A difficult Integral Question How to calculate $$\int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}}\exp\left(-\frac{t^2}{2x^2}-\frac{x^2}{2}\right)\,dx$$
Hi, this is a result of a characteristic function problem calculating the distribution of $X_1/X_2$ given that $X_1, X_2$ are independent $N(0,1)$ variables.
I have no idea how to proceed with the calculations here. My first thought was to create a derivative of $(-\frac{t^2}{2x^2}-\frac{x^2}{2})$ inside the integral but it obviously did not work. I would appreciate it if someone could give me a hint. Thanks in advance!
BTW, the final answer should be $\exp(-|t|)$
 A: This would be a good use of Glasser Master Theorem
$$\int_{-\infty}^\infty f\left(x-\frac{a}{x}\right)\:dx = \int_{-\infty}^\infty f(x)\:dx$$
for $a>0$. Here, we can complete the square in the exponential to get
$$\frac{t^2}{2x^2}+\frac{x^2}{2} = \frac{1}{2}\left(x-\frac{|t|}{x}\right)^2+|t|$$
which means
$$\frac{\exp(-|t|)}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac{1}{2}\left(x-\frac{|t|}{x}\right)^2\right)\:dx = \frac{\exp(-|t|)}{\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left(-\frac{x^2}{2}\right)\:dx $$
$$= \frac{\exp(-|t|)}{\sqrt{2\pi}}\sqrt{2\pi} = \boxed{\exp(-|t|)}$$
A: For now, and let's assume that $t\geq0$.
$$I(t) = \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\Big(-\frac{x^2}2-\frac{t^2}{2x^2}\Big)\,\mathrm dx $$
Noting that the integrand is even,
$$I(t) = \frac2{\sqrt{2\pi}}\int_{0}^\infty \exp\Big(-\frac{x^2}2-\frac{t^2}{2x^2}\Big)\,\mathrm dx $$
Substitute $x\mapsto \frac tx$.
$$I(t) = \frac2{\sqrt{2\pi}} \int_0^\infty  \exp\Big(-\frac{x^2}2-\frac{t^2}{2x^2}\Big)\frac t{x^2}\,\mathrm dx= -I'(t) $$
The solution for this differential equation is $I(t) =ce^{-t}$.
Noting that $I(0)=1$, we conclude that $I(t) =e^{-t}$. Also, noting that $I(t)=I(-t)$, we conclude that $\forall t\in\mathbb R$,
$$\boxed{\boxed{  \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\Big(-\frac{x^2}2-\frac{t^2}{2x^2}\Big)\,\mathrm dx =e^{-|t|}}}$$
