# Compute integral in unbounded area

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?

• convert the integral in the polar coordinate form – Euduardo Feb 11 at 3:04
• 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). – GReyes Feb 11 at 3:09
• @Euduardo I tried but found the calculation of this integral is too complicate, could you write down the process of calculation? Thank you. – Midas Hu Feb 11 at 10:23

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 $$\iint_G{\frac{|x-y|}{(x^2+y^2+1)^2}dxdy} = 2\times(\sqrt{2}-1)\times\frac{\pi}{4} = \frac{\pi}{2}(\sqrt{2}-1).$$