How should I calculate the following expression?
\begin{equation} T\sum_{n=-\infty}^\infty\frac{1}{\sqrt{k^2+(2\pi nT)^2}}-\int_{-\infty}^\infty\frac{d\omega}{2\pi}\frac{1}{\sqrt{k^2+\omega^2}} \end{equation}
Since neither of the two terms converges, so what this expression really means is \begin{equation} \lim_{N\rightarrow\infty}\left( T\sum_{n=-N}^N\frac{1}{\sqrt{k^2+(2\pi nT)^2}}-\int_{-2\pi NT}^{2\pi NT}\frac{d\omega}{2\pi}\frac{1}{\sqrt{k^2+\omega^2}} \right) \end{equation}
The answer to the above expression must be a function of $k/T$, say it is $f\left(\frac{k}{T}\right)$. I think this function $f$ has pretty good properties so that the following integral converges: \begin{equation} \int_0^\infty dk\cdot k\cdot f\left(\frac{k}{T}\right) \end{equation}
I suspect the final answer to the above integral is $\pi T^2\ln\left(\frac{1+\sqrt{5}}{2}\right)$ up to a factor of $2$.