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?
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$\begingroup$ convert the integral in the polar coordinate form $\endgroup$ – Euduardo Feb 11 at 3:04
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$\begingroup$ 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
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$\begingroup$ @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
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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 $\iint_G{\frac{|x-y|}{(x^2+y^2+1)^2}dxdy} = 2\times(\sqrt{2}-1)\times\frac{\pi}{4} = \frac{\pi}{2}(\sqrt{2}-1).$
