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- Thread starter chamilka
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- #1

- Feb 5, 2012

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Hi chamilka,Hi everyone!

I got two versions of one particular function and now I need to show those two versions are equivalent.

For that I need to show the follwing,

View attachment 223

Is it possible to show this by using the properties of Beta functions, Gaussian hypergeometric function etc?

Thanks in advance!!

I got the intuition of solving this problem from your previous problem. Consider the integral, \(\displaystyle\int_{0}^{1}x^{y+\alpha-1}(1-x)^{(n+\beta-y)-1}\,dx\).

\begin{eqnarray}

\int_{0}^{1}x^{y+\alpha-1}(1-x)^{(n+\beta-y)-1}\,dx&=&\int_{0}^{1}x^{y+\alpha-1}(1-x)^{n-y}(1-x)^{\beta-1}\,dx\\

&=&\int_{0}^{1}x^{y+\alpha-1}(1-x)^{n-y}\sum_{i=0}^{\infty} \; {\beta-1\choose i}\;(-x)^{i}\,dx

\end{eqnarray}

Proceed with the same method I have gone through in that post, and you will finally get,

\[\int_{0}^{1}x^{y+\alpha-1}(1-x)^{(n+\beta-y)-1}\,dx=\sum_{i=0}^{\infty}(-1)^{i}{\beta-1\choose i}B(y+\alpha+i,\,n-y+1)~~~~~~~~~~~(1)\]

provided, \(Re(y+\alpha+i)>0\mbox{ and }Re(n-y+1)>0\).

Also by the definition of the Beta function, if \(Re(y+\alpha)>0\mbox{ and }Re(n+\beta-y)>0\) we have,

\[\int_{0}^{1}x^{y+\alpha-1}(1-x)^{(n+\beta-y)-1}\,dx=B(y+\alpha,\,n+\beta-y)~~~~~~~(2)\]

From (1) and (2) we get,

\[B(y+\alpha,\,n+\beta-y)=\sum_{i=0}^{\infty}(-1)^{i}{\beta-1\choose i}B(y+\alpha+i,\,n-y+1)\]

provided, \(Re(y+\alpha)>0,\,Re(n-y+1)>0\mbox{ and }Re(n+\beta-y)>0\)

Kind Regards,

Sudharaka.

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- Jan 26, 2012

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I think you should be able to send PM's now so I've removed your replies to Sudharaka from public view. I can copy the posts to you through PM if you would like. The minimum post counts can be annoying I know, but they stop others from doing a lot of things that waste MHB's time so they are there for a reason. You should be able to use most of the functionality of the site already and if not only need a couple more posts. Let me know if I can help you in any way.

Jameson

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