How do I calculate this Fourier Transform defined by
$$ \int_{-\infty}^{\infty} d k \frac{\sinh ((\pi-2a) k)}{k(1+e^{-\frac{\pi k}{2}})\sinh (\pi k)}e^{ikx} =\int_{-\infty}^{\infty} d k \frac{e^{\frac{\pi k}{4}}\sinh ((\pi-2a) k)}{2k\sinh (\pi k) \cosh (\frac{\pi k}{4})}e^{ikx}=\int_{-\infty}^{\infty} d k \frac{e^{(\pi k)}\sinh (4(\pi-2a) k)}{2k\sinh^2 (\pi k) \cosh^3 (\pi k)}e^{4ikx} ? $$ I tried these manipulations to set-up a contour integral but I failed. I could not find an appropriate contour. Any help would be appreciated. This function appears, as a driving term, in some integral equations, which I can solve numerically. I would like to find explicitly this function to predict somehow the behavior of the solutions of these equations.
-------My second attempt ---- $$ \int_{-\infty}^{\infty} d k \frac{\sinh ((\pi-2a) k)}{k(1+e^{-\frac{\pi k}{2}})\sinh (\pi k)}e^{ikx} =\int_{-\infty}^{\infty} d k e^{ikx} \{\frac{e^{-2a k}}{k(1+e^{-\frac{\pi k}{2}})(1-e^{-2\pi k})}-\frac{e^{-2(\pi-a) k}}{k(1+e^{-\frac{\pi k}{2}})(1-e^{-2\pi k})} \} $$ I define these two integrals: $$J=\int_{-\infty}^{\infty} d k \frac{e^{ikx} e^{\alpha k}}{k(1+e^{-\frac{\pi k}{2}})(1-e^{-2\pi k})}$$ and $$I= -i\partial_x J=\int_{-\infty}^{\infty} d k \frac{e^{ikx} e^{\alpha k}}{(1+e^{-\frac{\pi k}{2}})(1-e^{-2\pi k})}$$ then I consider the rectangular closed contour integral enclosing the poles i, 2i and 3i inside . In addition, I avoid the poles 0 and 4i by drawing two infinitesimal semi-circles (counter clockwise in both cases): $$ \oint\limits_{\Gamma}=I-e^{4(i\alpha-x)}I+\int\limits_{C_1}+\int\limits_{C_2}=2\pi i\{Res_{k=i}(...)+Res_{k=2i}(...)+Res_{k=3i}(...)\}$$ then $$\int\limits_{C_1}=-\frac{i}{4}$$ $$\int\limits_{C_2}=\frac{i}{4}e^{4(i\alpha-x)}$$ $\oint\limits_{\Gamma}=\frac{e^{i\alpha-x}}{2}(i-1)+\frac{e^{3(i\alpha-x)}}{2}(i+1)+\frac{ie^{2(i\alpha-x)}}{2\pi}(4ix+5\pi+4\alpha)=2\pi i \sum{Res(...)}$
which gives $$ I=\frac{i}{4}+\frac{2\pi i \sum{Res(...)}}{1-e^{4(i\alpha-x)}}$$ $$ J=i\int I d x=i \int d x \{\frac{i}{4}+\frac{2\pi i \sum{Res(...)}}{1-e^{4(i\alpha-x)}}\}$$ Finally, the Fourier Transform is given by $$\int_{-\infty}^{\infty} d k \frac{\sinh ((\pi-2a) k)}{k(1+e^{-\frac{\pi k}{2}})\sinh (\pi k)}e^{ikx}= i\int d x \{\frac{2\pi i \sum{Res(...)}}{1-e^{4(i\alpha-x)}}\bigg\vert_{\alpha=-2a}-\frac{2\pi i \sum{Res(...)}}{1-e^{4(i\alpha-x)}}\bigg\vert_{\alpha=-2(\pi-a)}\} $$