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
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Two answers posted below explain that you can factor the double integral in a certain way. I decided to post an answer explaining why you can do that, rather than merely that you can do that. That is a point that I think deserves more attention than it usually gets. – Michael Hardy Jun 08 '15 at 01:48
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If you see $x^2+y^2$, use polar coordinates – Michael Galuza Jun 08 '15 at 05:23
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The obvious next thing is to write down the integral you get from substituting; what did you get for it? Or where did you have trouble writing it down? (also, what you wrote after the word "since" doesn't make any sense) – Jun 08 '15 at 05:30
3 Answers
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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.
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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: $$ \int_0^\infty\int_0^\infty xye^{-(x^2+y^2)}\,dxdy=\lim_{a,b\to\infty}\int_0^a\int_0^bxye^{-(x^2+y^2)}\,dxdy=\lim_{a,b\to\infty}\frac{(1-e^{-a^2})(1-e^{-b^2})}{4}=\frac14. $$
HorizonsMaths
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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:
$$\int_0^\infty \int_0^\infty xye^{-(x^2+y^2)} \, dx\,dy=\left ( \int_0^\infty xe^{-x^2} \, dx \right ) \left ( \int_0^\infty ye^{-y^2} \, dy \right )=\frac{1}{4}$$
Matematleta
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