In reference to this answer I gave the other day, I came across a very interesting function whose IFT would be nice to evaluate as part of completing the solution to the problem I answered. The function is this:

$$\hat{g}(k) = \frac{1}{k^2}\left (1 - 2 \cosh{\left [ k \left (y-\frac12 \right) \right]} \frac{\sinh{(k/2)}}{\sinh{k}} \right )$$

where $y \in [0,1]$.

All the king's horses, and all the king's men, as well as Mathematica, have been unable as of today to produce a closed-form inverse Fourier transform. Rest assured, I am trying. That said, I put it out there to anyone who may have seen this before or who knows a trick or two of which I am unaware.

Please note that, yes, $\hat{g}(k)$ is finite at $k=0$. Also note that I use the definition of IFT as

$$g(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \: \hat{g}(k) e^{-i k x}$$

One other thing: note the delicate balance of convergence at $\infty$ - the restriction on $y$ is critical.


1 Answer 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.

  • $\begingroup$ (+1) for a very interesting approach. Good stuff. I want to see more about how you computed this numerically. $\endgroup$
    – Ron Gordon
    Apr 27, 2013 at 15:22
  • $\begingroup$ I am not sure I understand what you mean. The point that the integral was in fact well-defined even for $s=-1$ (in fact, for $\mathrm{Re}s>-2$) manifests itself later in the cancellation of poles of $\Gamma(s)$ by zeros of the appropriate combination of $\zeta$'s. The answer for $I(s,a,x)$ is holomorphic near $s=-1$. $\endgroup$ Apr 27, 2013 at 15:32
  • $\begingroup$ I was just wondering about the numerical confirmation. $\endgroup$
    – Ron Gordon
    Apr 27, 2013 at 15:34
  • $\begingroup$ I just took random values of $x$ and $y$, and Mathematica knows $\zeta(s,a)$ (Zeta[s,a]). $\endgroup$ Apr 27, 2013 at 15:38
  • $\begingroup$ Ok, thanks. You will forgive me for waiting a bit before I accept, just because I want to see if someone else has another approach. $\endgroup$
    – Ron Gordon
    Apr 27, 2013 at 15:40

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