Fourier transform of $\frac{x}{\sinh(x)}$

Question

I was given to calculate the Fourier Transform of $\frac{x}{\sinh(x)}$.

So, the problem is to calculate the integral

$$ \int_\mathbb{R} \frac{x}{\sinh(x)}e^{-i \omega x} dx $$

I know such an integrals can be evaluated using complex analysis, but I don't know how to take the proper contour.

Function under integral has a removable singularity at $x=0$ and poles at $x=\pi i k, 0 \ne k \in \mathbb{Z}$

I've tried this contour:

But it doesn't work since the $x/\sinh(x)$ is not even bounded on the half-circle, so Jordan Lemma is inapplicable.

And this one:

But I don't quite see how to integrate over any side of a rectangle.

Any ideas?

Answer 1

$$I(\omega)=\int_{-\infty}^\infty\frac{x\,e^{-i\omega x}}{\sinh x}dx=i\frac{d}{d \omega}\int_{-\infty}^\infty\frac{e^{-i\omega x}}{\sinh x}dx=i\frac{d}{d \omega}J(\omega)$$ where $J(\omega)$ is understood in the principal value sense. To evaluate $J(\omega)$ let's consider the following contour:

where we added small half-circles around $z=0$ and $z=\pi i$, and also two paths $[1]$ and $[2]$ - to close the contour. Given that $\,\sinh(z+\pi i)=-\sinh z$, and $\,e^{-i\omega (z+\pi i)}=e^{-i\omega z}e^{\pi \omega}$ $$\oint \frac{e^{-i\omega z}}{\sinh z}dz=J(\omega)+I_{1r}+[1]+J(\omega)e^{\pi \omega}+I_{2r}+[2]$$ where $I_{1,2\,r}$ are the integrals along the half-circles. We can show that integrals $[1,2]$ tend to zero at $R\to\infty$. There are no poles inside the contour, therefore $\displaystyle \oint=0$. $$J(\omega)(1+e^{\pi\omega})=-I_{1r}-I_{2r}=\pi i\underset{z=0; \,e^{\pi i}}{\operatorname{Res}}\frac{e^{-i\omega z}}{\sinh z}=\pi i(1-e^{\pi \omega})$$ $$J(\omega)=-\pi i\tanh \frac{\pi \omega}{2}$$ $$I(\omega)=i\frac{d}{d\omega}J(\omega)=\frac{\pi^2}{2}\frac{1}{\cosh^2\frac{\pi \omega}{2}}$$

Answer 2

Jordan's lemma is not applicable, but under the assumption that $\omega <0$, the integral does vanish on the semicircle (or more obviously on the 3 sides of the rectangle) as $N \to \infty$.

This is because the magnitude of $e^{-i \omega z}$ decays exponentially fast to zero as $\Im(z) \to +\infty$, while the magnitude of $\frac{1}{\sinh(z)}$ decays exponentially fast to zero as $\Re(z) \to \pm \infty$.

Therefore, the value of the integral is simply $$ \begin{align} \int_{-\infty}^{\infty} \frac{xe^{-i \omega x}}{\sinh(x)} \, \mathrm dx &= 2 \pi i \sum_{n=1}^{\infty} \operatorname{Res}\left[ \frac{ze^{-i \omega z}}{\sinh(z)}, i n \pi\right] \\ &= 2 \pi i \sum_{n=1}^{\infty}\lim_{z \to i n \pi} \frac{ze^{-i \omega z}}{\cosh(z)} \\ &= 2 \pi i \sum_{n=1}^{\infty}\frac{in \pi e^{\omega n \pi}}{\cos(n \pi)} \\ &= -2 \pi^{2}\sum_{n=1}^{\infty} (-1)^{n}n e^{\omega n \pi} \\ &= 2 \pi^{2} \frac{e^{\omega \pi}}{(1+e^{\omega \pi})^{2}} \\ &= \frac{\pi^{2}}{2} \operatorname{sech}^{2} \left(\frac{\omega \pi}{2} \right). \end{align}$$

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EDIT:

If it's not clear that the integral vanishes on the upper side of the rectangle, notice that on the upper side of a rectangle with vertices at $z= \pm N , \pm N + i (N+1/2) \pi$, we have $$ \begin{align} \left| \int_{- N }^{N } \frac{\left(t+i(N+1/2) \pi \right)e^{-i \omega \left(t+i(N+1/2)\pi\right)}}{\sinh \left(t+i(N+1/2)\pi) \right)} \, \mathrm dt \right| &\le e^{\omega (N+1/2)\pi }\int_{-N}^{N} \frac{t+(N+1/2)\pi}{\cosh(t)} \, \mathrm dt \\ &= e^{\omega (N+1/2)\pi }\int_{-N }^{N } \frac{(N+1/2)\pi}{\cosh(t)} \, \mathrm dt \\ &\ < e^{\omega (N+1/2)\pi }\int_{-\infty}^{\infty} \frac{(N+1/2)\pi}{\cosh(t)} \, \mathrm dt \\ &= e^{\omega (N+1/2)\pi } (N+1/2)\pi^{2}, \end{align}$$ which goes to zero as $N \to \infty$ since $\omega < 0$.

Answer 3

Let $a=(N+1/2)\pi$, $N\in \mathbb{N}$. Then, on the circle $|z|=a$, we have

$$\begin{align} \left|\frac{z}{\sinh(z)}\right|&=\frac a{\sqrt{\sinh^2(a\cos(\phi))\cos^2(a\sin(\phi))+\cosh^2(a\cos(\phi))\sin^2(a\sin(\phi))}}\\\\ &=\frac a{\sqrt{\sinh^2(a\cos(\phi))+\sin^2(a\sin(\phi))}} \end{align}$$

Therefore, the integral over the semi-circles $|z|=a$, in the upper and lower half planes can be written

$$\begin{align} I&=2\int_0^{-\pi\text{sgn}(\omega)} \frac{ae^{i\phi}e^{-i\omega ae^{i\phi}}}{e^{ae^{i\phi}}-e^{-ae^{i\phi}}} \,iae^{i\phi}\,d\phi \end{align}$$

We have the estimate

$$\begin{align} |I|&\le 2\int_0^{\pi} \frac{a^2 e^{-|\omega| a\sin(\phi)}}{\left|e^{ae^{i\phi}}-e^{-ae^{i\phi}}\right|} \,d\phi\\\\ &=2\int_0^{\pi/2}\frac{a^2 e^{-|\omega| a\sin(\phi)}}{\sqrt{\sinh^2(a\cos(\phi))+\sin^2(a\sin(\phi))}}\,d\phi \end{align}$$

Note that bounded away from $\phi=\pi/2$, the denominator grows exponentially as $e^{a\cos(\phi)}$ and overwhelms the term $a^2$ in the numerator. And near $\phi =\pi/2$, the exponential in the numerator decays exponentially as $e^{-|\omega| a \sin(\phi)}$.

Can you convince yourself that as $a\to\infty$ that $I\to 0$?