}^{1}u^{x-1}\Big( 1-u\Big)^{y-1}du$

We conclude that

$B(x,y)=\int_{0}^{1}u^{x-1}\Big( 1-u\Big)^{y-1}du$

$\boxed{\int_{0}^{1}u^{x-1}\Big( 1-u\Big)^{y-1}du=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}}$ (7)

Lets consider another substitution in (7)

$u=\cos^2(\theta) \Rightarrow du=2 \cos(\theta)(-\sin(\theta))d\theta$

$\text{when} \,\,\,\, u=0 \Rightarrow \theta=\frac{\pi}{2}$

$\text{when} \,\,\,\, u=1 \Rightarrow \theta=0$

we get

$\int_{0}^{1}u^{x-1}\Big( 1-u\Big)^{y-1}du=2\int_{\pi/2}^{0} (\cos^2(\theta))^{x-1}(\sin^2(\theta))^{y-1}\cos(\theta) \sin(\theta)(-d \theta)$

$\boxed{ \int_{0}^{\pi/2}\cos^{2x-1}(\theta)\sin^{2y-1}(\theta)d \theta=\frac{1}{2}\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}}$ (8)

A last substitution in (8), i.e.

$2x-1=\alpha \qquad \text{and} \qquad 2y-1 = \beta$

gives us

$\int_{0}^{\pi/2}\cos^{\alpha}(\theta)\sin^{\beta}(\theta)d \theta=\frac{1}{2}\frac{\Gamma(\frac{\alpha+1}{2})\Gamma(\frac{\beta+1}{2})}{\Gamma(\frac{\alpha+\beta+2}{2})}$ (9)

This last form is extremely useful to evaluate a family of hard integrals. For instance, set $\alpha=0$ in (9)

$\int_{0}^{\pi/2}\sin^{\beta}(\theta)d \theta=\frac{1}{2}\frac{\Gamma(\frac{1}{2})\Gamma(\frac{\beta+1}{2})}{\Gamma(\frac{\beta+2}{2})}$

$\int_{0}^{\pi/2}\sin^{\beta}(\theta)d \theta=\frac{\sqrt{\pi}}{2}\frac{\Gamma(\frac{1}{2}+ \frac{\beta}{2})}{\Gamma(1+\frac{\beta}{2})}$

Now, differentiate the above expression with respect to $\beta$

$\frac{d}{d\beta}\int_{0}^{\pi/2}e^{\beta\log(\sin(\theta))}d \theta=\frac{\sqrt{\pi}}{2}\frac{d}{d\beta}\frac{\Gamma(\frac{1}{2}+ \frac{\beta}{2})}{\Gamma(1+\frac{\beta}{2})}$

$\int_{0}^{\pi/2}\sin^{\beta}(\theta)\log(\sin(\theta))d \theta=\frac{\sqrt{\pi}}{2}\frac{d}{d\beta}\frac{\Gamma(\frac{1}{2}+ \frac{\beta}{2})}{\Gamma(1+\frac{\beta}{2})}$

setting $\beta=0$

$\int_{0}^{\pi/2}\log(\sin(\theta))d \theta=\frac{\sqrt{\pi}}{2}\frac{d}{d\beta}\frac{\Gamma(\frac{1}{2}+ \frac{\beta}{2})}{\Gamma(1+\frac{\beta}{2})}\bigg|_{\beta=0}$

We can keep this process of differentiation further to get

$\boxed{\int_{0}^{\pi/2}\log^{k}(\sin(\theta))d \theta=\frac{\sqrt{\pi}}{2}\frac{d^{k}}{d\beta^{k}}\frac{\Gamma(\frac{1}{2}+ \frac{\beta}{2})}{\Gamma(1+\frac{\beta}{2})}\bigg|_{\beta=0}}$