# On line integral $\displaystyle\int_{a + i\infty}^{a – i\infty} \frac{\zeta^2(1-s)\zeta^2(s-4m-1)}{2 \cos\left(\frac{\pi s}{2}\right)} \,\mathrm{d}s$

complex-analysisriemann-zetasequences-and-series

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 $$4m+2 and $$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 $$I=2i\pi Res(f(s),0)$$