# Closed form for an integral involving the Riemann zeta function at the critical line

After seeing this question $$L_2$$ bounds for $$\zeta(1/2 + it)$$ and a related integral i became curious if/how the approach in the answer by reuns can be applied to evaluate

$$I_{a,b}=\int_{-\infty}^{\infty} \frac{\zeta(1/2 + it)}{(t-a)^2 + b^2} \mathrm{d}t$$

where $$\zeta$$ denotes the Riemann zeta function, $$a$$ and $$b$$ are constants. ?

For learning's sake, any other method for performing this integeral would be most welcome.

• Are you still reading/using this particular account? – Yemon Choi Dec 20 '18 at 8:27

We can rewrite the integral in the form $${1\over i}\int_{\Re(s)=1/2} {\zeta(s)\over (s-\alpha)(1-s-\overline{\alpha})} \;ds$$ for suitable complex $$\alpha$$. For $$\Re(\alpha)=1/2$$ some further regularization of the integral is necessary. The convexity bound (or related) on $$\zeta(s)$$ in the half-plane $$\Re(s)\ge 1/2$$ allow the contour to be moved to the right indefinitely, picking up (negatives of) residues at $$s=1$$ (the pole of $$\zeta(s)$$), and at either $$s=\alpha$$ or $$s=1-\overline{\alpha}$$, whichever is in the right half-plane. If it happens that either of the latter two coincides with $$s=1$$, it is (as usual) slightly more complicated to evaluate the residue at that point.
Specifically, with $$\Re(\alpha)>1/2$$, for example, we obtain $$-2\pi {1\over (1-\alpha)(-\overline{\alpha})} - 2\pi {\zeta(\alpha)\over 1-\alpha-\overline{\alpha}}$$