I was wondering if the integral $$I=\int_{-\infty}^{\infty}d\omega \omega e^{-(\omega/a)^2}\coth(\frac{b\omega}{2})\cos(\omega c)$$ where $a,b,c>0$ can be solved using complex countour integration methods. Well, the only part of the integrand that has poles in principle is the $\coth(\frac{b\omega}{2})$ which can be written as $$\coth\left(\frac{b\omega}{2}\right)=\frac{2}{b\omega}\left(1+2\sum_{n=1}^{\infty}\frac{\omega^2}{\omega^2+\nu_n^2}\right)$$. where $\nu_n = \frac{2\pi n}{b}$. Thus the integral splits in $$I=\frac{2}{b}\int_{-\infty}^{\infty}d\omega \;e^{-(\omega/a)^2}\cos(\omega c) + \frac{4}{b}\sum_{n=1}^{\infty}\int_{-\infty}^{\infty}d\omega \; e^{-(\omega/a)^2}\frac{\omega^2}{\omega^2+\nu_n^2}\cos(\omega c)$$ and the first one can be done integrating by parts. The result is $$\int_{-\infty}^{\infty}d\omega \; e^{-(\omega/a)^2}\cos(\omega c)=\frac{a\sqrt{\pi}}{b}e^{-\frac{a^2c^2}{4}}$$ but I have problems with the other part, up to prefactors $$\sum_{n=1}^{\infty}\int_{-\infty}^{\infty}d\omega \; e^{-(\omega/a)^2}\frac{\omega^2}{\omega^2+\nu_n^2}\cos(\omega c)$$ with at first view seem to be suitable for a contour integration approach, taking advantage of the poles of the $\coth$ at $z=\pm i\nu_n$ but the "straightforward" analytic continuation $\omega\rightarrow z\in \mathbb{C}$ gives problems, because the integrand do not vanishes in any path out of real axis because $\exp(-(z/a)^2)$ diverges at $\pm i\infty$. I'm guessing there should be "any other" analytic continuation for the integrand that allows to use the residue theorem enclosing either the positive or the negative imaginary poles $\pm i \nu_n$ or I'm missing something
Thanks in advance