Another interesting technique of proving $\int_{0}^{\infty}{(1+e^{-n^a{x}})^{n-1}\over [1+\cosh(n^a{x})]^n}dx={1\over n^{a+1}}$ I want to prove that

$$I=\int_{0}^{\infty}{(1+e^{-n^a{x}})^{n-1}\over [1+\cosh(n^a{x})]^n}dx={1\over n^{a+1}}\tag1$$

Elementary technique to prove (1):
Let $y=n^a$
$$1+\cosh(yx)={(1+e^{yx})^2\over 2e^{yx}}\tag2$$
$${1+e^{-yx}\over 1+e^{yx}}=e^{-yx}\tag3$$
Sub $(2)$ into $(1)\rightarrow (4)$
$$I=\int_{0}^{\infty}\left[{2e^{yx}\over (1+e^{yx})^2}\cdot{(1+e^{-yx}})\right]^n(1+e^{-yx})^{-1}dx\tag4$$
Rewrite $(4)$ as $(5)$ and  use $(3)$ to simplify to $(6)$
$$I=\int_{0}^{\infty}\left[{2e^{yx}\over 1+e^{yx}}\cdot{1+e^{-yx}\over 1+e^{yx}}\right]^n(1+e^{-yx})^{-1}dx\tag5$$
$$I=2^n\int_{0}^{\infty}{1\over (1+e^{yx})^n}\cdot{e^{yx}\over 1+e^{yx}}dx\tag6$$
Rewrite $(6)$ as
$$I=2^n\int_{0}^{\infty}{e^{yx}\over (1+e^{yx})^{n+1}}dx\tag7$$
$$I={2^n\over y}\int_{0}^{\infty}ye^{yx}(1+e^{yx})^{-n-1}dx\tag8$$
$$I=\left.-{2^n\over ny}\cdot{1\over (1+e^{yx})^n}\right|_{0}^{\infty}\tag9$$
$$I={1\over ny}={1\over n^{a+1}}\tag{10}$$
This method is elementary and not so interesting to prove (1). Anybody with an interesting method at approcaching (1)?
 A: There is a more elementary method. I don't know if it is "interesting" for you. We have $$I=\int_{0}^{\infty}\frac{\left(1+e^{-n^{\alpha}x}\right)^{n-1}}{\left(1+\cosh\left(n^{\alpha}x\right)\right)^{n}}dx\stackrel{n^{\alpha}x=u}{=}\frac{1}{n^{\alpha}}\int_{0}^{\infty}\left(\frac{1+e^{-u}}{1+\cosh\left(u\right)}\right)^{n-1}\frac{1}{1+\cosh\left(u\right)}du
 $$ now if we put $$\frac{1+e^{-u}}{1+\cosh\left(u\right)}=v,\,-\frac{1}{1+\cosh\left(u\right)}du=dv
 $$ we get $$I=\frac{1}{n^{\alpha}}\int_{0}^{1}v^{n-1}dv=\frac{1}{n^{\alpha+1}}.$$
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
\color{#f00}{I} & = \int_{0}^{\infty}{\pars{1 + \expo{-n^{^{a}}x}}^{n - 1}
\over \bracks{1 + \cosh\pars{n^{^{a}}x}}^{\,n}}\,\dd x\
\stackrel{n^{^{a}}x\ \to\ x}{=}\
n^{-a}\int_{0}^{\infty}{\pars{1 + \expo{-x}}^{n - 1}
\over \bracks{1 + \cosh\pars{x}}^{\,n}}\,\dd x
\\[3mm] & =
n^{-a}\int_{0}^{\infty}\bracks{\expo{x/2} + \expo{-x/2} \over
1 + \cosh\pars{x}}^{n- 1}
\,{\expo{-\pars{n - 1}x/2} \over 1 + \cosh\pars{x}}\,\dd x =
\half\,n^{-a}\int_{0}^{\infty}
\,{\expo{-\pars{n - 1}x/2} \over \cosh^{n + 1}\pars{x/2}}\,\dd x
\label{1}\tag{1}
\\[3mm] & =
2^{n}n^{-a}\int_{0}^{\infty}
{\expo{x} \over \pars{\expo{x} + 1}^{n + 1}}\,\dd x =
2^{n}n^{-a}\bracks{-\,{1 \over n}\,
{1 \over \pars{\expo{x} + 1}^{n}}}_{\ x\ =\ 0}^{\ x\ \to\ \infty}
\\[3mm] & =
2^{n}n^{-a}\pars{1 \over 2^{n}n}
= \color{#f00}{{1 \over n^{a + 1}}}
\end{align}

Note que $\ds{\quad 1 + \cosh\pars{x} = 2\cosh^{2}\pars{x \over 2}\quad}$ and
  $\ds{\quad\expo{x/2} + \expo{-x/2} = 2\cosh\pars{x \over 2}.\quad}$ See \eqref{1}.

