Fourier transform of hyperbolic function $\frac{d}{dx}\log({\sinh{x}})$

Question

I am trying to calculate the FT of $\frac{d}{dx}\log({\sinh{x}})$. I did not calculate it directly but I used the result of its derivative. This is a tentative: I found the following result in the literature but they did not mention how they get the result except adding a small positive $\epsilon$ which I believe is considered as the imaginary part of $x$ : $\hat{F}_k\Biggl\{\frac{d^2}{dx^2}\log({\sinh{x}})\Biggr\}=\frac{k}{1-e^{-\pi k}}$ then $\hat{F}_k\Biggl\{\frac{d}{dx}\log({\sinh{x}})\Biggr\}=\frac{1}{ik}\hat{F}_k\Biggl\{\frac{d^2}{dx^2}\log({\sinh{x}})\Biggr\}=\frac{i}{e^{-\pi k}-1}$ First, is my reasoning correct ? I want to understand how they get the result by adding $ i \epsilon$ and if it is possible how do I calculate the FT of $\frac{d}{dx}\log({\sinh{x}})$ without using FT properties.

Answer 1

I guess you should make use of the identity $$\coth(x)=\frac{1}{x} + \sum_{n=1}^\infty \frac{2x}{x^2+n^2\pi^2} \, .$$ The Fourier-Transform $${\cal F}\left[\coth(x)\right](k)=PV\int_{-\infty}^\infty {\rm d}x \, e^{-ikx} \coth(x)$$ (where $PV$ stands for the principal-value at $x=0$) is not convergent due to the behaviour of the cotangens hyperbolicus at $\pm \infty$. The integral can be regularized by introducing a dampening factor $e^{-\epsilon |x|}$ and eventually taking the limit $\epsilon \rightarrow 0^+$. This gives rise to the regularized Fourier-Transform $$ {\cal F}\left[\coth(x)\right](k) = \lim_{\epsilon \rightarrow 0} PV \int_{-\infty}^\infty {\rm d}x \, e^{-ikx-\epsilon|x|} \coth(x) = I_1(k) + I_2(k)$$ where $$I_1(k) = \lim_{\epsilon \rightarrow 0} PV\int_{-\infty}^\infty \frac{e^{-ikx-\epsilon|x|}}{x} \, {\rm d}x = PV\int_{-\infty}^\infty \frac{e^{-ikx}}{x} \, {\rm d}x = -i\pi \, {\rm sign}(k) \\ I_2(k) = \lim_{\epsilon \rightarrow 0}\int_{-\infty}^\infty {\rm d}x \sum_{n=1}^\infty \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx-\epsilon|x|} = \lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty \int_{-\infty}^\infty {\rm d}x \, \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx-\epsilon|x|} \\ = \sum_{n=1}^\infty \int_{-\infty}^\infty {\rm d}x \, \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx} = \sum_{n=1}^\infty -2\pi i \,{\rm sign}(k) \, e^{-\pi|k|n} = \frac{-2\pi i \, {\rm sign}(k)}{e^{\pi|k|}-1} \, .$$ The integral $I_1(k)$ converges even for $\epsilon=0$ so that the limit can be taken immediately. For $I_2(k)$ the principal-value is redundant as the integrand is well-behaved at $x=0$. Summation and integration can be interchanged by dom. convergence, since the RHS of $$\int_{-\infty}^\infty {\rm d}x \, \underbrace{\left|\sum_{n=1}^N \frac{2x}{x^2+n^2\pi^2}\right|}_{\leq 1 \,\forall N\in{\mathbb N}} \, e^{-\epsilon|x|} \leq \int_{-\infty}^\infty {\rm d}x \, e^{-\epsilon|x|}$$ converges for $\epsilon>0$. Since the integral and sum converge even for $\epsilon=0$, the limit can be taken immediately. The resulting integral can be calculated by means of the residue theorem for the 2 cases $k>0$ and $k<0$. ${\rm sign}(k)$ is the signum function.

Finally you can easily check that $$I_1(k)+I_2(k)=-i\pi \, {\rm sign}(k) \left(1+\frac{2}{e^{\pi|k|}-1}\right) = -i\pi \coth\left(\frac{\pi k}{2}\right) \, .$$