On line integral $\displaystyle\int_{a + i\infty}^{a - i\infty} \frac{\zeta^2(1-s)\zeta^2(s-4m-1)}{2 \cos\left(\frac{\pi s}{2}\right)} \,\mathrm{d}s$ I am curious on how one can evaluate the integral
$$\int_{a + i\infty}^{a - i\infty} \frac{\zeta^2(1-s)\zeta^2(s-4m-1)}{2 \cos\left(\frac{\pi s}{2}\right)} \,\mathrm{d}s$$
where $4m+2<a< 4m+3$ and $m \in \mathbb{N}$.
I think I should be pushing the contour left and collect poles as I move. Basically I am looking for a series representation of the line integral of the form $\sum_{n=1}^{\infty} f(n)$ if it exists. I am wondering whether the integral will be zero on horizontal strips as I move down. Any help would be appreciated.

The functional equation $\zeta(1-s) = {2\over (2\pi)^s}\,\cos\left(\frac{\pi s}2\right)\,\Gamma(s)\,\zeta(s)$ gives that your integral diverges due to the growth of $\Gamma(ai+t)$ as $t\to +\infty$.
If instead you meant $a\pm i\infty$ and $m\in \Bbb{Z}$ then letting $f(s)=\frac{\zeta^2(1-s)\zeta^2(s-4m-1)}{2 \cos\left(\frac{\pi s}{2}\right)}$
$$I=\int_{a -i\infty}^{a + i\infty} f(s)\,\mathrm{d}s$$
The change of variable $s=4m+2-z$ gives
$$I = \int_{4m+2-a+ i\infty}^{4m+2-a -i \infty} f(4m+2-z) \,\mathrm{d}(4m+2-z)
= -\int_{4m+2-a - i\infty}^{4m+2-a +i \infty} f(z) \,\mathrm{d}z
$$
On the other hand, the residue theorem and that $\zeta(-2k)=0$ gives for $m\ge 1$
$$I=\int_{4m+2-a -i\infty}^{4m+2-a + i\infty} f(s) \,\mathrm{d}s
 +2i\pi Res(f(s),0)+2i\pi Res(f(s),4m+2)$$
whence $$I = i\pi Res(f(s),0)+i\pi Res(f(s),4m+2)$$
Since $f(s)=-f(4m+2-s)$ and $Res(f(s),4m+2)=-Res(f(4m+2-s),0)$
we get that $I=2i\pi Res(f(s),0)$
