veksetz

2022-01-02

Double Integrals involving infinity
${\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}xy{e}^{-\left({x}^{2}+{y}^{2}\right)}dxdy$

hysgubwyri3

I will be explicit about something that I think people should more often be explicit about.
${\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}xy{e}^{-\left({x}^{2}+{y}^{2}\right)}dxdy={\int }_{0}^{\mathrm{\infty }}\left({\int }_{0}^{\mathrm{\infty }}\left(x{e}^{-{x}^{2}}\right)\left(y{e}^{-{y}^{2}}\right)dx\right)dy$
Look at the inside integral:
${\int }_{0}^{\mathrm{\infty }}\left(x{e}^{-{x}^{2}}\right)\left(y{e}^{-{y}^{2}}\right)dx$
As x goes from 0 to $\mathrm{\infty }$ in this integral, $y{e}^{-{y}^{2}}$ does not change. It is for that reason that it can be pulled out, getting
$y{e}^{-{y}^{2}}{\int }_{0}^{\mathrm{\infty }}x{e}^{-{x}^{2}}dx$
Now we have
${\int }_{0}^{\mathrm{\infty }}\left(y{e}^{-{y}^{2}}{\int }_{0}^{\mathrm{\infty }}x{e}^{-{x}^{2}}dx\right)dy$
${\int }_{0}^{\mathrm{\infty }}y{e}^{-{y}^{2}}dy\cdot {\int }_{0}^{\mathrm{\infty }}x{e}^{-{x}^{2}}dx$
The above is what should more often be made explicit, maybe in the form of an assigned exercise.
Then after that, you can write
${\int }_{0}^{\mathrm{\infty }}{e}^{-{x}^{2}}\left(2xdx\right)\cdot \frac{1}{2}$
and 2xdx becomes du, etc.

intacte87

For every $a,b>0$ we have
${\int }_{0}^{a}{\int }_{0}^{b}xy{e}^{-\left({x}^{+}{y}^{2}\right)}dxdy=\left({\int }_{0}^{a}x{e}^{-{x}^{2}}dx\right)\left({\int }_{0}^{b}y{e}^{-{y}^{2}}dy\right)$
$={\left[-\frac{1}{2}{e}^{-{x}^{2}}\right]}_{0}^{a}{\left[-\frac{1}{2}{e}^{-{y}^{2}}\right]}_{0}^{b}$
$=\frac{\left(1-{e}^{-{a}^{2}}\right)\left(1-{e}^{-{b}^{2}}\right)}{4}$
If you take the limit as $a,b\to \mathrm{\infty }$ you get:
${\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}xy{e}^{-\left({x}^{+}{y}^{2}\right)}dxdy=\underset{a,b\to \mathrm{\infty }}{lim}{\int }_{0}^{a}{\int }_{0}^{b}xy{e}^{-\left({x}^{+}{y}^{2}\right)}dxdy$
$=\underset{a,b\to \mathrm{\infty }}{lim}\frac{\left(1-{e}^{-{a}^{2}}\right)\left(1-{e}^{-{b}^{2}}\right)}{4}$
$=\frac{14}{}$

karton

The integrand is of the form $f\left(x\right)g\left(y\right)$ so the integral may be written as the product of two ordinary improper integrals and evaluated as follows:
${\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}xy{e}^{-\left({x}^{+}{y}^{2}\right)}dxdy=\left({\int }_{0}^{\mathrm{\infty }}x{e}^{-{x}^{2}}dx\right)\left({\int }_{0}^{\mathrm{\infty }}y{e}^{-{y}^{2}}dy\right)=\frac{1}{4}$

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