veksetz

2022-01-02

Double Integrals involving infinity

${\int}_{0}^{\mathrm{\infty}}{\int}_{0}^{\mathrm{\infty}}xy{e}^{-({x}^{2}+{y}^{2})}dxdy$

hysgubwyri3

Beginner2022-01-03Added 43 answers

I will be explicit about something that I think people should more often be explicit about.

${\int}_{0}^{\mathrm{\infty}}{\int}_{0}^{\mathrm{\infty}}xy{e}^{-({x}^{2}+{y}^{2})}dxdy={\int}_{0}^{\mathrm{\infty}}\left({\int}_{0}^{\mathrm{\infty}}\left(x{e}^{-{x}^{2}}\right)\left(y{e}^{-{y}^{2}}\right)dx\right)dy$

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.

Look at the inside integral:

As x goes from 0 to

Now we have

The above is what should more often be made explicit, maybe in the form of an assigned exercise.

Then after that, you can write

and 2xdx becomes du, etc.

intacte87

Beginner2022-01-04Added 42 answers

For every $a,b>0$ we have

${\int}_{0}^{a}{\int}_{0}^{b}xy{e}^{-\left({x}^{+}{y}^{2}\right)}dxdy=\left({\int}_{0}^{a}x{e}^{-{x}^{2}}dx\right)\left({\int}_{0}^{b}y{e}^{-{y}^{2}}dy\right)$

$={[-\frac{1}{2}{e}^{-{x}^{2}}]}_{0}^{a}{[-\frac{1}{2}{e}^{-{y}^{2}}]}_{0}^{b}$

$=\frac{(1-{e}^{-{a}^{2}})(1-{e}^{-{b}^{2}})}{4}$

If you take the limit as$a,b\to \mathrm{\infty}$ you get:

${\int}_{0}^{\mathrm{\infty}}{\int}_{0}^{\mathrm{\infty}}xy{e}^{-\left({x}^{+}{y}^{2}\right)}dxdy=\underset{a,b\to \mathrm{\infty}}{lim}{\int}_{0}^{a}{\int}_{0}^{b}xy{e}^{-\left({x}^{+}{y}^{2}\right)}dxdy$

$=\underset{a,b\to \mathrm{\infty}}{lim}\frac{(1-{e}^{-{a}^{2}})(1-{e}^{-{b}^{2}})}{4}$

$=\frac{14}{}$

If you take the limit as

karton

Expert2022-01-09Added 613 answers

The integrand is of the form

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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