Fourier transform of $\frac{x}{\sinh(x)}$ 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?
 A: $$I(\omega)=\int_{-\infty}^\infty\frac{x\,e^{-i\omega x}}{\sinh x}dx=i\frac{d}{d \omega}\int_{-\infty}^\infty\frac{e^{-i\omega x}}{\sinh x}dx=i\frac{d}{d \omega}J(\omega)$$
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 $\displaystyle \oint=0$.
$$J(\omega)(1+e^{\pi\omega})=-I_{1r}-I_{2r}=\pi i\underset{z=0; \,e^{\pi i}}{\operatorname{Res}}\frac{e^{-i\omega z}}{\sinh z}=\pi i(1-e^{\pi \omega})$$
$$J(\omega)=-\pi i\tanh \frac{\pi \omega}{2}$$
$$I(\omega)=i\frac{d}{d\omega}J(\omega)=\frac{\pi^2}{2}\frac{1}{\cosh^2\frac{\pi \omega}{2}}$$
A: 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 $$ \begin{align} \int_{-\infty}^{\infty} \frac{xe^{-i \omega x}}{\sinh(x)} \, \mathrm dx &= 2 \pi i \sum_{n=1}^{\infty} \operatorname{Res}\left[ \frac{ze^{-i \omega z}}{\sinh(z)}, i n \pi\right] \\ &= 2 \pi i \sum_{n=1}^{\infty}\lim_{z \to i n \pi} \frac{ze^{-i \omega z}}{\cosh(z)} \\ &= 2 \pi i \sum_{n=1}^{\infty}\frac{in \pi e^{\omega n \pi}}{\cos(n \pi)} \\ &= -2 \pi^{2}\sum_{n=1}^{\infty} (-1)^{n}n e^{\omega n \pi} \\ &= 2 \pi^{2} \frac{e^{\omega \pi}}{(1+e^{\omega \pi})^{2}} \\ &= \frac{\pi^{2}}{2} \operatorname{sech}^{2} \left(\frac{\omega \pi}{2} \right). \end{align}$$

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 $\omega < 0$.
A: 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 $I\to 0$?
