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


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

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

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

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

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

Help, where did I went wrong?

New Edit

From (3) we make a substitution

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


  • $\begingroup$ @AlexR. which is the same :) as the OP stated, it holds $1+\cosh(\pi x) = 1+\frac{e^{\pi x}+e^{-\pi x}}{2}={(e^{x\pi}+1)^2\over 2e^{x\pi}}$ $\endgroup$ – user190080 Jun 8 '16 at 22:11
  • 1
    $\begingroup$ You somehow introduced an $n$ into $(1+e^{n\pi x})^{2n}$. It should be $(1+e^{x\pi})^{2n}$ $\endgroup$ – Alex R. Jun 8 '16 at 22:16
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    $\begingroup$ The main issue is that $$ n x^{\color{red}{n}}\neq \frac{d}{dx} x^n.$$ $\endgroup$ – Jack D'Aurizio Jun 8 '16 at 22:25
  • $\begingroup$ @JackD'Aurizio which happens where exactly (I guess you refer to the substitution but this seems to be fine) $\endgroup$ – user190080 Jun 8 '16 at 22:32
  • $\begingroup$ @user190080: between $(4)$ and $(5)$. $\endgroup$ – Jack D'Aurizio Jun 8 '16 at 22:34

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 $\int_{0}^{\pi/2}\sin^{2n}(\theta)\,d\theta.$

  • $\begingroup$ +1, but I think you lack a $dt$ in the second integral of the second row $\endgroup$ – giobrach Jun 9 '16 at 11:42
  • $\begingroup$ @giobrach: thank you, now fixed. $\endgroup$ – Jack D'Aurizio Jun 11 '16 at 13:30

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

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    $\begingroup$ (+1) There is a nice side-effect: our answers together prove the Legendre duplication formula :D $\endgroup$ – Jack D'Aurizio Jun 8 '16 at 22:36

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!

  • $\begingroup$ Are you sure, Mark? Numerics don't match. After "reveals" we should have $I(n)=2^n\left(\ldots\right)$, for instance. $\endgroup$ – Jack D'Aurizio Jun 8 '16 at 22:29
  • $\begingroup$ @JackD'Aurizio Working on it ... yes there is an error. Thanks!! $\endgroup$ – Mark Viola Jun 8 '16 at 22:32
  • $\begingroup$ @chinacat Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Jun 11 '16 at 3:48

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