real analysis - Questions on the integration $int_0^infty e^{-(x^2+a^2/x^2)}dx$

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.

Answer

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