 Solution

The residue theorem states that a positively oriented simple closed contour enclosing isolated poles and none on the contour has integral $2\pi i$ times the sum of their residues, provided the function is holomorphic elsewhere inside. Near a simple zero of $\psi$, $\psi (z)={\psi }^{\prime }(a)(z-a)+O((z-a{)}^2)$. HenceFor $\cot z=\cos z/\sin z$, this gives $\boxed{{\mathrm{Res}}_{z=k\pi }\cot z=1}$ for every integer $k$.

To evaluate the real integral, integrate $F(z)=\mathrm{Log}z/(z^2+a^2)$ around an upper semicircle, with a small upper semicircular indentation excluding zero. Use the branch real on the positive axis and with argument between zero and $\pi$ in the upper half-plane. The large arc tends to zero as $O(\ln R/R)$ and the small arc as $O(ϵ|\ln ϵ|)$. The only enclosed pole is $ia$, whose contribution isOn the negative real axis the upper boundary value is $\mathrm{Log}(-x)=\ln x+i\pi$. The two real segments therefore contributeComparing real parts yieldsThe integral is absolutely convergent at both zero and infinity, so no principal-value interpretation is needed.