Double Integrals involving infinity Double Integrals involving infinity 
$$
\int_0^\infty\int_0^\infty xye^{-(x^2+y^2)}\,dx\,dy
$$
my work 
let $$t = x^2 +y^2$$
then $$ dt =2x\,dx$$
since $$ye^{-t}$$
$$[ye^{-t}]$$ from $\infty$ to $0$ 
what next stuck here 
 A: For every $a,b>0$ we have
$$
\int_0^a\int_0^bxye^{-(x^2+y^2)}\,dxdy=\left(\int_0^axe^{-x^2}\,dx\right)\left(\int_0^bye^{-y^2}\,dy\right)=\left[-\frac12e^{-x^2}\right]_0^a\left[-\frac12e^{-y^2}\right]_0^b=\frac{(1-e^{-a^2})(1-e^{-b^2})}{4}.
$$
If you take the limit as $a,b\to\infty$ you get:
$$
\int_0^\infty\int_0^\infty xye^{-(x^2+y^2)}\,dxdy=\lim_{a,b\to\infty}\int_0^a\int_0^bxye^{-(x^2+y^2)}\,dxdy=\lim_{a,b\to\infty}\frac{(1-e^{-a^2})(1-e^{-b^2})}{4}=\frac14.
$$
A: The integrand is of the form $f(x)g(y)$ so the integral may be written as the product of two ordinary improper integrals and evaluated as follows: 
$$\int_0^\infty \int_0^\infty xye^{-(x^2+y^2)} \, dx\,dy=\left ( \int_0^\infty xe^{-x^2} \, dx \right ) \left ( \int_0^\infty ye^{-y^2} \, dy \right )=\frac{1}{4}$$
A: I will be explicit about something that I think people should more often be explicit about.
$$
\int_0^\infty\int_0^\infty xye^{-(x^2+y^2)}\,dx\,dy = \int_0^\infty\left(\int_0^\infty (xe^{-x^2})(ye^{-y^2})\,dx \right) \,dy
$$
Look at the inside integral:
$$
\int_0^\infty (xe^{-x^2})(ye^{-y^2})\,dx
$$
As $x$ goes from $0$ to $\infty$ in this integral, $ye^{-y^2}$ does not change.  It is for that reason that it can be pulled out, getting
$$
ye^{-y^2}\int_0^\infty xe^{-x^2}\,dx.
$$
Now we have
$$
\int_0^\infty \left( ye^{-y^2} \underbrace{\int_0^\infty xe^{-x^2}\,dx} \right)\,dy.
$$
And now, as $y$ goes from $0$ to $\infty$, the expression over the $\underbrace{\text{underbrace}}$ above does not change.  For that reason, it can also be pulled out, getting
$$
\int_0^\infty ye^{-y^2} \,dy \cdot \int_0^\infty xe^{-x^2}\,dx.
$$
The above is what should more often be made explicit, maybe in the form of an assigned exercise.
Then after that, you can write
$$
\int_0^\infty e^{-x^2}\Big(2x\,dx\Big)\cdot\frac 1 2
$$
and $2x\,dx$ becomes $du$, etc.
