<s>[INST] Calculate double integral $\iint_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}dxdy$ over unbounded region $D$ Calculate
$$I = \iint\limits_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}\,dx\,dy$$
where the region of integration is:
$$D = \{(x,y) \in \mathbb{R}^2 \mid x+y - a\sqrt{2} \geq 0, -x+y +a\sqrt{2} \geq 0, a > 0\}$$
I tried using polar coordinates, but then, I don't know how to proceed because of the region $D$ and of the integrand:
$$I = \iint\limits_D\frac{\sqrt{r^2-a^2}\cos\theta}{r^2}\,dr\,d\theta$$
 [/INST] With the variable changes $x=au$ and $y=av$
\begin{align}
&\iint_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}\,dx\,dy\\
=&\int_0^\infty \int_{-v+\sqrt2}^{v+\sqrt2} \frac{\sqrt{u^2+v^2 - 1}\ u}{(u^2+v^2)^2}\,du\,dv\\
=& \int_0^\infty \frac12\bigg(\tan^{-1}{\sqrt{u^2+v^2 - 1}}-\frac{\sqrt{u^2+v^2 - 1}}{u^2+v^2}\bigg)\bigg|_{-v+\sqrt2}^{v+\sqrt2}dv\\
 =&\ \frac12\int_0^\infty \bigg(\tan^{-1}(1+\sqrt2v) -\tan^{-1}|1-\sqrt2v|\bigg)dv\\
&-\frac12\int_0^\infty \left(\frac{1+\sqrt2v}{(v+\sqrt2)^2+v^2}  -\frac{|1-\sqrt2v|}{(v-\sqrt2)^2+v^2}\right)dv\\
=& \ \frac12 \cdot \frac{\pi +2\ln2}{2\sqrt2}
-\frac12 \left(-\frac{\ln2}{\sqrt2}\right)\\
=&\ \frac1{\sqrt2}\left(\frac\pi4 +\ln2\right)
\end{align}
</s>