0
$\begingroup$

Compute the following integral:

$$\iint_D \frac { |x-y| } { (x^2 + y^2 + 1)^2 } \, dx dy , $$ where $D = \{ (x, y) \in \mathbb R^2 \mid x \geq 0 ,\, y \geq 0 \}$.

So is there an easy way to solve the integral?

$\endgroup$
  • $\begingroup$ convert the integral in the polar coordinate form $\endgroup$ – Euduardo Feb 11 at 3:04
  • $\begingroup$ Given the symmetry w.r.t. the line $x=y$ you can just compute half of it over $D\cap\{x>y\}$ which is nice in polar coordinates (a sector). $\endgroup$ – GReyes Feb 11 at 3:09
  • $\begingroup$ @Euduardo I tried but found the calculation of this integral is too complicate, could you write down the process of calculation? Thank you. $\endgroup$ – Midas Hu Feb 11 at 10:23
0
$\begingroup$

By the symmetry, it is sufficient to integrate over the sector S = $\{(x,y):x\ge y>0\}.$ That is, the integral $\iint_G{\frac{|x-y|}{(x^2+y^2+1)^2}dxdy} = 2\iint_S{\frac{x-y}{(x^2+y^2+1)^2}dxdy}.$ Substituting $x=r\cos\theta, y=r\sin\theta$, we have $\iint_S{\frac{x-y}{(x^2+y^2+1)^2}dxdy} = \iint_A{\frac{r(\cos\theta-\sin\theta)}{(r^2+1)^2}rdrd\theta}.$ Note that $dxdy$ is replaced by $rdrd\theta$ and the converted region is $A = \{(r,\theta):r\ge0, 0\le\theta\le\frac{\pi}{4}\}.$ Then one can calculate the integration as follows. $\iint_A{\frac{r(\cos\theta-\sin\theta)}{(r^2+1)^2}rdrd\theta} = \int_{0}^{\infty}{\frac{r^2}{(r^2+1)^2}dr}\int_{0}^{\frac{\pi}{4}}{\cos\theta-\sin\theta d\theta} = (\sqrt{2}-1)\int_{0}^{\infty}{\frac{r^2}{(r^2+1)^2}dr}.$ Put $r=\tan{t},$then $\int_{0}^{\infty}{\frac{r^2}{(r^2+1)^2}dr} = \int_{0}^{\frac{\pi}{2}}{\frac{\tan^2{t}}{(\tan^2{t}+1)^2}\sec^2{t}dt} = \int_{0}^{\frac{\pi}{2}}{\sin^2{t}dt} = \frac{\pi}{4}.$ Therefore, the value of the integral $\iint_G{\frac{|x-y|}{(x^2+y^2+1)^2}dxdy} = 2\times(\sqrt{2}-1)\times\frac{\pi}{4} = \frac{\pi}{2}(\sqrt{2}-1).$

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.