Cómo evaluar $\int_{0}^{+\infty}\exp(-ax^2-\frac b{x^2})\,dx$ para $a,b>0$

Question

¿Cómo puedo evaluar $$I=\int_{0}^{+\infty}\!e^{-ax^2-\frac b{x^2}}\,dx$$ para $a,b>0$ ?

Mis métodos:

Dejemos que $a,b > 0$ y que $$I(b)=\int_{0}^{+\infty}e^{-ax^2-\frac b{x^2}}\,dx.$$ Entonces $$I'(b)=\int_{0}^{\infty}-\frac{1}{x^2}e^{-ax^2-\frac b{x^2}}\,dx.$$

¿Qué otros métodos puedo utilizar para evaluarlo? Gracias.

Answer 1

$$\begin{align} I = & \int_0^{\infty} e^{-ax^2 - bx^{-2}} dx\\ \stackrel{\color{blue}{[1]}}{=} & \left(\frac{b}{a}\right)^{1/4}\int_0^{\infty} e^{-\sqrt{ab}(y^2 + y^{-2})} dy\\ = & \left(\frac{b}{a}\right)^{1/4}\left[ \int_0^{1} + \int_1^{\infty} \right] e^{-\sqrt{ab}(y^2 + y^{-2})} dy\\ \stackrel{\color{blue}{[2]}}{=} & \left(\frac{b}{a}\right)^{1/4} \int_1^{\infty} e^{-\sqrt{ab}(y^2 + y^{-2})} \left(\frac{1}{y^2} + 1\right) dy\\ = & \left(\frac{b}{a}\right)^{1/4} \int_1^{\infty} e^{-\sqrt{ab}((y-y^{-1})^2+2)} d\left( y - \frac{1}{y}\right)\\ \stackrel{\color{blue}{[3]}}{=} & \left(\frac{b}{a}\right)^{1/4} e^{-2\sqrt{ab}} \int_0^{\infty} e^{-\sqrt{ab}\,z^2} dz\\ = & \left(\frac{b}{a}\right)^{1/4} e^{-2\sqrt{ab}} \frac{\sqrt{\pi}}{2(ab)^{1/4}}\\ = & \sqrt{\frac{\pi}{4a}} e^{-2\sqrt{ab}} \end{align} $$ Notas

• $\color{blue}{[1]}$ sustituto $x$ por $y = \sqrt{\frac{a}{b}} x$ .
• $\color{blue}{[2]}$ sustituto $y$ por $\frac{1}{y}$ en el intervalo $[0,1]$ .
• $\color{blue}{[3]}$ sustituto $y$ por $z = y - \frac{1}{y}$ .

Answer 2

Antes de utilizar una diferenciación bajo el signo integral es conveniente hacer el siguiente intercambio de variables: $x=\frac{t}{\sqrt{a}}$

$$I=\frac{1}{\sqrt{a}}\int_{0}^{+\infty}\!e^{(-t^2-\frac{s^2}{t^2})}\,dt;s^2=ab$$ Ahora, considérelo como una función de $s$ y diferenciarlo con respecto a $s$ :

$$\frac{dI}{ds}=\frac{-2}{\sqrt{a}}\int_{0}^{+\infty}\frac{e^{(-t^2-\frac{s^2}{t^2})}}{t^2}sdt=\frac{-2}{\sqrt{a}}\int_{0}^{+\infty}e^{(-t^2-\frac{s^2}{t^2})}dt=-2I$$

Por lo tanto, para obtener una respuesta tenemos que resolver la ecuación diferencial

$$\frac{dI}{ds}=-2I$$ y utilizar el hecho de que

$$I(0)=\frac{1}{\sqrt{a}}\int_{0}^{+\infty}\!e^{-t^2}\,dt=\frac{1}{\sqrt{a}}\frac{\sqrt{\pi}}{2}$$

Answer 3

La integral se puede evaluar de la siguiente manera $$ \begin{align} \int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx&=2\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\ &=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dx\\ &=e^{\large-2\sqrt{ab}}\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx. \end{align} $$ El truco para resolver la última integral es poner $$ I=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx. $$ Dejemos que $t=-\frac{1}{x}\sqrt{\frac{b}{a}}\;\rightarrow\;x=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dx=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$ entonces $$ I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt. $$ Dejemos que $t=x\;\rightarrow\;dt=dx$ entonces $$ I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt. $$ Sumando los dos $I_t$ s rendimientos $$ 2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt. $$ Dejemos que $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ y para $0<t<\infty$ corresponde a $-\infty<s<\infty$ entonces $$ I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}. $$ Así, $$ \begin{align} \int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx&=e^{\large-2\sqrt{ab}}\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx\\ &=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{\large-2\sqrt{ab}}. \end{align} $$

Answer 4

La integral es $$\frac{1}{2}e^{-2ab}\int_{-\infty}^{\infty}e^{-a^2(x-b/ax)^2}dx=\frac{1}{2}e^{-2ab}\int_{-\infty}^{\infty}e^{-a^2x^2}dx=\frac{\sqrt{\pi}}{2a}e^{-2ab}.$$ (Véase M.L. Glasser, A Remarkable Property of Definite Integrals, Math.Comp.Vol 40, p.561 (1983).

Answer 5

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

Vamos a $\ds{x \equiv A\expo{\theta}}$ tal que $\ds{-ax^{2} - {b \over x^{2}} = -aA^{2}\expo{2\theta} - {b \over A^{2}}\, \expo{-2\theta}}$ .
Elegimos $$ A\ \mbox{such that}\quad aA^{2} = {b \over A^{2}}\quad\imp\quad A = \pars{b \over a}^{1/4}. \mbox{Then,}\ -ax^{2} - {b \over x^{2}} = -2\root{ab}\cosh\pars{2\theta} $$

\begin{align} I&=\int_{-\infty}^{\infty}\!\!\expo{-2\root{ab}\cosh\pars{2\theta}}\pars{b \over a}^{1/4} \expo{\theta}\,\dd\theta =2\pars{b \over a}^{1/4}\int_{0}^{\infty}\expo{-2\root{ab}\cosh\pars{2\theta}} \cosh\pars{\theta}\,\dd\theta \\[5mm]&=2\pars{b \over a}^{1/4}\ \overbrace{\int_{0}^{\infty}\expo{-2\root{ab}\bracks{2\sinh^{2}\pars{\theta} + 1}} \cosh\pars{\theta}\,\dd\theta}^{\ds{\mbox{Set}\quad t \equiv \sinh\pars{\theta}}} \\[5mm]&=2\pars{b \over a}^{1/4}\expo{-2\root{ab}} \int_{0}^{\infty}\expo{-4\root{ab}t^{2}}\,\dd t \\[5mm]&=2\pars{b \over a}^{1/4}\expo{-2\root{ab}} \bracks{{1 \over 2\pars{ab}^{1/4}}}\ \overbrace{\int_{0}^{\infty}\expo{-t^{2}}\,\dd t}^{\ds{=\ {\root{\pi} \over 2}}}\ =\ \color{#00f}{\large \half\,\root{\pi \over a}\expo{-2\root{ab}}} \end{align}