MHB A very interesting complex beta integral:

[alyafey22]

This is one of the most interesting integrals I've ever seen

$$\frac{1}{2\pi i }\int^{c+i\infty}_{c-i\infty}t^{-a} (1-t)^{-b-1}\, dt = \frac{1}{b\,\beta(a,b)}$$

Does anybody have any idea how to prove it ?

 

[Sudharaka]

This is one of the most interesting integrals I've ever seen

$$\frac{1}{2\pi i }\int^{c+i\infty}_{c-i\infty}t^{-a} (1-t)^{-b-1}\, dt = \frac{1}{b\,\beta(a,b)}$$

Does anybody have any idea how to prove it ?

Hi ZaidAlyafey, :)

Is \(a\) and/or \(b\) integers?

 

[alyafey22]

Hi ZaidAlyafey, :)

Is \(a\) and/or \(b\) integers?

Hi, when first I saw this equality there didn't seem to be this restriction , but let us
assume for simplicity that a and b are integers.

 

[alyafey22]

I guess you are thinking about using residues in a way similar to the broomwich integral.

 

[Sudharaka]

I guess you are thinking about using residues in a way similar to the broomwich integral.

Yeah, but I don't think I can find a way to get the required answer using that method. Where did you find this integral?

 

[alyafey22]

Yeah, but I don't think I can find a way to get the required answer using that method. Where did you find this integral?

Well, after searching I got that which is called the Third(Cauchy's) beta integral :

$$\int^{\infty}_{-\infty}\frac{dt}{(1-it)^a(1+it)^b}= \frac{\pi 2^{2-a-b}\Gamma{(a+b-1)}}{\Gamma(a)\Gamma(b)}$$

Clearly our integral can be derived by doing a substitution , so this is a general formula
I found this in a paper SPECIAL FUNCTIONS AND THEIR SYMMETRIES by Vadim KUZNETSOV.