I am trying to prove this relation, which I think can only be evaluated via contour integration: $$ \int_{-\infty}^{\infty}dk\;e^{ikx}\frac{k\sinh\left[k(y-\frac{w}{2})\right]}{\cosh\left(\frac{kw}{2}\right)}=\frac{\pi^{2}}{w^{2}}Re\left[\frac{\cosh[\frac{\pi}{w}(x+iy)]}{\sinh^{2}[\frac{\pi}{w}(x+iy)]}\right] $$ This is equation 18 (see also below Eq 16) in the paper "Linking Spatial Distributions of Potential and Current in Viscous Electronics" (arXiv PDF).

Any help will be highly appreciated.

  • $\begingroup$ @metmorphy, This is the reference "arxiv.org/pdf/1607.00986.pdf". Eq: 18 and also see below Eq:16. $\endgroup$ Commented Aug 10, 2019 at 20:07
  • $\begingroup$ @metamorpy Thanks I have fixed it. I think the contour integral is the only possible route to solve it. $\endgroup$ Commented Aug 10, 2019 at 20:25
  • $\begingroup$ In fact the absolute value of the integrand in the LHS $k \to f(k)=k.*\sinh(A*k)/\cosh(B*k)$ where $A$ and $B$ are constants is not integrable : $|f|$ tends to $+\infty$ when $k$ tends to $\infty$ !). Thus this integral cannot be used in any elementary meaning (using or not contour integration). It can have only a meaning in the theory of distributions. $\endgroup$
    – Jean Marie
    Commented Aug 10, 2019 at 20:26
  • $\begingroup$ @JeanMarie: It converges when $|A|<|B|$. $\endgroup$
    – metamorphy
    Commented Aug 10, 2019 at 20:31

1 Answer 1


Let $U=\{u\in\mathbb{C} : |\Re u|<1\}$, $u\in U$, and consider $$I_R=\int_{C_R}\frac{e^{uz}\,dz}{\sinh z},$$ where $R>0$, and $C_R$ is the rectangular contour (closed, ccw-oriented) with vertices at $z=\pm R\pm\pi i/2$. By the residue theorem, it is equal to $2\pi i$ times the residue of the integrand at $z=0$: $I_R=2\pi i$. On the other hand, with $R\to\infty$, the integrals over vertical sides vanish, and we have $$\lim_{R\to\infty}I_R=\int_{-\infty}^{\infty}\left(\frac{\exp[u(t-\pi i/2)]}{\sinh(t-\pi i/2)}-\frac{\exp[u(t+\pi i/2)]}{\sinh(t+\pi i/2)}\right)\,dt=2i\cos\frac{\pi u}{2}\int_{-\infty}^{\infty}\frac{e^{ut}\,dt}{\cosh t}.$$ Thus, $\int_{-\infty}^{\infty}\frac{e^{ut}\,dt}{\cosh t}=\frac{\pi}{\cos(\pi u/2)}$. Now take the derivative w.r.t. $u$ (which is admissible under the integral sign too), put $u=1+2i(x+iy)/w$ and substitute $t=wk/2$.

(You may happen to know the last integral already; it is related to the $\mathrm{B}$-function.)

  • $\begingroup$ Thanks. Can you share any reference book or website for such integrals? $\endgroup$ Commented Aug 10, 2019 at 22:29
  • $\begingroup$ @metamorphy You are right. My bad. I did the same error as before. $\endgroup$
    – Jean Marie
    Commented Aug 12, 2019 at 2:30
  • 1
    $\begingroup$ As you say, the last integral is known. I was used to it under the following form (see math.stackexchange.com/q/799616) that the Fourier Transform of $1/\cosh(t)$ is $\pi/\cosh(\pi \omega/2)$, which, expressed with integrals, is : $\int_{-\infty}^{+\infty}e^{-i \omega t}\dfrac{dt}{\cosh(t)}=\dfrac{\pi}{\cosh(\pi \omega/2)}$. The intuitive thing behind this is that the inverse of $\cosh$ is (very approximately) Gaussian, therefore has an (approximative again) Gaussian Fourier Transform. $\endgroup$
    – Jean Marie
    Commented Aug 12, 2019 at 2:49

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