TITLE: Integral$\int_0^\infty \ln x\,\exp(-\frac{1+x^4}{2\alpha x^2}) \frac{x^4+\alpha x^2- 1}{x^4}dx$?
QUESTION [13 upvotes]: I am trying to prove
$$
I:=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+\alpha x^2- 1}{x^4}dx=\frac{\sqrt{2\alpha^3 \pi}}{2\sqrt[\alpha]e},\qquad \alpha>0.
$$
Note: The proof below shows how this is just a Gaussian integral!
I am not sure how to start this one.  It seems very difficult to me However the answer is very nice.
I thought maybe trying to write $I(\alpha)$ and $I'(\alpha)$ to try and simplify things but it didn't help much.  at $x=0$ there seems to be a problem with the integrand also however I am not sure how to go about using this.  Perhaps we could try and use a series expansion for $e^x=\sum_{n=0}^\infty x^n / n!$ however the function $e^{-1/x^2}$ is well known that its taylor series is zero despite the function not being. The factor of $x^4+\alpha x^2-1$ has been giving me trouble with simplifying the integrand.  Thanks.
To those who just made an edit:  If you are looking for a +2, please edit something worthwhile.  I edited it back to what I had considering you didn't fix anything as is shown in the Edit History.

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{I\equiv\int_{0}^{\infty}\ln\pars{x}
     \exp\pars{-\,{1 + x^{4} \over 2\alpha x^{2}}}\,
     {x^4+\alpha x^2- 1 \over x^4}\,\dd x=
     {\root{2\alpha^{3}\pi} \over 2\root[\alpha]{\expo{}}}:\ {\large ?},
     \qquad \alpha > 0}$.

From @Chen Wang answer $\ds{\pars{~\mbox{line}\ 4~}}$:
  $$
I=\alpha\int_{0}^{\infty}{1 \over x^{2}}
\exp\pars{-\,{1 + x^{4} \over 2\alpha x^{2}}}\,\dd x
$$

With $\ds{\expo{\theta} = x}$:
\begin{align}
I&=\alpha\int_{-\infty}^{\infty}\expo{-2\theta}
\exp\pars{-\,{\cosh\pars{2\theta} \over \alpha}}\,\expo{\theta}\,\dd\theta
=2\alpha\int_{0}^{\infty}\cosh\pars{\theta}
\exp\pars{-\,{\cosh\pars{2\theta} \over \alpha}}\,\dd\theta
\end{align}

Since
  $\ds{\cosh\pars{2\theta} = \cosh^{2}\pars{\theta} + \sinh^{2}\pars{\theta}
= 2\sinh^{2}\pars{\theta} + 1}$ and
  $\ds{\totald{\sinh\pars{\theta}}{\theta} = \cos\pars{\theta}}$ we'll have:
  \begin{align}
\color{#44f}{\large I}&=
2\alpha\expo{-1/\alpha}\
\overbrace{\int_{0}^{\infty}\cosh\pars{\theta}
\exp\pars{-\,{2\sinh^{2}\pars{\theta} \over \alpha}}\,\dd\theta}
^{\ds{\mbox{Lets}\ u\ \equiv\ \sinh\pars{\theta}}}
={2\alpha \over \root[\alpha]{\expo{}}}\int_{0}^{\infty}\expo{-2u^{2}/\alpha}
\,\dd u
\\[3mm]&={2\alpha \over \root[\alpha]{\expo{}}}\,\root{\alpha \over 2}\
\underbrace{\int_{0}^{\infty}\expo{-u^{2}}\,\dd u}_{\ds{=\ {\root{\pi} \over 2}}}
=\color{#44f}{\large{\root{2\alpha^{3}\pi} \over 2\root[\alpha]{\expo{}}}}
\end{align}