Showing that $\int_{\mathbb{R}} e^{x(t+1)} \left(1 + \frac{e^x}{k} \right)^{-(k+1)}dx = k^{t+1} \frac{\Gamma(k-t)\Gamma(1+t)}{\Gamma{(k+1)}}$

Question

I want to show that

$$ f(t) = \int_{\mathbb{R}} e^{x(t+1)} \left(1 + \frac{e^x}{k} \right)^{-(k+1)}dx = k^{t+1} \frac{\Gamma(k-t)\Gamma(1+t)}{\Gamma{(k+1)}}$$

Here, $k,t \in \mathbb{R}$.

I have tried a few things, but none worked out. First, I noticed that the quotient with all the gamma terms can arises from the beta function with $x=k-t$, $y=1+t$, and $x+y=k+1$. But I haven't been able to transform my integral into anything of the required form.

Next, I tried using the fact that we can write $\Gamma(x)\Gamma(y)$ as a double integral with limits of integration $[0,\infty)$, but that did not get me anywhere.

I would greatly appreciate any advice on how to solve this problem.

Answer 1

Consider the change of variables $\frac{e^x}{k}=z , dx=k\frac{dz}{z}$ , the integral becomes

$$ I(k,t)=k^{t+1}\int_{\mathbb{R\geq0}}\frac{z^{t}}{(1+z)^{k+1}} $$

which equals a integral representation of the Beta function

$$ I(k,t)=k^{t+1}B(t+1,k-t)=k^{t+1}\frac{\Gamma(k-t)\Gamma(t+1)}{\Gamma(k+1)} $$

as long as $t+1>0$ and $k-t>0$