Evaluating $\int_{-\infty}^\infty \frac{dx}{\cosh(x-a)\cos(2x)}$ I have been asked to evaluate $$\int_{-\infty}^\infty \frac{dx}{\cosh(x-a)\cos(2x)}$$. I'm deliberating on whether this indefinite integral exists or not. The integrand diverges when $x=\frac{1}{2}(n+\frac{1}{2})\pi$ but the $\cosh(x-a)$ term relaxes these singularities exponentially as $x$ goes to infinity. 
If it does exist, then I'm left with the problem of computing it. This is for a complex variables class, so I was using residue methods. However, there are countably many simple poles along the real axis as well as for each $z=i(n+\frac{1}{2})\pi +a$ in the complex plane, so I don't even know what contour to use. It looks like rectangles and semicircles are out. Any suggestions?

So it turns out that the integral is definitely not convergent (as expected). Consider a small interval about $2x=\frac{\pi}{2}$, ie $x=\frac{\pi}{4}$. Then $\frac{1}{\cos(2x)}$ behaves like $\frac{1}{x}$ in this small interval. Also the $\cosh(x-a)$ term is non-zero so has little influence; it can be treated like a constant. But $\int \frac{1}{x} dx = \log x$, which diverges in this small interval so the integral diverges over the real line also.
