2
$\begingroup$

Let $F(a):=\int_0^\infty e^{-(x^2+a^2/x^2)}dx$ with $a>0$. My questions are as follows:

(1) Calculate $\lim_{a \to 0^+}F(a)$.

(2) Show that $F'(a)=-2F(a)$.

(3) Calculate $F(a)$.

It seems to me that the well-known Gaussian integral $\int_0^\infty e^{-x^2}dx=\sqrt \pi/2$ plays some role here. However, if I set $t=x+a/x$ then $dt=(1-a/x^2)dx$. I'm stuck in finding a function $f$ so that $1-a/x^2=f(t).$ For part (2), I don't quite understand how to find the derivative of $F(a)$. It seems that we need to find the closed-form of $F(a)$ first. Assume that I am given part (2) to solve part (3). Is it true that $F(a)=e^{-2a+\ln(\sqrt \pi/2)}$?

Any help would be much appreciated.

$\endgroup$
2
2
$\begingroup$

Consider \begin{align} x^{2} + \frac{a^2}{x^{2}} = \left( x - \frac{a}{x} \right)^{2} +2a \end{align} for which \begin{align} I = \int_{0}^{\infty} e^{-\left(x^{2} + \frac{a^2}{x^{2}}\right)} \, dx = e^{-2a} \, \int_{0}^{\infty} e^{-\left(x - \frac{a}{x}\right)^{2}} \, dx. \end{align} Now make the substitution $t=\frac{a}{x}$ to obtain \begin{align} e^{2a} I = a\int_{0}^{\infty} e^{- \left( t - \frac{a}{t} \right)^{2}} \, \frac{dt}{t^{2}}. \end{align} Adding the two integral form leads to \begin{align} 2 e^{2a} I = \int_{0}^{\infty} e^{- \left( t - \frac{a}{t} \right)^{2}} \left(1 + \frac{a}{t^{2}} \right) \, dt = \int_{-\infty}^{\infty} e^{- u^{2}} \, du = 2 \int_{0}^{\infty} e^{- u^{2}} \, du = \sqrt{\pi}, \end{align} where the substitution $u = t - \frac{a}{t}$ was made. It is now seen that \begin{align} \int_{0}^{\infty} e^{-\left(x^{2} + \frac{a^2}{x^{2}}\right)} \, dx = \frac{\sqrt{\pi}}{2}e^{-2a}. \end{align}

Note: Above is a modified version of Evaluate $\int_{0}^{\infty} \mathrm{e}^{-x^2-x^{-2}}\, dx$ 's accepted answer.

Answers :

(1) $\lim_{a \to 0^+}F(a) = \lim_{a \to 0^+}\frac{\sqrt{\pi}}{2}e^{-2a} = \frac{\sqrt{\pi}}{2}.$

(2) $F'(a)=\frac{\sqrt{\pi}}{2}e^{-2a}(-2)=-2F(a).$

(3) $F(a)=\frac{\sqrt{\pi}}{2}e^{-2a}$.

$\endgroup$
5
$\begingroup$

We have $$ I(a)= e^{-2a}\int_{0}^{+\infty}\exp\left[-\left(x-\frac{a}{x}\right)^2\right]\,dx $$ hence by Glasser's master theorem $$ I(a) = e^{-2a}\int_{0}^{+\infty}\exp(-x^2)\,dx = \color{red}{\frac{\sqrt{\pi}}{2}e^{-2a}} $$ and everything follow nicely.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.