I want to know if the following relation is true
$$ \frac{1}{2\pi i} \int_{-\infty}^{\infty} dz \tanh{\pi z} \left[\frac{1}{\sqrt{1-z^2}} - \frac{1}{\sqrt{2-z^2}} \right] = \sum_{n>0} \left[\frac{1}{\sqrt{1+(n+1/2)^2}} - \frac{1}{\sqrt{2+(n+1/2)^2}} \right]. $$ I will name the last series $S$. I consider a complex contour that begins in the negative part of the real axis, goes around the points $\pm 1, \pm \sqrt{2}$ with half circle, then closes with a half circle in the upper half plane. The small half circles on the real axis won't give any contribution to the full integral because the residues at $z=\pm 1, \pm \sqrt{2}$ will be zero. Same for the half circle in the upper half plane when we took its radius to be large (the integral for the $\frac{1}{\sqrt{1-z^2}}$ term will be canceled by the $\frac{1}{\sqrt{2-z^2}}$ term for radius that tends to infinity). Now the contour enclosed poles of $\tanh{\pi z}$ which are at the points $i(n+1/2)$, $n>0$. By the residue theorem I get that the integral written above $I= 2\pi i S$.
I tried to numerically evaluate $I$ in Mathematica to compare it to the numerical value of $S$. I found significative differences between the two.
So I would like to know if my calculation is flawed.