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

veksetz

veksetz

Answered question

2022-01-02

Double Integrals involving infinity
00xye(x2+y2)dxdy

Answer & Explanation

hysgubwyri3

hysgubwyri3

Beginner2022-01-03Added 43 answers

I will be explicit about something that I think people should more often be explicit about.
00xye(x2+y2)dxdy=0(0(xex2)(yey2)dx)dy
Look at the inside integral:
0(xex2)(yey2)dx
As x goes from 0 to in this integral, yey2 does not change. It is for that reason that it can be pulled out, getting
yey20xex2dx
Now we have
0(yey20xex2dx)dy
0yey2dy0xex2dx
The above is what should more often be made explicit, maybe in the form of an assigned exercise.
Then after that, you can write
0ex2(2xdx)12
and 2xdx becomes du, etc.
intacte87

intacte87

Beginner2022-01-04Added 42 answers

For every a,b>0 we have
0a0bxye(x+y2)dxdy=(0axex2dx)(0byey2dy)
=[12ex2]0a[12ey2]0b
=(1ea2)(1eb2)4
If you take the limit as a,b you get:
00xye(x+y2)dxdy=lima,b0a0bxye(x+y2)dxdy
=lima,b(1ea2)(1eb2)4
=14
karton

karton

Expert2022-01-09Added 613 answers

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:
00xye(x+y2)dxdy=(0xex2dx)(0yey2dy)=14

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