How to calculate $ \int_0^\infty \exp(-\frac{a^2}{x^2}-x^2)~\mathrm{d}x $

Question

suppose $a>0$, how to integrate:

$$ \int_0^\infty e^{-a^2/x^2}e^{-x^2}~\mathrm{d}x $$

Answer 1

In general,

\begin{align} \int_{-\infty}^{\infty}f\left(x-\frac{1}{x}\right) dx &= \int_{-\infty}^{0}f\left(x-\frac{1}{x}\right) dx+ \int_{0}^{\infty}f\left(x-\frac{1}{x}\right) dx\\ & \overset{t=-\frac1x} = \int^{\infty}_{0}f\left(t-\frac{1}{t}\right) \frac {dt}{t^2}+ \int^{0}_{-\infty}f\left(t-\frac{1}{t}\right) \frac {dt}{t^2}\\ &= \frac12\int_{-\infty}^{0}f\left(t-\frac{1}{t}\right)\left( 1+\frac1{t^2}\right)dt + \frac12\int_{0}^{\infty}f\left(t-\frac{1}{t}\right)\left( 1+\frac1{t^2}\right)dt\\ & \overset{x=t-\frac1t } =\int_{-\infty}^{\infty}f(x)dx \end{align}

Thus,

$$ \int_0^\infty e^{-a^2/x^2}e^{-x^2}dx \overset{ t=\frac x{\sqrt a}}=\frac12\sqrt{a}e^{-2a }\int_{-\infty}^\infty e^{-a(t-\frac1t)^2 }dt\\ = \frac12\sqrt{a}e^{-2a }\int_{-\infty}^\infty e^{-a t^2 }dt = \frac12\sqrt{a}e^{-2a } \cdot \sqrt{\frac\pi a}= \frac12\sqrt{\pi}e^{-2a } $$

Answer 2

We can rewrite the integrand as

$$I = \frac{e^{-2a}}{2}\int_{-\infty}^\infty e^{-\left(x - \frac{a}{x}\right)^2}\:dx$$

because the integrand is even and we can complete the square. Then we can use Glasser's master theorem to assert that

$$I = \frac{e^{-2a}}{2}\int_{-\infty}^\infty e^{-x^2}\:dx = \frac{e^{-2a}\sqrt{\pi}}{2}$$