An improper integral relates to Gamma function

Question

Let $p\in(1,2)$. Consider the following integral: for $x>0$, $$ \int_{0}^{\infty}\left(\frac{1}{1+e^{-(x+y)}}-\frac{1}{1+e^{-x}}\right)y^{-p}dy. $$ Is it possible to find the value of this integral? Or, to prove that it's divergent for all $x>0$ since the integral in the part that $y\in(0,1)$ seems to be too large. Any hint, references or help would be much appreciated. Thank you very much.

Answer 1

The integral is convergent; the difference in exponentials easily has order $y$ as $y\to 0$ which is sufficient since $y^{1-p}$ is integrable around zero. In fact, the bounds on $p$ are perfect for some contour integration. To elaborate, notice the exponential difference is: $$e^{-x}\cdot\frac{e^{-y}-1}{(1+e^{-x-y})(1+e^{-x})}=\frac{e^{-x}}{(1+e^{-x})^2}\cdot(-y)+\text{higher order terms}$$

If we extend the symbol "$y^{-p}$" to the complex plane via the logarithm induced from $0\le\arg<2\pi$ we get ourselves an integrand which is meromorphic on $\Bbb C\setminus[0,\infty)$. A standard Hankel contour setup (closed with a sensible box contour) suffices here; all the things vanish that need to vanish. We pick up the residues at $z_n=-x+\pi i\cdot(2n+1)$, $n\in\Bbb Z$ and find: $$(1-e^{-2\pi ip})\cdot J=2\pi i\sum_{n\in\Bbb Z}z_n^{-p}$$Where $J$ is our integral.

With this branch of the argument we compute: $$z_n^{-p}=\exp\left(-\pi p i+ip\cdot\arctan\left(\frac{\pi(2n+1)}{x}\right)\right)\cdot(x^2+\pi^2(2n+1)^2)^{-p/2}$$Summing over all integers, observing the even-odd symmetries in the expression for $z_n^{-p}$, we are reduced to: $$J=2\pi\csc(\pi p)\cdot\sum_{n=1}^\infty\cos\left(p\cdot\arctan\left(\frac{\pi(2n-1)}{x}\right)\right)\cdot(x^2+\pi^2(2n-1)^2)^{-p/2}$$

In my view, this is quite intractable as a series. I would be happy to be corrected on this point.

The only interesting thing is that we catch the asymptotics as $x\to 0^+$; we see $J$ converges to: $$J(0^+)=\pi^{1-p}\csc(\pi p/2)\cdot\left(1-2^{-p}\right)\zeta(p)=(2^p-1)\cdot\Gamma(1-p)\zeta(1-p)$$

Which isn't especially obvious at first glance but is not too hard to obtain using contour integration.