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

Solution 1:

$$\begin{align*} I&=\int_0^\infty \ln x\,\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) \frac{x^4+\alpha x^2- 1}{x^4}dx\\ &=\int_0^\infty \ln x\, d\left(-\alpha x^{-1}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right)\\ &=-\alpha\left(\left.\frac{\ln x}{x}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)\right|_0^\infty-\int_0^\infty \frac{1}{x}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) d\,\ln x\right)\\ &=\alpha\int_0^\infty \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right) dx\\ &=\alpha\left(\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\underbrace{\int_1^\infty \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx}_{x\to1/x}\right) \\ &=\alpha\left(\int_0^1 \frac{1}{x^2}\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx+\int_1^0 -\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\right) \\ &=\alpha\int_0^1 (1+\frac{1}{x^2})\exp\left(-\frac{1+x^4}{2\alpha x^2}\right)dx\\ &=\alpha\int_0^1 \exp\left(-\frac{1}{\alpha}-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\ &=\alpha e^{-1/\alpha}\int_0^1 \exp\left(-\frac{(x-1/x)^2}{2\alpha }\right)d(x-1/x)\\ &=\alpha e^{-1/\alpha}\int_{-\infty}^0 \exp\left(-\frac{y^2}{2\alpha }\right)dy\\ &=\alpha e^{-1/\alpha}\sqrt{\frac{\alpha\pi}{2}}. \end{align*}$$

Solution 2:

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

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}