Evaluate the following integral $$ \int_{-i\infty}^{+i\infty}\frac{dq}{2\pi i}\frac{e^{q\lambda}}{\cosh \sqrt{q}-\frac{1}{2}}.$$Substituting $q=ix$ has not lead me anywhere. Nothing is specified about $\lambda$, so one would assume it is some complex constant. Any leads as to how to go about this would be very helpful.
Thanks
For $\lambda \gt 0$, close a semicircular contour to the left of the imaginary axis. You can use the residue theorem to evaluate. Note that the poles of the integrand are all along the negative real axis only at $q=-(6 k+1)^2 \pi^2/9$ and $q=-(6 k+5)^2 \pi^2/9$ for $k$ a nonnegative integer. Note that there are no branch points in the integrand! Thus, we may simply write down the integral for $\lambda \gt 0$ as
$$\int_{-i \infty}^{i \infty} \frac{dq}{i 2 \pi} \frac{e^{q \lambda}}{\cosh{\sqrt{q}}-\frac12} = \frac{4 \pi}{3 \sqrt{3}} \sum_{k=0}^{\infty} \left [(6 k+1) e^{-(6 k+1)^2 \pi^2 \lambda/9} - (6 k+5) e^{-(6 k+5)^2 \pi^2 \lambda/9} \right ]$$
For $\lambda \lt 0$, one closes to the right. As there are no poles there, the integral is zero.