# Integral$\int_0^\infty \ln x\,\exp(-\frac{1+x^4}{2\alpha x^2}) \frac{x^4+\alpha x^2- 1}{x^4}dx$?

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.

• Is $a=\alpha {}$? Apr 30, 2014 at 14:11
• Okay, how was this problem constructed? Perhaps this could give us insight into how to solve the problem. I doubt the right approach is contour integration despite the $\sqrt{\pi}$ and I think it will involve the $\Gamma$ function? Apr 30, 2014 at 14:20
• @RonGordon Thank you I fixed it. $\alpha$ Apr 30, 2014 at 14:29
• @ChrisK Yes I was thinking contour integration but wasn't sure of how to go about that, it seems tough that way. Thanks!. What do you mean how was this constructed? This is a definite integral that I saw years ago and I still have. It is not from another problem, or any kind of engineering/physics application. It is an integral that came from a list of many other integrals. Hope that helps! Apr 30, 2014 at 14:30
• I see no one has answered yet. Maybe I should get out some paper and work on it? May 1, 2014 at 23:06

\begin{align*} 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\\ &=\int_0^\infty \ln x\, d\left(-\alpha x^{-1}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right)\\ &=-\alpha\left(\left.\frac{\ln x}{x}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right|_0^\infty-\int_0^\infty \frac{1}{x}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) d\,\ln x\right)\\ &=\alpha\int_0^\infty \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) dx\\ &=\alpha\left(\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\underbrace{\int_1^\infty \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx}_{x\to1/x}\right) \\ &=\alpha\left(\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\int_1^0 -\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\right) \\ &=\alpha\int_0^1 (1+\frac{1}{x^2})\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\\ &=\alpha\int_0^1 \exp\left(-\frac{1}{\alpha}-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\ &=\alpha e^{-1/\alpha}\int_0^1 \exp\left(-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\ &=\alpha e^{-1/\alpha}\int_{-\infty}^0 \exp\left(-\frac{y^2}{2\alpha }\right)dy\\ &=\alpha e^{-1/\alpha}\sqrt{\frac{\alpha\pi}{2}}. \end{align*}
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{I\equiv\int_{0}^{\infty}\ln\pars{x} \exp\pars{-\,{1 + x^{4} \over 2\alpha x^{2}}}\, {x^4+\alpha x^2- 1 \over x^4}\,\dd x= {\root{2\alpha^{3}\pi} \over 2\root[\alpha]{\expo{}}}:\ {\large ?}, \qquad \alpha > 0}$.
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}