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So, I came across the following integral

$\tag{1}\Gamma(\omega) = \int_{0}^{\infty}dse^{i\omega s}G^{+}(s)$

where $G^{+}(s) = \langle \phi(t)\phi(t - s)\rangle = \left[-16\pi\alpha^2\sinh^2(\frac{s}{2\alpha}-\frac{i\epsilon}{\alpha})\right]^{-1}$, $\epsilon > 0$, is a Wightman function that doesn't really depend on $t$. Also, ${G^{+}(s)}^{*} = G^{+}(-s)$. I decomposed it in the form $\Gamma = \frac{\gamma}{2} + iS$, so that $\frac{\gamma}{2}$ is the real part while $S$ is the imaginary one. The real part, $\gamma = \Gamma + \Gamma^{*}$, takes on the simples form

$\tag{2}\gamma(\omega) = \int_{-\infty}^{\infty}dse^{i\omega s}G^{+}(s)$

Using the series

$\tag{3}\text{cosec}(\pi x) = \frac{1}{\pi^2}\sum_{k = -\infty}^{\infty}(x - k)^{-2}$

I can rewrite the integrand in (2) and do the contour integration - a semicircle in the upper-half plane + the whole real line, with the integral vanishing in the semicircle - obtaining

$\tag{4}\gamma(\omega) = \frac{\omega}{2\pi}\frac{1}{1 - e^{-2\pi\omega\alpha}}$

Now I want to evaluate $S = (\Gamma - \Gamma^{*})/2i$. The problem is that, unlike in the previous case, the two integrals won't merge into one from $-\infty$ to $\infty$, meaning I can't use the Cauchy theorem, since now there's a contribution from the integration over the imaginary axis from $0$ to $\infty$. How am I supposed to perform this integration? I've seen some similar integrals - where none is performed explicitly - and in some cases $S$ seems to be a Hilbert transform of $\Gamma$. Something like

$\tag{5}S(\lambda) = \frac{P}{\pi i}\int_{-\infty}^{\infty}d\omega\frac{\Gamma(\omega)}{\omega - \lambda}$

Not that I know how to evaluate (5) either, I'm just trying to get my head around this, and it might help someone help me. Thanks in advance.

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