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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?

Olivier Oloa
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user2506893
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  • Try converting to polar coordinates: $u = r\cos\theta$, $v = r\sin\theta$; then integrate with respect to $r$. This leaves you a tractable integral with respect to $\theta$. – rogerl May 22 '16 at 16:05

2 Answers2

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Hint. Assume $-1<t<1$. One may just integrate with respect to $u$, using the classic gaussian 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

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

Community
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Olivier Oloa
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Lemma. Let $q(x,y)=Ax^2+2Bxy+Cy^2$ a positive definite quadratic form, associated with the symmetric matrix $M=\begin{pmatrix}A & B \\ B & C \end{pmatrix}$. We have: $$ \iint_{\mathbb{R}^2}\exp\left(-q(x,y)\right)\,dx dy = \frac{\pi}{\sqrt{\det M}}.$$

In our case, the quadratic form has coefficients $A=C=\frac{1}{2}$ and $B=-\frac{t}{2}$, hence:

$$ \frac{1}{2\pi}\iint_{\mathbb{R}^2}\exp\left(-\frac{u^2}{2}-\frac{v^2}{2}+tuv\right)\,du\,dv = \color{red}{\frac{1}{\sqrt{1-t^2}}}$$

as soon as $|t|<1$.

Jack D'Aurizio
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