calculus - Evaluating $dfrac{1}{2 pi} int_{-infty}^infty int_{-infty}^infty e^{tuv} e^{-u^2/2} e^{-v^2/2} du dv$

Solving a probability problem I came across this integral:

$$\dfrac{1}{2 \pi} \int_{-\infty}^\infty \int_{-\infty}^\infty e^{tuv} e^{-u^2/2} e^{-v^2/2}\ du\ dv$$

Can you explain how to integrate this?

Answer

Hint. Assume $-1gaussian result,

$$\int_{-\infty}^\infty e^{tuv} e^{-u^2/2} \ du=\sqrt{2\pi} \:e^{t^2v^2/2}$$ then with respect to $v$,
$$\int_{-\infty}^\infty e^{t^2v^2/2} e^{-v^2/2} \ dv=\int_{-\infty}^\infty e^{-(1-t^2)v^2/2} \ dv=\frac{\sqrt{2\pi}}{\sqrt{1-t^2}}$$ obtaining

$$\dfrac{1}{2 \pi} \int_{-\infty}^\infty \int_{-\infty}^\infty e^{tuv} e^{-u^2/2} e^{-v^2/2}\ du\ dv=\frac1{\sqrt{1-t^2}}.$$