# Integrating $\frac{x^k }{1+\cosh(x)}$

In the course of solving a certain problem, I’ve had to evaluate integrals of the form:

$$\int_0^\infty \frac{x^k}{1+\cosh(x)} \mathrm{d}x$$

for several values of k. I’ve noticed that that, for k a positive integer other than 1, the result is seemingly always a dyadic rational multiple of $\zeta(k)$, which is not particularly surprising given some of the identities for $\zeta$ (k=7 is the first noninteger value).

However, I’ve been unable to find a nice way to evaluate this integral. I’m reasonably sure there’s a way to change this expression into $\int \frac{x^{k-1}}{e^x+1} \mathrm{d}x$, but all the things I tried didn’t work. Integration by parts also got too messy quickly, and Mathematica couldn’t solve it (though it could calculate for a particular value of k very easily).

So I’m looking for a simple way to evaluate the above integral.

#### Solutions Collecting From Web of "Integrating $\frac{x^k }{1+\cosh(x)}$"

Just note that
$$\frac{1}{1 + \cosh x} = \frac{2e^{-x}}{(1 + e^{-x})^2} = 2 \frac{d}{dx} \frac{1}{1 + e^{-x}} = 2 \sum_{n = 1}^{\infty} (-1)^{n-1} n e^{-n x}.$$
Thus we have
$$\begin{eqnarray*}\int_{0}^{\infty} \frac{x^k}{1 + \cosh x} \, dx & = & 2 \sum_{n = 1}^{\infty} (-1)^{n-1} n \int_{0}^{\infty} x^{k} e^{-n x} \, dx \\ & = & 2 \sum_{n = 1}^{\infty} (-1)^{n-1} \frac{\Gamma(k+1)}{n^k} \\ & = & 2 (1 – 2^{1-k}) \zeta(k) \Gamma(k+1). \end{eqnarray*}$$
This formula works for all $k > -1$, where we understand that the Dirichlet eta function $\eta(s) = (1 – 2^{1-s})\zeta(s)$ is defined, by analytic continuation, for all $s \in \mathbb{C}$.

sos440 Added it to my collection $$\int\limits_0^\infty \frac{x^{s-1}}{e^x-1}dx=\zeta(s)\Gamma(s)$$

$$\int\limits_0^\infty \frac{x^{s-1}}{e^x+1}dx=\eta(s)\Gamma(s)$$

$$\int\limits_0^\infty \frac{x^{s}e^x}{(e^x-1)^2}dx=\zeta(s)\Gamma(s+1)$$

$$\int\limits_0^\infty \frac{x^{s}e^x}{(e^x+1)^2}dx=\eta(s)\Gamma(s+1)$$

$$\int\limits_0^1 \frac{x^s}{1-x}\log x dx = \sum\limits_{k=0}^\infty \frac{1}{(k+s)^2}$$
I’m guessing my memory is all right, but feel free to correct any mistakes!