Fourier transform of $\frac{1}{\sinh(x \pm i\epsilon)\cosh(x \pm i\epsilon)}$

Question

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 : $$ \mathcal{F}_k\left(\frac{1}{\sinh(x + i\epsilon)\cosh(x + i\epsilon)}\right) = -\frac{i}{1+exp(-\frac{\pi k}{2})} $$

Answer 1

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 $i$. $$I_{-}\Big(1+e^{\pi k/2}\Big)=i\,\,\, \Rightarrow \,\,\, I_{-}=\frac{i}{1+e^{\pi k/2}}$$

Answer 2

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.