I want to integrate the function
$$f(x,y)= \frac{1}{2 \pi (t-s)s} e^{- \frac{(y-x)^2}{2(t-s)}- \frac{x^2}{2s}}$$
on $[0,\infty)^2$ or in other words. I am looking for
$$\int_{0}^{\infty} \int_0^{\infty} f(x,y) dx dy.$$
Does anybody know if it is possible to do this analytically (thereby I mean not with a computer...I think you know what I mean)?
Thank you in advance.
Start with switching the order of integration. Relabeling constants gives: $$ I=\frac{1}{ab\pi}\int_0^{\infty} dx \left( \int_0^{\infty} dy e^{-\frac{(x-y)^2}{2a}}\right)e^{-\frac{x^2}{2b}} $$
The inner intgral yields after a subsitution $x-y=q$ $$ I=\frac{1}{2b\sqrt{a\pi}} \int_0^{\infty} dx\text{Erfc}\left(\frac{-x}{\sqrt{a}}\right)e^{-\frac{x^2}{2b}}=\frac{1}{2\sqrt{\pi a b}} \underbrace{ \int_0^{\infty} dz\text{Erfc}(cz)e^{-z^2}}_{J} $$
with $c=-\sqrt{\frac{b}{a}}$ and the complementary Errorfunction $\text{Erfc}(x)$
Differentiating the last integral with respect to $c$ gives a standard integral:
$$ J'= \frac{-2}{\sqrt{\pi}}\int_0^{\infty} dz z e^{-z^2(1+c^2)}= \frac{-2}{\sqrt{\pi}}\frac{1}{1+c^2} $$
integrating back is also pleasingly easy $$ J=\int J'(c)dc= \frac{-2}{\sqrt{\pi}}\arctan(c)+d $$
now putting everything together
$$ I=-\frac{1}{\pi \sqrt{a b}}\arctan\left(-\sqrt{\frac{b}{a}}\right)+d=\frac{1}{\pi \sqrt{a b}}\arctan\left(\sqrt{\frac{b}{a}}\right)+d $$
to fix $d$ we may observe that our orinal integral should vanish as $a\rightarrow \infty$ which fixes $d=0$ and therefore: