Specifically, I'm interested in the real part of this integral:
$$\int\limits_{\Lambda}^\infty \frac{dx}{x}\frac{ e^{-x \alpha^2}}{\cos(x)} $$
where $\Lambda$ is a real, positive constant that serves as a cutoff to regulate the pole at $x=0$. Note that $\alpha$ is also real. Of course, there are infinitely many poles on the real axis at $x = n \pi /2$ for odd $n$. These can be regulated by an analytic continuation in which we get a sum in $n$ of imaginary contributions (one for each pole), so I'm able to solve the imaginary part. The real part is then the Cauchy principal value of the integral:
$$\mathcal{P}\int\limits_{\Lambda}^\infty \frac{dx}{x}\frac{e^{-x \alpha^2}}{\cos(x)} $$
though I'm stuck when I try to evaluate this by the definition of Cauchy principal value. To simplify, I've tried expanding the $\sec(x)$ but run into problems with the expansion's radius of convergence (I have made progress with an asymptotic expansion in the limit of large $\alpha$, but this limit is not ideal in my case).
I've also tried a contour integral with the residue theorem in which I take a half semi-circle contour in the first quadrant of the complex plane. This leaves me to solve the real part of:
$$\int\limits_{\Lambda}^\infty \frac{dx}{x}\frac{ e^{-i x\alpha^2}}{\cosh(x)} $$
This appears simplified as we're no longer dealing with infinitely many poles, but I'm having difficulty solving it as well. (Note that there will be an $\Lambda$-dependent quarter-arc integral that does not vanish when doing the contour integration, but I'm specifically able to ignore this in my use-case.)
Any ideas appreciated. Thanks!