Evaluating $\int_{0}^{\infty}{\frac{1}{\cosh^{2k+1}(x)} dx}$

Question

I tried :

$$\begin{align}\int_{0}^{\infty}{\frac{1}{\cosh^{2k+1}(x)} dx}&=2^{2k+1}\int_{0}^{\infty}{(e^{x}+e^{-x})^{-(2k+1)}dx}\\&=2^{2k+1}\int_{0}^{\infty}{\frac{1}{u}\left(u+\frac{1}{u}\right)^{-(2k+1)}du}\\&=2^{2k+1}\int_{0}^{\infty}{u^{-2k}(u^2+1)^{-(2k+1)}du}\\&=2^{2k}\int_{0}^{\infty}{t^{-(k+1/2)}(t+1)^{-(2k+1)}dt}\end{align}$$

I know this has answers here but I'm wondering if what I have so far can be continued to arrive at one of those answers.

Answer 1

Consider the substitution $\cos t =\frac1{\cosh x}$. Then, $dt = \frac{dx}{\cosh x}$ and

\begin{align}\int_{0}^{\infty}{\frac{1}{\cosh^{2k+1}x} dx}&=\int_{0}^{\frac\pi2}\cos^{2k}t\ dt\\ =& \ \sum_{j=0}^{2k} \frac{\binom {2k}j}{2^{2k}}\int_0^{\frac\pi2}\cos[2(k-j)t]dt =\frac{\pi \binom {2k}k}{2^{2k+1}} \end{align} where only the term of $j=k$ survives the integration.