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Here's my problem. Let $$ f(z) = \frac{z}{\sqrt{z^2 + m^2}} \frac{2z + p}{(2z + p)^2 + \omega^2} \frac{1}{e^{L \sqrt{z^2 + m^2} - i a} - 1} $$ where $m, p, \omega, L$ and $a$ are real, $L$ is positive. I want to compute $$ \lim_{\omega \to 0} \int_{-\infty}^{+\infty} dx f(x). $$ Actually, the full expression I'm looking for is the sum of four such integrals, with $\pm p$ and $\pm a$, but I tried to compute the above single expression first. Here is my reasoning. We first notice that $f(x)$ has no singularity on the real axis and decreases exponentially, so the integral must be convergent. In order to compute it with complex analysis methods, we identify the poles of $f(z)$ : $$ \frac{1}{2}(-p \pm i \omega) \hspace{1cm} \text{and} \hspace{1cm} \pm i \sqrt{m^2 + m_{n, a}^2} $$ for $n \in \mathbb{Z}$ where I defined $m_{n, a}^2 = \frac{2 \pi n + a}{L}$. This set of poles comes from the exponential, that vanishes whenever its exponent is equal to $2 \pi i$. The function also has branch points at $$ \pm i m. $$ Now comes the time to choose an integration contour. Since most of the poles lie on the imaginary axis, taking the branch cut as $[-\infty, -im]$ and $[im, i\infty]$ is not ideal. I therefore define it to be $[-im, im]$ instead. My integration contour is thus a keyhole contour in the upper-half plane, with the "keyhole part" going around $im$ from the origin. My contour skips the $[-\delta, \delta]$ interval on the real axis because of the branch cut, but I suppose the limit $\delta \to 0$ really gives the integral over the whole real axis in the end. On the infinite half-circle as well as on the small circle going around $im$, $f(z)$ vanishes. For the first pole, I have $$ \lim_{\omega \to 0} {\rm Res}_f \left( \frac{1}{2}(-p + i \omega) \right) = \frac{-p}{4 \sqrt{p^2 + 4 m^2}} \frac{1}{e^{\frac{L}{2} \sqrt{p^2 + 4 m^2} - ia} - 1}. $$ For the set of poles along the imaginary axis, I believe the residue of the exponential is $$ \frac{1}{L} \sqrt{\frac{m_{n, a}^2}{m^2 + m_{n, a}^2}} $$ and therefore $$ \lim_{\omega \to 0} {\rm Res}_f \left( i \sqrt{m^2 + m_{n, a}^2} \right) = \frac{1}{L} \frac{1}{2 i \sqrt{m^2 + m_{n, a}^2} + p}. $$

So, specifically, my questions are :

  1. Did I get anything wrong so far?

  2. Could the sum over $n \in \mathbb{Z}$ of the above residues be expressed in a nice way? Or at least for the case $m = 0$? For the latter, I'm expecting to get something like a digamma function.

  3. How do I have to compute the remaining integral on both sides of the branch cut? I'm pretty sure I got it wrong.

Remember that I sum the four pieces corresponding to the possible combinations of $\pm p$, $\pm a$ if this helps for any of the answers.

P. S.: this is my first post, feel free to tell me how to improve for the next time =)

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    $\begingroup$ Beautiful first post, welcome to Math SE! $\endgroup$
    – user409521
    Mar 19, 2017 at 1:56

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