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- #1

- Jan 17, 2013

- 1,667

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

Does anybody have any idea how to prove it ?

- Thread starter ZaidAlyafey
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- Thread starter
- #1

- Jan 17, 2013

- 1,667

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

Does anybody have any idea how to prove it ?

- Feb 5, 2012

- 1,621

Hi ZaidAlyafey,

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

Does anybody have any idea how to prove it ?

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

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- #3

- Jan 17, 2013

- 1,667

Hi, when first I saw this equality there didn't seem to be this restriction , but let usHi ZaidAlyafey,

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

assume for simplicity that a and b are integers.

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- #4

- Jan 17, 2013

- 1,667

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

- Feb 5, 2012

- 1,621

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

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- #6

- Jan 17, 2013

- 1,667

Well, after searching I got that which is called theYeah, but I don't think I can find a way to get the required answer using that method. Where did you find this 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