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?
 A: My calculation shows that
$$ \mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{dx}{\cosh(x-a)\cos x} = 2\pi \sum_{k=0}^{\infty} \frac{(-1)^{k}}{\cosh \left( \pi k + \frac{\pi}{2} \right)} \cos(2k+1)a. \tag{1} $$
Indeed, what I derived is that
$$ \int_{-\infty}^{\infty} \frac{dx}{\cosh(x-a)\cos (x+i\epsilon)} = 2\pi \sum_{k=0}^{\infty} \frac{(-1)^{k}e^{-(2k+1)\epsilon}}{\cosh \left( \pi k + \frac{\pi}{2} \right)} e^{i(2k+1)a},
\quad \epsilon > 0. $$
I have no idea how to simplify $\text{(1)}$, except for the following obvious special case:
$$ \mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{dx}{\cosh\left(x-\frac{\pi}{2}\right)\cos x} = \mathrm{PV}\!\!\int_{-\infty}^{\infty} \frac{dx}{\cosh x \sin x} = 0. $$
A: 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.
