Showing $\pi\int_{0}^{\infty}[1+\cosh(x\pi)]^{-n}dx={(2n-2)!!\over (2n-1)!!}\cdot{2\over 2^n}$

Question

Showing

$$\pi\int_{0}^{\infty}[1+\cosh(x\pi)]^{-n}dx={(2n-2)!!\over (2n-1)!!}\cdot{2\over 2^n}\tag1$$

Recall

$$1+\cosh(x\pi)={(e^{x\pi}+1)^2\over 2e^{x\pi}}\tag2$$

$$I_n=2^n\pi\int_{0}^{\infty}{e^{xn\pi}\over (1+e^{x\pi})^{2n}}dx\tag3$$

$$I_n={2^n\pi\over n\pi}\int_{0}^{\infty}n\pi e^{nx\pi}(1+e^{x\pi})^{-2n}dx\tag4$$

$$I_n=\left.{2^n\over n}\cdot{1\over (1+e^{x\pi})^{2n-1}}\right|_{0}^{\infty}\tag5$$

$$I_n={1\over n(2n-1)2^{n-1}}\tag6$$

Help, where did I went wrong?

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New Edit

From (3) we make a substitution

let $u=x\pi \rightarrow du=dx$

$$I_n=2^n\int_{0}^{\infty}e^{un}(1+e^u)^{-2n}du\tag{3a}$$

Answer 1

First: remove the useless constant by setting $x=\frac{z}{\pi}$. Then, through $z=\log u$ and $v=\frac{1}{u}$:

$$ \int_{0}^{+\infty}(1+\cosh(z))^{-n}\,dz = 2^n\int_{1}^{+\infty}\frac{\left(2+u+\frac{1}{u}\right)^{-n}}{u}\,du=2^n\int_{0}^{1}\frac{\left(2+v+\frac{1}{v}\right)^{-n}}{v}\,dv$$ so the LHS equals: $$ 2^{n-1}\int_{0}^{+\infty}\frac{u^{n-1} du}{(u+1)^{2n}}\,du = 2^n\int_{0}^{+\infty}\frac{t^{2n-1}\,dt}{(1+t^2)^{2n}}= 2^{n-1} B(n,n) = 2^{n-1}\frac{\Gamma(n)^2}{\Gamma(2n)}.$$

As an alternative, just apply IBP multiple times.
It leads to a recursion similar to the one for $\int_{0}^{\pi/2}\sin^{2n}(\theta)\,d\theta.$

Answer 2

Take $u=\tanh\left(x\right) $, then $\cosh\left(x\right)=\frac{1+u^{2}}{1-u^{2}} $ and $dx=\frac{2du}{1-u^{2}} $ so $$\int_{0}^{\infty}\frac{1}{\left(\cosh\left(x\right)+1\right)^{n}}dx=\int_{0}^{1}\frac{2}{\left(1-u^{2}\right)\left(\frac{1+u^{2}}{1-u^{2}}+1\right)^{n}}du $$ $$=\frac{1}{2^{n-1}}\int_{0}^{1}\left(1-u^{2}\right)^{n-1}du=\frac{1}{2^{n}}\int_{0}^{1}\left(1-v\right)^{n-1}v^{-1/2}dv=\frac{B\left(1/2,n\right)}{2^{n}}$$ $$=\frac{1}{2^{n}}\frac{\sqrt{\pi}\left(n-1\right)!}{\Gamma\left(n+1/2\right)}.$$

Answer 3

Let $I(n)$ be the integral given by

$$\begin{align} I(n)&=\pi\int_0^\infty\frac{1}{\left(1+\cosh(\pi x)\right)^n}\,dx\\\\ &=\int_0^\infty\frac{1}{\left(1+\cosh(x)\right)^n}\,dx\tag 1 \end{align}$$

Then, using the identity $1+\cosh(x)=\frac12 e^{x}\left(1+e^{-x}\right)^2$ in $(1)$ reveals

$$\begin{align} I(n)&=2^n\int_0^\infty \frac{e^{-nx}}{\left(1+e^{-x}\right)^{2n}}\,dx\\\\ &=2^n\int_0^1 \frac{x^{n-1}}{(1+x)^{2n}}\,dx\tag 2 \end{align}$$

Now, making the substitution $x\to 1/x$ in $(2)$, we obtain

$$\begin{align} I(n)&=2^n\int_1^\infty \frac{x^{n-1}}{(1+x)^{2n}}\,dx \tag 3 \end{align}$$

Combining $(2)$ and $(3)$ yields

$$I(n)=2^{n-1}\int_0^\infty \frac{x^{n-1}}{(1+x)^{2n}}\,dx \tag 4$$

Then, enforcing the substitution $x\to \frac{x}{1-x}$ in $(4)$, we find that

$$\begin{align} I(n)&=2^{n-1}\int_0^1 x^{n-1}(1-x)^{n-1}\,dx\\\\ &=2^{n-1}B(n,n)\\\\ &=2^{n-1}\frac{\Gamma^2(n)}{\Gamma(2n)}\\\\ &=\frac{(n-1)!}{(2n-1)!!}\\\\ &=\frac{(2n-2)!!}{2^{n-1}(2n-1)!!} \end{align}$$

as was to be shown!