For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. Throughout the rest of this question we assume $x > 1$ and $y > 1$. We denote $\Gamma ( x ) = \int _ { 0 } ^ { + \infty } t ^ { x - 1 } e ^ { - t } \mathrm {~d} t$, $F _ { x , y }$ the antiderivative on $\mathbb { R } ^ { + }$ of $t \mapsto e ^ { - t } t ^ { x + y - 1 }$ which vanishes at 0, and $G ( a ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } F _ { x , y } ( ( 1 + u ) a ) \mathrm { d } u$. Express for $a > 0$, $G ^ { \prime } ( a )$ as a function of $\Gamma ( x ) , e ^ { - a }$ and $a ^ { y - 1 }$.
For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. Throughout the rest of this question we assume $x > 1$ and $y > 1$. We denote $\Gamma ( x ) = \int _ { 0 } ^ { + \infty } t ^ { x - 1 } e ^ { - t } \mathrm {~d} t$, $F _ { x , y }$ the antiderivative on $\mathbb { R } ^ { + }$ of $t \mapsto e ^ { - t } t ^ { x + y - 1 }$ which vanishes at 0, and $G ( a ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } F _ { x , y } ( ( 1 + u ) a ) \mathrm { d } u$.
Express for $a > 0$, $G ^ { \prime } ( a )$ as a function of $\Gamma ( x ) , e ^ { - a }$ and $a ^ { y - 1 }$.