Gaussian-like Integral (show equation is true) Show the following integration is true.
$$\int_0^\infty e^{-x^{2}-\frac{\alpha^2}{x^2}}dx = \frac{\sqrt{\pi}}{2}e^{-2\alpha}$$
I know this is eerily similar to the half-space Gaussian Integral:
$$\int_0^\infty e^{-\alpha x^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{\alpha}}$$
But I am unsure where to begin.  Any hints or steps would be much appreciated.
 A: Note that $x^2 + a^2/x^2 = (x-a/x)^2 + 2a$. Hence, we have $$\exp\left(-x^2 - \dfrac{a^2}{x^2}\right) = \exp\left(-\left(x- \dfrac{a}x\right)^2-2a\right)$$
Hence, it suffices to show
$$\int_0^{\infty} \exp\left(-\left(x- \dfrac{a}x\right)^2\right)dx = \dfrac{\sqrt{\pi}}2$$
WLOG, we can assume $a\geq 0$ and let $$I(a) = \int_0^{\infty} \exp\left(-\left(x- \dfrac{a}x\right)^2\right)dx \tag{$\star$}$$
Letting $y = \dfrac{a}x$, we get
$$I(a) = \int_0^{\infty} \exp\left(-\left(y- \dfrac{a}y\right)^2\right)\dfrac{a}{y^2}dy \tag{$\dagger$}$$
We then have
$$I'(a) = \int_0^{\infty} \exp\left(-\left(x- \dfrac{a}x\right)^2\right)\left( -2\left(x-\dfrac{a}x\right)\left(-\dfrac1x\right)\right)dx = \int_0^{\infty} \exp\left(-\left(x- \dfrac{a}x\right)^2\right)\left( 2\left(1-\dfrac{a}{x^2}\right)\right)dx = 0$$ using $\star$ and $\dagger$. Hence, $I(a) = I(0) = \dfrac{\sqrt{\pi}}2$.
A: At $x=\sigma^{2/3}$ the integrand switches from increasing to decreasing.  So split the integral into two parts and carry out the substitution $t=(x^4+\sigma^2)/x^2$ on both parts separately.  You then get two integrals which have $e^{-t}$ times some ratio of polynomials of $t$.  Since this is homework I'll leave it to you to fill in the details.
