# Thread: [SOLVED] Improper double integral

1. ## [SOLVED] Improper double integral

D is the positive quadrant.

$\displaystyle \int \int_D \frac{xydxdy}{(1+x^2+y^2)^3}$

Switching to polar coordinates doesn't help much.

2. Hello,
Originally Posted by Spec
D is the positive quadrant.

$\displaystyle \int \int_D \frac{xydxdy}{(1+x^2+y^2)^3}$

Switching to polar coordinates doesn't help much.
Using polar coordinates:

\displaystyle \begin{aligned}\iint_D \frac{xy}{(1+x^2+y^2)^3}\,\mathrm{d}x\,\mathrm{d}y &=\int_0^{\frac{\pi}{2}}\int_0^{\infty} \frac{\rho \cos \theta \rho \sin \theta}{(1+\rho^2)^3}\,\rho\,\mathrm{d}\rho\,\math rm{d}\theta\\ &=\int_0^{\frac{\pi}{2}}\cos \theta \sin\theta \,\mathrm{d}\theta \int_0^{\infty}\rho^2 \frac{\rho}{(1+\rho^2)^3}\,\mathrm{d}\rho \end{aligned}

To compute $\displaystyle \int_0^{\infty}\rho^2 \frac{\rho}{(1+\rho^2)^3}\,\mathrm{d}\rho$ you can try integration by parts (let $\displaystyle u'=\frac{\rho}{(1+\rho^2)^3}$ and $\displaystyle v=\rho ^2$).

3. Originally Posted by flyingsquirrel

To compute $\displaystyle \int_0^{\infty}\rho^2 \frac{\rho}{(1+\rho^2)^3}\,\mathrm{d}\rho$ you can try integration by parts (let $\displaystyle u'=\frac{\rho}{(1+\rho^2)^3}$ and $\displaystyle v=\rho ^2$).
Just let $\displaystyle z=1+\rho^2,$ no need to apply partial integration.

4. It was easy when you integrated in polar coordinates using $\displaystyle z = 1 + r^2$

I was just put off by the looks of that integral, so I didn't really attempt to solve it initially.