Computing the integral $\int_0^\infty e^{ -t- \frac{1}{t}} \frac{dt}{\sqrt{t}}$

Question

I am trying to Compute the integral,

$$I=\int_0^\infty e^{ -t- \frac{1}{t}} \frac{dt}{\sqrt{t}}$$

My attempt. Enforcing the change of variables $x= \sqrt{t}$ it becomes

$$I=2\int_0^\infty e^{-t^2- \frac{1}{t^2}} dt\\ =2e^{±2}\int_0^\infty e^{ -\left(t\pm\frac{1}{t}\right)^2} dt$$

How do I move from here?

Answer 1

You're on the right track. We have

$$\begin{align} \int_0^\infty \frac{e^{-(t+1/t)}}{\sqrt t}\,dt&\overbrace{=}^{t\mapsto t^2}\\\\ &=2\int_0^\infty e^{-(t^2+1/t^2)}\,dt\\\\ &=2e^{-2}\int_{0}^\infty e^{-(t-1/t)^2}\,dt\\\\ &\overbrace{=}^{t\mapsto 1/t}2e^{-2}\int_0^\infty e^{-(t-1/t)^2}\frac1{t^2}\,dt\\\\ &=2e^{-2}\int_0^\infty e^{-(t-1/t)^2} \left(1+\frac1{t^2}\right)\,dt\\\\ &\overbrace{=}^{t-1/t\mapsto t}2e^{-2}\int_0^\infty e^{-t^2}\,dt\\\\ &= e^{-2}\sqrt \pi \end{align}$$

Answer 2

There is a formula which says that for $f\in L^{1}(-\infty,\infty)$, \begin{align*} \int_{-\infty}^{\infty}f\left(x-\dfrac{1}{x}\right)dx=\int_{-\infty}^{\infty}f(u)du. \end{align*} This can be done by substituting $u=x-1/x$. So the integral in question is somehow integrating with $e^{-(\cdot)^{2}}$ along the line.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & \equiv \int_{0}^{\infty}\expo{-t - 1/t}{\dd t \over \root{t}} \,\,\,\stackrel{t\ =\ \exp\pars{\theta}}{=}\,\,\, \int_{-\infty}^{\infty}\expo{-2\cosh\pars{\theta}}\expo{\theta/2}\dd\theta = 2\int_{0}^{\infty}\expo{-2\cosh\pars{\theta}}\cosh\pars{\theta \over 2} \,\dd\theta \\[5mm] & = 2\int_{0}^{\infty}\expo{-4\sinh^{2}\pars{\theta/2} - 2} \cosh\pars{\theta \over 2}\,\dd\theta = 4\expo{-2}\ \underbrace{\int_{\theta\ =\ 0}^{\theta\ \to\ \infty}\expo{-4\sinh^{2}\pars{\theta/2}}\,\dd\sinh\pars{\theta \over 2}} _{\ds{\root{\pi} \over 4}} \\ & = \bbx{\root{\pi} \over \expo{2}} \approx 0.2399 \end{align}