Infinite sum involving beta functions

Question

I am trying to prove the following identity

$\sum_{k=0}^{\infty}\frac{\Gamma(k+c)}{k!}B(k+a, b)=\Gamma(c)B(a, b-c)$, where a and c are non-negative integers.

My first approach was to use the definitions of the Gamma and Beta functions. That is

$\sum_{k=0}^{\infty}\frac{\Gamma(k+c)}{k!}B(k+a, b)=\sum_{k=0}^{\infty}\frac{1}{k!}\int_{0}^{\infty}z^{k+c+1}\exp(-z)dz\int_{0}^{1}x^{k+a-1}(1-x)^{b-1}dx$.

Then, exchanging the sum and the integrals and rearranging i get

$\sum_{k=0}^{\infty}\frac{\Gamma(k+c)}{k!}B(k+a, b)=\int_{0}^{\infty}\int_{0}^{1}\exp(-z)(1-x)^{b-1}z^{c}x^{a}\left[\sum_{k=0}^{\infty}\frac{(xz)^{k-1}}{k!}\right]$dxdz.

And i got stuck at this point. Any help will be appreciated.

Answer 1

\begin{align} \sum_{k=0}^\infty\frac{\Gamma(k+c)}{k!}\mathrm{B}(k+a,\color{red}{b})&=\sum_{k=0}^\infty\frac{1}{k!}\int_0^\infty z^{k+c\color{red}{-}1}e^{-z}\,dz\int_0^1 x^{k+a-1}(1-x)^{b-1}\,dx\\&=\int_0^\infty\!\!\!\int_0^1 z^{c\color{red}{-1}}x^{a\color{red}{-1}}(1-x)^{b-1}e^{-z}\underbrace{\sum_{k=0}^\infty\frac{(xz)^k}{k!}}_{=e^{xz}}\,dx\,dz\\&=\int_0^1 x^{a-1}(1-x)^{b-1}\underbrace{\int_0^\infty z^{c-1}e^{-(1-x)z}\,dz}_{=\Gamma(c)/(1-x)^c}\,dx\\&=\Gamma(c)\int_0^1 x^{a-1}(1-x)^{b-c-1}\,dx=\Gamma(c)\mathrm{B}(a,b-c). \end{align} [Valid (at least) for real $a>0$, $b>c>0$.]

Answer 2

hint:

$\frac{\Gamma (k+c)}{k!}=\Gamma(c) \frac{\Gamma(k+c)}{k! \Gamma(c)} = \Gamma(c) \binom{k+c-1}{k}$