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.

  • $\begingroup$ Are you still reading/using this particular account? $\endgroup$ – 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 $$ -2\pi {1\over (1-\alpha)(-\overline{\alpha})} - 2\pi {\zeta(\alpha)\over 1-\alpha-\overline{\alpha}} $$


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