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I'm studying on an article related to complex analysis, and at a certain point this integral comes out. My belief is to proceed with contour integration, however I'm stuck so I'm asking you to kindly help me. We have to evaluate the following:

$$\frac{1}{2\pi}\int_{\mathbb R}\frac{\xi}{\sinh{2\beta \xi}}e^{i(z-\bar w)\xi}\mathrm d\xi, \quad\beta>0,$$ with $w\in \mathcal S_\beta:=\{z=x+iy\in\mathbb C:|y|<\beta \}$. I think (but i admit I'm not sure about it), that even the $z$ inside the integral belongs to $\mathcal S_\beta$, so if it is needed, for example to ensure absolute convergence of the integral, you can assume it.

The result reported on the paper is that the value is $$\frac{\pi}{16\beta^2}\left(\cosh\frac{\pi(z-\bar w)}{4\beta}\right)^{-2}.$$ If you can write down all the steps, this would be very helpful. Thanks in advance.

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up vote 4 down vote accepted

Since $z$ and $w$ always occur in the same combination, let's write $y=z-\bar w$. The integrand has simple poles at $\xi_n=(n\pi\mathrm i)/(2\beta)$ for $n\in\mathbb Z\setminus\{0\}$, and since $\sinh (n\pi\mathrm i+\xi)=(-1)^n\xi+O(\xi^3)$, the residues at the poles are

$$r_n=(-1)^n\frac{\xi_n}{2\beta}\mathrm e^{\mathrm iy\xi_n}\;.$$

Since $\omega,z\in\mathcal S_\beta$ ensures $y\in\mathcal S_{2\beta}$, the integrand goes to $0$ exponentially on both sides of the real axis, and depending on whether $y$ has positive or negative real part, we can close the contour of integration with a half-circle at infinity in the upper or lower half-plane, respectively. Assuming $y$ has positive real part, this yields

$$ \begin{eqnarray} \frac1{2\pi}\int_{\mathbb R}\frac{\xi}{\sinh{2\beta \xi}}e^{\mathrm iy\xi}\mathrm d\xi &=& \frac{2\pi\mathrm i}{2\pi}\sum_{n\in\mathbb N}(-1)^n\frac{\xi_n}{2\beta}\mathrm e^{\mathrm iy\xi_n} \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\sum_{n\in\mathbb N}(-1)^n\mathrm e^{\mathrm iy\xi_n} \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\sum_{n\in\mathbb N}(-1)^n\mathrm e^{-n\pi y/(2\beta)} \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\sum_{n\in\mathbb N}\left(-\mathrm e^{-\pi y/(2\beta)}\right)^n \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\frac{-\mathrm e^{-\pi y/(2\beta)}}{1+\mathrm e^{-\pi y/(2\beta)}} \\ &=& \frac1{2\beta}\frac{\partial}{\partial y}\frac{-1}{1+\mathrm e^{\pi y/(2\beta)}} \\ &=& \frac1{2\beta}\frac{\pi}{2\beta}\frac{\mathrm e^{\pi y/(2\beta)}}{(1+\mathrm e^{\pi y/(2\beta)})^2} \\ &=& \frac\pi{4\beta^2}\frac1{(\mathrm e^{\pi y/(4\beta)}+\mathrm e^{-\pi y/(4\beta)})^2} \\ &=& \frac{\pi}{16\beta^2}\left(\cosh\frac{\pi y}{4\beta}\right)^{-2}\;. \end{eqnarray} $$

If $y$ has negative real part, the sign from the clockwise integration cancels the sign in $\xi_n$, and the sign in the exponent ends up in the argument of the even function $\cosh$, so the result is the same.

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@ Joriki... simply... AWESOME.. i would upvote more than once if it would be possible. No words to express my gratitude. Thanks. –  uforoboa Mar 28 '12 at 18:29
    
@uforoboa: You're quite welcome :-) –  joriki Mar 28 '12 at 18:34
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