integral with hyperbolic cosines

Question

I want to show that $\displaystyle \int_{0}^{\infty} \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} \ dx = \frac{\cos \left( \frac{a}{2} \right) \cos \left( \frac{b}{2} \right)} {\cos (a) + \cos (b)} \ |a|+|b| < \pi$

I was certain I could evaluate the integral by integrating in the complex plane around the rectangle with vertices at $N+0i, N+i, -N+i, \text{and} -N+0i$.

I tried $\displaystyle f(z) = \frac{e^{(a+b)z}}{\cosh \pi z}, f(z) = \frac{e^{az} \cosh bz}{\cosh (\pi z)}$, and $\displaystyle f(z) = \frac{\cosh az \cosh bz}{\cosh \pi z}$. I was unable to extract the above integral using any of these three choices for $f(z)$.

Answer 1

I have a way of showing this without contour integration in the complex plane. There is a bit of a trick involved and, frankly, Mathematica misleads. It should be noted that the condition $|a|+|b| < \pi$ is needed for the integral to converge. Basically, rewrite the $\cosh$'s as exponentials:

$$\begin{align} \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} &= 2 \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{1+e^{-2 \pi x}} e^{-\pi x} \\ &= \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \int_{0}^{\infty} dx \: (e^{a x}+e^{-a x}) (e^{b x}+e^{-b x}) e^{-(2 k+1) \pi x} \\ \end{align} $$

Evaluating the integrals, we get

$$= \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{(2 k+1)\pi -(a+b)} + \frac{1}{(2 k+1)\pi +(a+b)}\right ] $$ $$ + \frac{1}{2} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{(2 k+1)\pi -(a-b)} + \frac{1}{(2 k+1)\pi +(a-b)} \right ] $$

Here I note that $a+b$ and $a-b$ should not be some multiple of $\pi$, so that the above sums behave properly.

To get the sums into a somewhat familiar form, I rearrange them a bit to get

$$= \frac{1}{4 \pi} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{k +\left (\frac{1}{2} - \frac{a+b}{2 \pi} \right )} + \frac{1}{k +\left (\frac{1}{2} + \frac{a+b}{2 \pi}\right )}\right ] $$ $$ + \frac{1}{4 \pi} \sum_{k=0}^{\infty} (-1)^k \left [ \frac{1}{k +\left (\frac{1}{2} - \frac{a-b}{2 \pi} \right )} + \frac{1}{k +\left (\frac{1}{2} + \frac{a-b}{2 \pi}\right )}\right ] $$

Now, here is the interesting part (at least to me). Let

$$f(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k+z} $$

This looks like it should be a trig function of some sort. It is not; rather, it is something called a Hurwitz-Lerch transcendent, which does not look like it will be much help. That said, it almost looks like a trig function, so I instead considered the following:

$$\begin{align} f(z) + f(1-z) &= \sum_{k=0}^{\infty} \frac{(-1)^k}{k+z} + \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1-z}\\ &= \sum_{k=0}^{\infty} \frac{(-1)^k}{z+k} + \sum_{k=0}^{\infty} \frac{(-1)^{k+1}}{z-(k+1)}\\ &= \sum_{k=-\infty}^{\infty} \frac{(-1)^k}{z+k} \\ &= \frac{\pi}{\sin{\pi z}}\\ \end{align}$$

This is very helpful, because we have precisely this functional form above, e.g.,

$$\frac{1}{2} - \frac{a+b}{2 \pi} = 1 - \left ( \frac{1}{2} + \frac{a+b}{2 \pi} \right ) $$ $$\frac{1}{2} - \frac{a-b}{2 \pi} = 1 - \left ( \frac{1}{2} + \frac{a-b}{2 \pi} \right ) $$

So we get for the integral:

$$\begin{align} \int_{0}^{\infty} dx \: \frac{\cosh (ax) \cosh (bx)}{\cosh (\pi x)} &= \frac{1}{4} \left [ \frac{1}{\sin{\left ( \frac{\pi}{2} - \frac{a+b}{2} \right )}} + \frac{1}{\sin{\left ( \frac{\pi}{2} - \frac{a-b}{2} \right )}} \right ] \\ &= \frac{1}{4} \left [ \frac{1}{\cos{\left ( \frac{a+b}{2} \right )}} + \frac{1}{\cos{\left ( \frac{a-b}{2} \right )}} \right ] \\ &= \frac{\cos{\frac{a}{2}} \cos{\frac{b}{2}}}{\cos{a} + \cos{b}} \end{align}$$

QED