Hyperbolic tangent integral $ \int_{-\infty} ^{\infty} e^{i x \frac h k} \tanh x \ dx$ Find the integral of
$$ \int \limits _{-\infty} ^{\infty} e^{\Bbb i x \frac h k} \tanh x \ \Bbb dx$$
I tried to expand this but that didn't help.
 A: This is the Fourier transform of the hyperbolic tangent, evaluated at $-h/k$, at least if we agree to work on non-integrable functions (I guess the integral has to be taken in some principal value sense). 
I'll sketch a way to proceed, below.
The general rule that the Fourier transform of a derivative essentially corresponds to multiplication by the dual variable holds here as well. So, what you could do is to try to calculate the Fourier transform of $D\tanh x=\text{sech}^2x$, which happens to be a nice function, and then divide the result by $i\xi$.
I don't see immediately how to calculate the Fourier transform of $\text{sech}^2x$, but the transform of $\text{sech}\,x$ can be handled like this. With our constants, we have
$$
\mathcal{F}(\text{sech}\,x)(\xi)=\pi\text{sech}\,\bigl(\pi \xi/2\bigr).
$$
So, we have to convolve the hyperbolic secant with itself, since $\mathcal{F}(f^2)=\frac{1}{2\pi}\mathcal{F}(f)*\mathcal{F}(f)$. This means that we should calculate the integral
$$
\frac{\pi}{2}\int_{-\infty}^{+\infty}\text{sech}\,\bigl(\pi(\xi-t)/2\bigr)\text{sech}\,\bigl(\pi t/2\bigr)\,dt.
$$
Luckily, we have a primitive function,
$$
\text{csch}\,\bigl(\pi \xi/2\bigr)\log\frac{\text{csch}\,\bigl(\pi t/2\bigr)}{\text{csch}\,\bigl(\pi (t-\xi)/2\bigr)}.
$$
Inserting the limits we get
$$
\pi\xi\,\text{csch}\,\bigl(\pi\xi/2\bigr).
$$
Division by $i\xi$ gives
$$
\mathcal{F}\bigl(\tanh x\bigr)(\xi)=-i\pi\text{csch}\,\bigl(\pi\xi/2\bigr).
$$
You had $\xi=-h/k$, so your integral equals (in some sense, be cautious with convergence)
$$
-i\pi\text{csch}\,\Bigl(-\frac{\pi h}{2k}\Bigr)=i\pi\text{csch}\,\Bigl(\frac{\pi h}{2k}\Bigr).
$$
I leave to you to find some small errors, but in principle I think it should go like this.
Edit As kindly commented by @JanG (privately), when dividing by $i\xi$ one should be a bit careful, since, in principal, a term $c\delta_0(\xi)$ should be added. In this case, however, both the original hyperbolic tangent and the hyperbolic cosecant are odd, and $\delta_0(\xi)$ is even, so $c$ must be zero.
