Calculating integral using residue theorem I want to show that

$$\int_{-\infty}^{\infty}e^{-2\pi ix\xi}\frac{\sin(\pi a)}{\cosh(\pi x) + \cos(\pi a)}dx= 2\frac{\sinh(2\pi a \xi)}{\sinh(2\pi \xi)}$$

for $\xi \in \mathbb R$ and $a\in ]0,1[$ using the residue theorem.
But I fail trying to find the poles of this function:
The poles of the function $f(z) := e^{-2\pi iz\xi}\frac{\sin(\pi a)}{\cosh(\pi z) + \cos(\pi a)}$ are the $z\in \mathbb C$ where $\cosh(\pi z) = -\cos(\pi a)$.
Splitting this up in imaginary and real part and writing $z = x + iy$ I get 
$$\Im(\cosh(\pi z)) = \sinh(\pi x)\sin(\pi y) = \Im(-\cos(\pi a)) = 0$$ 
where solving for $x$ and $y$ yields $x = 0$ or $y = k\pi, k\in\mathbb Z$.
But now looking the real part I get $$\Re(\cosh(\pi z)) = \cosh(\pi x)\cos(\pi y) = \Re(-\cos(\pi a)) = -\cos(\pi a)$$
Is there a good way to solve $\cosh(\pi x)\cos(\pi y) = -\cos(\pi a)$ or is my attempt going in the wrong direction?  
Any help appreciated.
 A: In order to prove such identity, it is probably wiser to go in the opposite direction and find the Fourier transform of $g(\xi)=\frac{\sinh(2\pi a \xi)}{\sinh(2\pi \xi)}$ under the assumption $a\in(0,1)$. We have
$$ g(\xi) = \frac{e^{2\pi a\xi}-e^{-2\pi a \xi}}{e^{2\pi \xi}-e^{-2\pi \xi}} =\frac{e^{2\pi(a-1)\xi}-e^{-2\pi(a+1)\xi}}{1-e^{-4\pi\xi}}\tag{1}$$
and $g(\xi)$ is an even function with the following expansion:
$$ g(\xi) = \left(e^{2\pi(a-1)\xi}-e^{-2\pi(a+1)\xi}\right)\cdot\left(1+e^{-4\pi \xi}+e^{-8\pi \xi}+e^{-12\pi \xi}+\ldots\right)\tag{2} $$
for any $\xi>0$. If $k$ is a negative number we have:
$$ \int_{0}^{+\infty} e^{k\xi} e^{2\pi i \xi x}\,d\xi = -\frac{1}{k+2\pi i x}\tag{3}$$
hence:
$$ \int_{0}^{+\infty}g(\xi)e^{2\pi i \xi x}\,d\xi =\sum_{m\geq 0}\left(-\frac{1}{2\pi(a-1)-4\pi m+2\pi i x}-\frac{1}{-2\pi(a+1)-4\pi m+2\pi i x}\right)$$
and the inverse Fourier transform of $g(\xi)$ is related with a tangent, since by considering the logarithmic derivative of the Weierstrass product for the sine function we have:
$$ \tan(z) = \frac{1}{z}+\sum_{m\geq 1}\left(\frac{1}{z+m\pi}+\frac{1}{z-m\pi}\right) $$
hence:
$$ (\mathcal{F}^{-1}g)(x) = \int_{-\infty}^{+\infty}g(\xi)e^{2\pi i x \xi}\,d\xi = \frac{1}{4}\left[\tan\left(\frac{\pi}{2}(a-ix)\right)+\tan\left(\frac{\pi}{2}(a+ix)\right)\right]\tag{4}$$
and by recalling $\tan\frac{z}{2}=\frac{\sin z}{1+\cos z}$ we get:
$$ (\mathcal{F}^{-1}g)(x) = \frac{1}{4}\left[\frac{\sin(\pi a-\pi i x)}{1+\cos(\pi a-\pi i x)}+\frac{\sin(\pi a+\pi i x)}{1+\cos(\pi a+\pi i x)}\right] \tag{5}$$
and the claim follows by Fourier inversion.
