How to find the fourier sine and cosine transform of :
$$\frac{e^{ax}+e^{-ax}}{e^{\pi x}-e^{-\pi x}}$$
Comments:
I found out partial derivative of the transform w.r.t $a$ and $b$ (I generalized the denominator as $b$ instead of $\pi$).
Also, i tried to compute the transform using integration by parts with the given function as the first function and $cos(\alpha x)$ as the second function.
Then, i could rewrite the transform $I$ as
$$a\frac{\partial I}{\partial a} + b\frac{\partial I}{\partial b} - \alpha\frac{\partial I}{\partial \alpha} = I + \frac{2}{b}$$
By far i only know solving partial differential equations (PDEs) in two variables. This PDE appears to be simple but i can't solve it as of now.
The Fourier transform of $\operatorname{csch} \pi x$ (understood in the p.v. sense) can be found in the same way as here, giving $$F(w) = \mathcal F[\operatorname{csch} \pi x] = i \tanh \frac w 2.$$ Then, since the Fourier integral and $\tanh(w/2)$ are analytic in the strip $-\pi < \operatorname{Im} w < \pi$, $$\frac 1 2 \mathcal F[(e^{a x} + e^{-a x}) \operatorname{csch} \pi x] = \frac 1 2 (F(w - i a) + F(w + i a)) = \\ \frac {i \sinh w} {\cosh w + \cos a}, \quad -\pi < a < \pi.$$