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

Solution 1:

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 $s \in \mathbb{C}$.

Solution 2:

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!