In the course of my research I have found a few integrals that I would like to have closed-form answers to:
$$\int_{c- i \infty}^{c+ i \infty} \frac{1}{z-1} \frac{8 \pi^4 \cot{ \big( \frac{\pi}{6} z \big)}}{\sqrt{3} (\cosh{2 \pi a} - \cos{2 \pi z})} \ dz $$ and
$$\int_{c- i \infty}^{c+ i \infty} \frac{1}{z-1} \frac{16 \pi^4}{\big(1+2 \cos{ \big(\frac{\pi}{3} z\big)} \big) (\cosh{2 \pi a} - \cos{2 \pi z})} \ dz $$
I am concerned with the answer when $1 > Re(c) >0$ and $a > 0$ is a known numerical constant.
In both cases, the rational function ruins the periodicity, but its still simple enough that it seems like there might be a way to compute these if you pick the correct contour. Sadly, neither function decays fast enough along the real line to be able to close the contour with a semi-circle. Thanks to the exponential decay away form the real axis, I have been trying to work with box contours and find 2 verticals lines with a known relationship between the integrals along those lines (eg they're equal up to some known factor) but now I've run out of good ideas.
I also need to compute a similar integral that contains only trig functions:
$$\int_{c- i \infty}^{c+ i \infty} \frac{2 \pi^4 \sec{ \big( \frac{\pi}{2} z \big)} \cos{ \big( \frac{\pi}{6} z \big)}}{-\cosh{2 \pi a} + \cos{2 \pi z}} \ dz $$
with the same restrictions. It has a period of 12 but I can't figure out how to use this to compute the integral.
Shortly after asking this question I had a new idea; multiply by a 'convergence factor' inside the integral that reduces the contribution from the semi-circular arc. Sum the residues and calculate the value of this new integral, then take the limit that the convergence factor approaches $1$, yielding the original integral.
In this case, $$\lim_{b->0} \int_{c- i \infty}^{c+ i \infty} \frac{e^{-b z}}{z-1} \frac{8 \pi^4 \cot{ \big( \frac{\pi}{6} z \big)}}{\sqrt{3} (\cosh{2 \pi a} - \cos{2 \pi z})} \ dz $$ can be computed. The sum of the residues appears to be logarithmically divergent, but these divergences cancel when all of the residues are included. The answer is somewhat messy so I won't post it until I have simplified it further, but this is the main idea. It should work for all 3 integrals.