Double Integrals involving infinity ∫0∞∫0∞xye−(x2+y2)dxdy

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Double Integrals involving infinity  ∫0∞∫0∞xye−(x2+y2)dxdy

hysgubwyri3 · Accepted Answer

I will be explicit about something that I think people should more often be explicit about.  ∫0∞∫0∞xye−(x2+y2)dxdy=∫0∞(∫0∞(xe−x2)(ye−y2)dx)dy  Look at the inside integral:  ∫0∞(xe−x2)(ye−y2)dx  As x goes from 0 to ∞ in this integral, ye−y2 does not change. It is for that reason that it can be pulled out, getting  ye−y2∫0∞xe−x2dx  Now we have  ∫0∞(ye−y2∫0∞xe−x2dx)dy  ∫0∞ye−y2dy⋅∫0∞xe−x2dx  The above is what should more often be made explicit, maybe in the form of an assigned exercise.  Then after that, you can write  ∫0∞e−x2(2xdx)⋅12  and 2xdx becomes du, etc.

intacte87 · Accepted Answer

For every a,b&amp;gt;0 we have  ∫0a∫0bxye−(x+y2)dxdy=(∫0axe−x2dx)(∫0bye−y2dy)  =[−12e−x2]0a[−12e−y2]0b  =(1−e−a2)(1−e−b2)4  If you take the limit as a,b→∞ you get:  ∫0∞∫0∞xye−(x+y2)dxdy=lima,b→∞∫0a∫0bxye−(x+y2)dxdy  =lima,b→∞(1−e−a2)(1−e−b2)4  =14

karton · Accepted Answer

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: ∫0∞∫0∞xye−(x+y2)dxdy=(∫0∞xe−x2dx)(∫0∞ye−y2dy)=14

Double Integrals involving infinity \int_0^\infty\int_0^{\infty}xye^{-(x^2+y^2)}dxdy

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