Euler's Integral

There are two types of Euler's Integral :

$1.$ Euler's integral of first kind. It is the also known as Beta Function and is defined as

$B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \mathrm{d}t$

for all $x,\ y \in \mathbb{C}$ such that $\Re(x),\ \Re(y) > 0$

For some positive integers $m,\ n$ we can define the beta function as

$B(m,n) = \dfrac{(m-1)!(n-1)!}{(m+n-1)!}$

$2.$ Euler's integral of second kind. It is the also known as Gamma Function and is defined as

$\Gamma(z) = \int_0^\infty t^{z-1}e^{-t} \mathrm{d}t$

for all $z \in \mathbb{C}$ such that $\Re(z) > 0$

For some positive integer $n$, we can define the gamma function as

$\Gamma(n) = (n-1)!$

Both of the Euler's integral can be related to each other by the following formula

$B(x,y)\Gamma(x+y) = \Gamma(x)\Gamma(y)$

Recall the definition of gamma function that :

$\displaystyle \Gamma(s) = \int _{ 0 }^{ \infty }{ { x }^{ s-1 }{ e }^{ -x }dx }$

Now one can write :

$\displaystyle \Gamma(m)\Gamma(n)=\int _{ 0 }^{ \infty }{ { x }^{ m-1 }{ e }^{ -x }dx } \int _{ 0 }^{ \infty }{ { y }^{ n-1 }{ e }^{ -y }dy }$

We can rewrite it as a double integral :

$\displaystyle \Gamma(m)\Gamma(n)\int _{ 0 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { x }^{ m-1 }{ y }^{ n-1 }{ e }^{ -(x+y) }dx dy} }$

Apply the substitution $x=vt , y=v(1-t)$ we have :

$\displaystyle \Gamma(m)\Gamma(n) = \int _{ 0 }^{ \infty }{ { t }^{ m-1 }{ (1-t) }^{ n-1 }dt\int _{ 0 }^{ 1 }{ v^{ m+n-1 }{ e }^{ -v }dv } }$

Using definition of gamma and beta function we have :

$\Gamma(m)\Gamma(n) = B(m,n)\Gamma (m+n)$

Hence proved $_\square$

For further details you can read the following wikis

• Beta Function

• Gamma Function