Tough Inverse Fourier Transform

Solution 1:

Part 1

First rewrite $g(x)$ as $$ g(x)=\frac{1}{2\pi}\int_{0}^{\infty}\frac{\cosh\frac{k}{2}-\cosh k\left(y-\frac12\right)}{k^2\cosh\frac{k}{2}}2\cos kx\,dk.\tag{1}$$ What I will do next is not entirely rigorous but I think can be made so relatively easily. Namely, let us consider instead of (1) a two-parameter deformation $$ I(s,a,x)=\int_{0}^{\infty}k^{s-1}\frac{\cosh\frac{ak}{2}-\cosh \frac{kb}{2}}{\cosh \frac{k}{2}}2\cos kx\,dk,\tag{2}$$ where $b=2y-1$. In subsequent calculations I will assume that $0<a<1$ and $s>0$, though obviously we are interested in the values $a=1$, $s=-1$. They will be obtained by analytic continuation in the final answer.

We can thus write \begin{align} I(s,a,x)=\sum_{\epsilon,\epsilon'=\pm1}\int_0^{\infty}k^{s-1}\frac{e^{-\left(\frac{1}{2}(1+\epsilon ' a)+i\epsilon x\right)k}-e^{-\left(\frac{1}{2}(1+\epsilon ' b)+i\epsilon x\right)k}}{1+e^{-k}}dk. \end{align} This can be expressed in terms of Hurwitz zeta function $\zeta(s,\alpha)$, since $$ \int_{0}^{\infty}k^{s-1}\frac{e^{-2\alpha k}}{1+e^{-k}}dk=\frac{\Gamma(s)}{2^s}\left[\zeta(s,\alpha)-\zeta\left(s,\alpha+\frac12\right)\right].$$ So we have \begin{align} I(s,a,x)=\frac{\Gamma(s)}{2^s}\sum_{\epsilon,\epsilon'=\pm1}\Bigl[\zeta\left(s,\frac{1+\epsilon ' a+2i\epsilon x}{4}\right)-\zeta\left(s,\frac{3+\epsilon ' a+2i\epsilon x}{4}\right)\\ \Bigl.-\zeta\left(s,\frac{1+\epsilon ' b+2i\epsilon x}{4}\right)+\zeta\left(s,\frac{3+\epsilon ' b+2i\epsilon x}{4}\right)\Bigr]. \end{align} It is not immediately obvious but the sum $\sum_{\epsilon,\epsilon'}$, as a function of $s$, has a simple zero at $s=-1$, which compensates the pole of the gamma function. Then, taking appropriate limits, we get $$ g(x)=\frac{2}{\pi}\sum_{\epsilon=\pm1}\mathrm{Re}\Bigl[\zeta'_{-1}\left(\frac{1+\epsilon b+2i x}{4}\right)-\zeta'_{-1}\left(\frac{3+\epsilon b+2i x}{4}\right)-\zeta'_{-1}\left(\frac{1+\epsilon +2i x}{4}\right)+\zeta'_{-1}\left(\frac{3+\epsilon +2ix}{4}\right)\Bigr],\tag{3}$$ where I denote $\zeta'_{-1}(a)=\left[\frac{\partial}{\partial s}\zeta(s,a)\right]_{s=-1}$. This was confirmed numerically.

Maybe the result written in such a form is not useful anyway, but I don't exclude that it can be further simplified. For this, one should look if there exists a closed formula for $\zeta'_{-1}(a)$ (for example, it does exist for $\zeta(-1,a)=-\frac12\left(a^2-a+\frac16\right)$).

Part 2

Apparently (formula (4) here) the derivative $\zeta'_{-1}(a)$ can be expressed in terms of Barnes $G$-function: $$ \zeta'_{-1}(a)-\zeta'_{-1}(1)=(a-1)\ln\Gamma(a)-\ln G(a).$$ This is quite good as I understand $G(z)$ much better (in fact I've already thought about this function since the structure of your IFT looks somewhat similar to its integral representation) and I am now sure that the answer is more or less non-simplifiable for generic $x,y$. One thing that does simplify is the contribution of the last two terms in (3): $$ \frac{2}{\pi}\sum_{\epsilon=\pm1}\mathrm{Re}\Bigl[-\zeta'_{-1}\left(\frac{1+\epsilon +2i x}{4}\right)+\zeta'_{-1}\left(\frac{3+\epsilon +2ix}{4}\right)\Bigr]=-\frac{|x|}{2}.$$ Introducing the notation $\xi=\frac{y+ix}{2}$, $\bar{\xi}=\frac{y-ix}{2}$, the whole answer can be rewritten as \begin{align} g(x)=\frac{1}{\pi}\left[\xi\ln\cot\pi\xi+\bar{\xi}\ln\cot\pi\bar{\xi}-\ln\Gamma(\xi)\Gamma(\bar{\xi})+\frac{1}{2}\ln\frac{\Gamma\left(\frac12+\xi\right)\Gamma\left(\frac12+\bar{\xi}\right)}{ \Gamma\left(\frac12-\xi\right)\Gamma\left(\frac12-\bar{\xi}\right)}\right]-\frac{|x|}{2}+\\+\frac{1}{\pi}\ln\frac{G(1-\xi)G(1-\bar{\xi})G\left(\frac12+\xi\right)G\left(\frac12+\bar{\xi}\right)}{G(\xi)G(\bar{\xi}) G\left(\frac12-\xi\right)G\left(\frac12-\bar{\xi}\right)}.\end{align} The combinations of type $\ln\frac{G(z)}{G(1-z)}$ can also be written in terms of Clausen function, but this makes the result less symmetric without considerable simplification.