Using the residue theorem, I want to evaluate the integral along the entire real line,
\begin{equation} \int_{-\infty}^{\infty} \dfrac{1}{\sqrt{1-q^2}\cos(\sqrt{1-q^2})}\ \mathrm{d}q, \end{equation} I can evaluate this numerically ($\approx 4.3023-2.2343 i$) but I really want to get the solution using the residue theorem. Analytic continuation gives, \begin{equation} \oint_\mathcal{C} \dfrac{1}{\sqrt{1-z^2}\cos(\sqrt{1-z^2})}\ \mathrm{d}z, \end{equation} where $\mathcal{C}$ is a closed contour in the complex plane containing the entire real line $\Gamma$ and then some contour $\cap$ in the upper complex plane at $|z|\gg 1$ which I assume I can choose such that the contour integral along $\cap$ vanishes.
I have fond that there are infinitely many residues at $z_n=i\sqrt{(\frac{\pi}{2}+n\pi)^2-1}$ with the value, \begin{equation} \mathrm{Res}_n= \dfrac{(-1)^{n}}{z_n}, \quad n\in \mathbb{Z}. \end{equation} But if I calculate $2\pi i\sum_{n=0}^{\infty} \mathrm{Res}_n\approx 4.3023$, only the real part of the numeric solution is reproduced while the imaginary part vanishes.
Can you spot my mistake?
I believe the problem is about my contour around the branch points at $z=\pm 1$. The imaginary part of the integrand above is even for a contour exactly at the real line but odd for a contour which is shifted infinitesimally upwards (or downwards) in the imaginary direction. Maybe I need some clever contour in the complex plane to catch the imaginary part?
N.B. This is physics, so I am satisfied with truncated or approximated solutions.
Coming back to this problem, I think there are two things you can do, the first is to notice that the integrand is even, and that it is only complex valued for $|x|>1$. Denote $I$ by the original integral, then we have
$$I = \int_{-1}^{1} \frac{1}{\sqrt{1-t^2}\cos{(\sqrt{1-t^2}})}\, dt + 2 i \int_{1}^{\infty}\frac{1}{\sqrt{t^2-1}\cosh{(\sqrt{t^2-1}})}\, dt$$
You have, with a pretty loose argument, shown the real part of the integral correctly is the sum you've written, however your contour only accounts for the real part of the integral, not any imaginary contributions.
Both integrals can be attacked with a rectangular contour, avoiding the branch cut on [-1, 1]
Edit: I've drawn the contour:
Technically you would be within epsilon above the interval on the real axis, $[-1,1]$ since that is the branch cut and is not a region you can integrate through. Further, the contributions to the integral as the two semi-circles approach zero would also need to be computed.