Approach to an improper integral involving sinh

Question

How do I calculate an integral $\int_{-\infty}^{\infty} \frac{z-a}{\sinh(z-a)}\frac{z+a}{\sinh(z+a)} dz$ for $a>0$? Expanding the integrand and integrating term-by-term (a-la polylog) gives rise to an ugly looking double series. Edit: it surely can be done this way, but what is the most straightforward way to calculate this?

Answer 1

Using the Plancherel tehorem, we can write $$I(a,b)=\int_{-\infty}^\infty \frac{z-a}{\sinh(z-a)}\frac{z-b}{\sinh(z-b)} dz=\frac1{2\pi}\int_{-\infty}^\infty\hat{f}(a,k)\hat{f}^{*}(b,k)dk$$ where we define the Fourier transform as $$\hat{f}(a,k)=\int_{-\infty}^\infty \frac{z-a}{\sinh(z-a)}e^{ikz}dz$$ Making the substitution $z-a=t$ $$\hat{f}(a,k)=e^{ika}\int_{-\infty}^\infty \frac t{\sinh t}e^{ikt}dt=-ie^{ika}\frac{\partial}{\partial k}P.V.\int_{-\infty}^\infty \frac {e^{ikx}}{\sinh x}dx=-ie^{ika}\frac{\partial}{\partial k}J(k)$$ To evaluate the last integral we consider the closed contour $C$ in the complex plane $R\to R\to R+\pi i\to -R+\pi i\to -R$, with added small arches around $z=0$ and $z=\pi i$, clockwise. There are no poles inside the contour; we only have to evaluate the integrals along these small arches: $$J(k)\big(1+e^{-\pi k}\big)=\pi i\underset{z=0; z=\pi i}{\operatorname{Res}}\frac {e^{ikz}}{\sinh z}=\pi i\big(1-e^{-\pi k}\big)$$ $$J(k)=\pi i\tanh\frac{\pi k}2$$ $$\hat{f}(a,k)=e^{ika}\frac{\pi^2}2\frac1{\cosh^2\frac{\pi k}2}$$ $$I(a,b)=\left(\frac{\pi^2}2\right)^2\frac1{2\pi}\int_{-\infty}^\infty\frac{e^{ik(a-b)}}{\cosh^4\frac{\pi k}2}dk=\frac{\pi^3}4\int_{-\infty}^\infty\frac{e^{2ix(a-b)}}{\cosh^4(\pi x)}dx$$ To evaluate the last integral, we again use the rectangular contour ($-R\to R\to R+i\to-R+i\to-R$). In this case we have a single pole of order $4$ inside the contour (at $z=\frac i2$) $$I(a,b)\big(1-e^{-2(a-b)}\big)=\frac{\pi^3i}2\underset{z=\frac i2}{\operatorname{Res}}\frac{e^{2iz(a-b)}}{\cosh^4(\pi z)}=\frac23e^{-(a-b)}\big(a-b\big)\big(\pi^2+(a-b)^2\big)$$ $$\boxed{\,\,I(a,b)=\frac{a-b}{3\sinh(a-b)}\big(\pi^2+(a-b)^2\big)\,\,}$$ In the case of the initial integral ($b=-a$) $$I(a,-a)=\int_{-\infty}^\infty \frac{z-a}{\sinh(z-a)}\frac{z+a}{\sinh(z+a)} dz=\frac23\frac{a(\pi^2+4a^2)}{\sinh (2a)}$$

Answer 2

If you ask WolframAlpha, it says "Standard computation time exceeded...". If you use the free download version of Wolfram Engine you get an answer. If it's a homework problem you should of course not ask it here.

As for the ugly looking series you mention, the primitive function is according to the engine: $$ \frac{1}{{e^{4 a}-1}} \left\{e^{2 a} \left(-2 a^2 \log (\sinh (a-z) \text{csch}(a+z))-2 z \text{Li}_2\left(e^{2 a-2 z}\right)+2 z \text{Li}_2\left(e^{-2 (a+z)}\right)-\text{Li}_3\left(e^{2 a-2 z}\right)+\text{Li}_3\left(e^{-2 (a+z)}\right)+2 z^2 \left(\log \left(1-e^{2 a-2 z}\right)-\log \left(1-e^{-2 (a+z)}\right)\right)\right)\right\} $$ which is in closed form, not a (double) series at all. And for the definite integral it gives an even simpler result where no polylogs at all are present any more.

Answer 3

Another contour integration approach:

Let $N$ be a positive integer greater than $a$, and integrate the function $$f(z) = \frac{z-a}{\sinh(z-a)} \frac{z+a}{\sinh(z+a)} \, e^{ipz}, \quad p>0,$$ around a rectangular contour with vertices at $-N$, $N$, $N+ i \pi (2N+1)/2$, and $-N+i\pi (2N+1)/2$.

For any positive integer $N$, the top of the contour passes halfway between adjacent poles of $f(z)$.

As $N \to \infty$, the integral vanishes on the left and right sides of the contour because $|\sinh(z)|$ grows exponentially as $\Re(z) \to \pm \infty$.

And the integral vanishes on the top the contour because $|e^{i p z}|$ decays exponentially to zero as $\Im(z) \to \infty$.

We therefore have $$ \begin{align} \int_{-\infty}^{\infty} f(x) \, \mathrm dx &= 2 \pi i \left(\sum_{n=1}^{\infty} \operatorname*{Res}_{z=a+ i \pi n} f(z) + \sum_{n=1}^{\infty} \operatorname*{Res}_{z = -a + i \pi n } f(z)\right) \\ &= 2 \pi i \sum_{n=1}^{\infty} \lim_{z \to a+ i \pi n} \frac{(z-a)(z+a)}{\cosh(z-a)\sinh(z+a)+\sinh(z-a)\cosh(z+a)} \, e^{i p z} \\ &+ 2 \pi i \sum_{n=1}^{\infty} \lim_{z \to -a+ i \pi n} \frac{(z-a)(z+a)}{\cosh(z-a)\sinh(z+a)+\sinh(z-a)\cosh(z+a)} \, e^{i p z} \\ &= 2 \pi i \left(\sum_{n=1}^{\infty}\frac{i \pi n(2a + i \pi n)e^{ipa}e^{- n \pi p}}{\sinh(2a)} + \sum_{n=1}^{\infty} \frac{(-2a + i \pi n)(i \pi n)e^{-ipa}e^{- n \pi p}}{-\sinh(2a)}\right) \\ &= -\frac{4 \pi^{2}}{\sinh(2a)} \left( 2a \cos(p a) \sum_{n=1}^{\infty} n e^{- n \pi p } - \pi \sin(pa) \sum_{n=1}^{\infty} n^{2} e^{- n \pi p }\right) \\ &= -\frac{4 \pi^{2}}{\sinh(2a)} \left(2a \cos(p a) \, \frac{e^{-\pi p}}{(1-e^{- \pi p})^{2}}- \pi \sin(p a) \, \frac{e^{-\pi p}(1+e^{- \pi p})}{(1-e^{- \pi p})^{3}} \right) \\ &= \frac{\pi^{2}}{\sinh(2a)} \frac{1}{\sinh^{2} (\tfrac{\pi p }{2})} \left(\pi \sin(pa) \coth \left(\frac{\pi p}{2}\right) - 2a \cos(pa) \right). \end{align}$$

For $p \ge 0$, $f(x)$ is dominated by the integrable function $\frac{(x-a)(x+a)}{\sinh(x-a) \sinh(x+a)}$.

We may therefore use the dominated convergence theorem to conclude that

$$\begin{align} &\int_{-\infty}^{\infty} \frac{x-a}{\sinh(x-a)} \frac{x+a}{\sinh(x+a)} \, \mathrm dx \\ &= \lim_{p \to 0^{+}} \frac{\pi^{2}}{\sinh(2a)} \frac{1}{\sinh^{2} (\tfrac{\pi p}{2})} \left(\pi \sin(pa) \coth \left(\frac{\pi p }{2}\right) - 2a \cos(pa) \right) \\ &= \small \frac{\pi^{2}}{\sinh(2a)} \lim_{p \to 0^{+}} \left(\frac{4}{\pi^{2} p^{2}} +O(1) \right) \left(\pi \left(ap- \frac{a^{3}p^{3}}{6}+ O(p^{5}) \right)\left(\frac{2}{\pi p } + \frac{\pi p }{6} + O(p^{3}) \right) -2a+a^{3}p^{2}+ O(p^{4})\right) \\ &= \frac{\pi^{2}}{\sinh(2a)} \lim_{p \to 0^{+}} \left(\frac{4}{\pi^{2} p^{2}}+O(1) \right) \left(\frac{\pi^{2} a p^{2}}{6} - \frac{a^{3}p^{2}}{3} + a^{3}p^{2}+ O(p^{3}) \right) \\ &= \frac{4}{\sinh(2a)} \left(\frac{\pi^{2}a}{6}+\frac{2 a^{3}}{3} \right). \end{align}$$