Deriving the Beta Function Integral Using Residue Theorem

[mkbh_10]

Will some one help me to prove this identity

G(n)+G(1-n)= pi/ sin npi 0<n<1

B(m,n) = (m-1)! / n(n+1)...(n+m+1) ,for beta function

 

[Rainbow Child]

You mean
$$\Gamma(n)*\Gamma(1-n)=\frac{\pi}{\sin(n\,\pi)}$$

First of all use the identity
$$B(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$$
with $x=n,\,y=1-n$ to arrive to $B(n,1-n)=\Gamma(n)\,\Gamma(1-n)$, i.e.

$$\Gamma(n)\,\Gamma(1-n)=\int_0^\infty\frac{u^{n-1}}{u+1}\,d\,u$$

which can be calculated with the use of residues.

 

[mkbh_10]

by residue it will give limit u tending to -1 [(-1)^n-1] Integral = 2pi i * Residue

which =2pi i *(-1)^n-1 ,how to proceed further

 

[Rainbow Child]

by residue it will give limit u tending to -1 [(-1)^n-1] Integral = 2pi i * Residue

which =2pi i *(-1)^n-1 ,how to proceed further

I cann't understand that you are saying. In order to calculate the integral choose a keyhole contour like this