# How to evaluate the imaginary part of this one-sided fourier transform?

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.