1
$\begingroup$

I am interested in the following integral $$\int_{-\infty}^{\infty}\mathrm{d}v_1\int_{-\infty}^{\infty}\mathop{\mathrm{d}v_2}v_1v_2\exp\left(-\frac{(v_1-v_2)^2}{2a}\right).$$ Does any one have any idea? My idea was to use the transformation $$v=v_1-v_2,\qquad v_+=\frac{v_1+v_2}{2}.$$ But I am stuck.

$\endgroup$

1 Answer 1

2
$\begingroup$

Your integral diverges. $$ \int_{-\infty}^{\infty}\mathop{\mathrm{d}v_2}v_2\exp\left(-\frac{(v_1-v_2)^2}{2a}\right)=\sqrt{2 \pi a}\nu_1\ . $$ Hence $$ \int_{-\infty}^\infty d\nu_1 \nu_1 (\sqrt{2 \pi a}\nu_1) = \infty\ . $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .