Let $\Gamma$ be the Gamma function defined as the pointwise limit on $]0, +\infty[$ of $\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}$.
Show that for all $a \in ]0, +\infty[$ and $x \in ]0, +\infty[$:
$$\int_0^{+\infty} \frac{t^{x-1}}{(1+t)^{x+a}} dt = \frac{\Gamma(x)\Gamma(a)}{\Gamma(x+a)}.$$
Hint: you may set, for $x \in ]0, +\infty[$, $f(x) = \frac{\Gamma(x+a)}{\Gamma(a)} \int_0^{+\infty} \frac{t^{x-1}}{(1+t)^{x+a}} dt$.