Is it Possible to show this? Beta function

Sudharaka

Well-known member
MHB Math Helper
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?

Hi chamilka, 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.

• Chris L T521 and chamilka

Jameson

Hi chamilka! Welcome to MHB! 