First of all consider

\begin{align} \displaystyle \int_{0}^{\infty}\frac{\sin(ax)}{\sinh(bx)}dx & = \int_{0}^{\infty}2\frac{\sin(ax)}{e^{bx}-e^{-bx}} \\ & = 2\int_{0}^{\infty}\frac{\sin(ax)e^{-bx}}{1-e^{-2bx}} \\ & = 2\int_{0}^{\infty}\sin(ax)\sum_{n=0}^{\infty}e^{-(bx+2bnx)} \\ & = 2\sum_{n = 0}^{\infty} \int_{0}^{\infty}\sin(ax)e^{-x(b+2bn)} dx \\ & = 2\sum_{n=0}^{\infty}\frac{a}{a^{2}+(b+2bn)^{2}} \\ & = \frac{2}{a}\sum_{n=0}^{\infty}\frac{1}{1+\left(\frac{b+2bn}{a}\right)^{2}}. \end{align}

Now, how can we calculate this sum? My teacher said me about Poisson's method? But I haven't seen it yet. Any ideas?


With Weierstrass product of hyperbolic cosine function $$\cosh x=\prod_{n=1}^\infty\left(1+\dfrac{4x^2}{\pi^2(2n-1)^2}\right)$$ and differentiate of $\ln\cosh x$ then find $$\tanh x=2x\sum_{n=1}^\infty \dfrac{1}{x^2+\frac14(2n-1)^2\pi^2}$$ Now let $x=\dfrac{\pi}{2}\dfrac{a}{b}$.

  • $\begingroup$ It looks nice, but it help us if we know the answer. It's harder to get it. $\endgroup$ – openspace Dec 13 '17 at 6:56
  • $\begingroup$ Here is the same as you've done! math.stackexchange.com/questions/2543862/… $\endgroup$ – Nosrati Dec 13 '17 at 7:01
  • $\begingroup$ @MyGlasses (+1) for this approach. $\endgroup$ – Mark Viola Dec 13 '17 at 15:21

Herein, we present an approach that uses contour integration. To that end we now proceed.

Without loss of generality, we will assume that $a>0$ and $b>0$.

First, note that the integral of interest can be written

$$\int_0^\infty \frac{\sin(ax)}{\sinh(bx)}\,dx=\frac12\text{Im}\left(\text{PV}\int_{-\infty}^\infty \frac{e^{iaz}}{\sinh(bz)}\,dz\right)$$

where $\text{PV}$ denotes the Cauchy Principal Value.

Next, we analyze the contour integral $\displaystyle \oint_C \frac{e^{iaz}}{\sinh(bz)}\,dz$, where $C$ is the contour comprised of $(1)$ the real line segment from $-(N+1/2)\frac\pi b$ to $-\epsilon$, $(2)$ the semi-circular arc $z=\epsilon e^{i\phi}$, $\phi\in[-\pi,0]$, $(3)$ the real line segment from $\epsilon$ to $(N+1/2)\frac\pi b$, and $(4)$ the semi-circular arc $z=(N+1/2)\frac\pi b e^{i\phi}$, $\phi\in[0,\pi]$.

Using the residue theorem, we have

$$\begin{align} \oint_C \frac{e^{iaz}}{\sinh(bz)}\,dz&=2\pi i \left(\sum_{n=1}^N \frac1b(-1)^n e^{-n\pi b/a}\right)\\\\ &=-2\pi i \left(\frac1b\,\frac{e^{-\pi a/b}}{1+e^{-\pi a/b}}\right) \end{align}$$

As $N\to \infty$ and $\epsilon \to 0$ we find that

$$\begin{align} \text{PV}\int_{-\infty }^\infty \frac{e^{iax}}{\sinh(bx)}\,dx&=i\frac{\pi}{b}\left(1-2\frac{e^{-\pi a/b}}{1+e^{-\pi a/b}}\right)\\\\ &=i\frac{\pi}{b}\left(\tanh\left(\frac{\pi a}{2b}\right)\right)\tag 1 \end{align}$$

Taking the imaginary part of $(1)$ and dividing by $(2)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{\sin(ax)}{\sinh(bx)}\,dx=\frac{\pi }{2b}\tanh\left(\frac{\pi a}{2b}\right)}$$

  • $\begingroup$ @MyGlasses Thank you! Much appreciated. $\endgroup$ – Mark Viola Dec 13 '17 at 15:18

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