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Fact 1. ($B$ is beta-function) $$\int_{0}^{1}\frac{z^{-(n+1)}}{(1+\frac{1}{z})^{2n}}=\frac{1}{2}B(n,n)$$

I can find above fact by using MATLAB.

But i like to show above fact using beta function property.

To do this, i try to use "$B(m,n)=2\int_{0}^{1}{x^{2m-1}(1-x^{2})^{n-1}dx}$".

But i fail to show above fact.

Is there any other related beta function property related with my case?

Or how i can show above fact?

Thank you!

Kim
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1 Answers1

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Let $I_n$ your integral. First, the change of variable $z=1/t$ give that $\displaystyle I_n=\int_1^{+\infty}\frac{t^{n-1}}{(1+t)^{2n}} dt$. Hence $\displaystyle 2I_n=\int_0^{+\infty}\frac{t^{n-1}}{(1+t)^{2n}} dt$. Now in the last integral, we put $\displaystyle u=\frac{1}{1+t}$. This gives $\displaystyle 2I_n=\int_0^1 u^{n-1}(1-u)^{n-1}du=B(n,n)$.

Kelenner
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  • Thank you for your comment. um... I think in second integral equation $(1+t)^{n}$ is corrected to $(1+t)^{2n}$. Is it correct? And what is property used to reach $2I_{n}$ – Kim Jul 11 '16 at 07:57
  • Thank you for your remark, I have corrected the misprint. To show the formula for $2I_n$,( I do not know is this is your question), I have used $\int_0^{+\infty}=\int_0^1+\int_1^{+\infty}$ – Kelenner Jul 11 '16 at 08:20