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