Fourier transform of $\frac{1}{\sinh(x \pm i\epsilon)\cosh(x \pm i\epsilon)}$ I would like to calculate the Fourier transform of the following functions where epsilon is considered as a small positive number:
$$
\mathcal{F}_k\left(\frac{1}{\sinh(x \pm i\epsilon)\cosh(x \pm i\epsilon)}\right) := \frac{1}{2\pi} \int dx \ \frac{1}{\sinh(x \pm i\epsilon)\cosh(x \pm i\epsilon)} e^{-ikx} \,, \qquad \epsilon>0  \,.
$$
I tried to used the properties of the FT and some well known FTs then I get :
$$ \mathcal{F}_k\left(\frac{1}{\sinh(x + i\epsilon)\cosh(x + i\epsilon)}\right) = -\frac{i}{1+exp(-\frac{\pi k}{2})} $$
 A: To obtain the results presented by @HJames and @Sangchul Lee we can use contour integration.
First of all, $i\epsilon$ is needed only for setting the path of bypassing the poles.
($x+i\epsilon$) means that we go above $x=0$, and ($x-i\epsilon$) - below $x=0$.
Let's define
$$I_{+}= \frac{1}{2\pi} \int  \ \frac{e^{-ikx}}{\sinh(x + i\epsilon)\cosh(x + i\epsilon)} dx=\frac{1}{\pi} \int  \ \frac{e^{-ikx}}{\sinh(2x + i\epsilon)}dx=\frac{1}{2\pi} \int dx \ \frac{e^{-ikx/2}}{\sinh(x + i\epsilon)}$$ and $$I_{-}=\frac{1}{2\pi} \int  \ \frac{e^{-ikx/2}}{\sinh(x - i\epsilon)}dx$$
We can write $I_{+}=\frac{1}{2\pi}P.V.\int_{-\infty}^\infty  \ \frac{e^{-ikx/2}}{\sinh x}dx+\frac{1}{2\pi}\int_{C_{1+}}=J+\frac{1}{2\pi}\int_{C_{1+}}$, where $J$ - integral in the principal value sense, and $\frac{1}{2\pi}\int_{C_{1+}}$ - integral along small half-circle above $x=0$.
Let's first consider $J$ and the following rectangular closed contour ($R\to\infty$):

It can be shown that integrals along path $[1]$ and $[2]$ $\to0$ as $R\to\infty$.
We also have $\frac{e^{-ik(x+\pi i)/2}}{\sinh (x+\pi i)}=-\frac{e^{-ik\pi i/2}}{\sinh x}e^{-ikx/2}$. Our contour "catches" one pole at $x=\pi i$, and we get
$$J+\frac{1}{2\pi}\int_{C_{1+}}+Je^{-ik\pi i/2}+\frac{1}{2\pi}\int_{C_{2+}}=2\pi iRes_{x=\pi i}\frac{1}{2\pi}\frac{e^{-ikx/2}}{\sinh x}$$
Integrals along half-circles above $x=0$ and $x=\pi i$ can be easily evaluated:
$\int_{C_{1+}}=-\pi i$ and $\int_{C_{2+}}=-\pi i e^{-ik\pi i/2}=\int_{C_{1+}}e^{-ik\pi i/2}$
Finally we get:
$$\Big(J+\frac{1}{2\pi}\int_{C_{1+}}\Big)\Big(1+e^{\pi k/2}\Big)=I_{+}\Big(1+e^{\pi k/2}\Big)=-i\,e^{\pi k/2}$$
$$I_{+}=\frac{-i}{1+e^{-\pi k/2}}$$
In the same way $I_{-}$ can be evaluated - we have to integrate along the following contour:

In this case the pole is at $x=0$, and residual is simply $i$.
$$I_{-}\Big(1+e^{\pi k/2}\Big)=i\,\,\, \Rightarrow \,\,\, I_{-}=\frac{i}{1+e^{\pi k/2}}$$
A: Couple of identities that might be useful for things like this:
$$\sinh(2x)=2\sinh(x)\cosh(x)$$
$$\cosh(2x)=\cosh^2(x)+\sinh^2(x)$$
$$\tanh(2x)=\frac{2\tanh(x)}{1+\tanh^2(x)}$$
obviously in this case the top one is the important one. Another thing to take into account is the definition of the hyperbolic functions:
$$\sinh(x)=\frac{e^x-e^{-x}}{2}$$
and so when we have the $\sinh$ of a complex number the imaginary part will be an imaginary exponential, which can be represented in terms of trig functions.
