Calculate the following improper integral: double
Calculate the following improper integral: $$ \iint_D \frac{\arctan(x+y)}{(x^2+y^2)^2}\,dx\,dy, $$ where $D$ is the region $x>0$, $y>0$, $x+y>1$.
Explanation of Method
Since the numerator depends only on $x+y$, the substitution $u=x+y,\,v=x-y$ separates the integral into a $u$-dependent factor and a $v$-only piece: with $x^2+y^2=\frac{u^2+v^2}2$, the inner $v$-integral (over the fixed strip $-u\leqslant v\leqslant u$) reduces to a pure function of $u$, after which the outer integral is a standard $\arctan u/u^3$-type integral handled by parts.
- Change variables to $u=x+y$, $v=x-y$.
- Evaluate the inner integral over $v$ using the substitution $v=u\tan\theta$.
- Evaluate the outer integral over $u$ via integration by parts, then a partial-fraction split.
Step 1 · Change of variables
Let $u=x+y$, $v=x-y$, so $x=\frac{u+v}2$, $y=\frac{u-v}2$, with Jacobian $\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|=\dfrac12$. The region becomes $u>1$, $-u<v<u$ (from $x>0\Leftrightarrow u+v>0$ and $y>0\Leftrightarrow u-v>0$), and:
$$ x^2+y^2 = \frac{(u+v)^2+(u-v)^2}4 = \frac{u^2+v^2}2, \qquad (x^2+y^2)^2=\frac{(u^2+v^2)^2}4. $$
So:
$$ I = \int_1^\infty\int_{-u}^{u} \frac{\arctan u}{(u^2+v^2)^2/4}\cdot\frac12\,dv\,du = \int_1^\infty 2\arctan u\left[\int_{-u}^{u}\frac{dv}{(u^2+v^2)^2}\right]du. $$
Step 2 · Evaluate the inner integral
Let $v=u\tan\theta$, $dv=u\sec^2\theta\,d\theta$, $u^2+v^2=u^2\sec^2\theta$; the limits $v=\mp u$ correspond to $\theta=\mp\frac\pi4$:
$$ \int_{-u}^{u}\frac{dv}{(u^2+v^2)^2} = \int_{-\pi/4}^{\pi/4}\frac{u\sec^2\theta}{u^4\sec^4\theta}\,d\theta = \frac1{u^3}\int_{-\pi/4}^{\pi/4}\cos^2\theta\,d\theta. $$
Using $\cos^2\theta=\frac{1+\cos2\theta}2$:
$$ \int_{-\pi/4}^{\pi/4}\cos^2\theta\,d\theta = \frac12\left[\theta+\frac{\sin2\theta}2\right]_{-\pi/4}^{\pi/4} = \frac12\left(\frac\pi2+1\right) = \frac{\pi+2}4, $$
so:
$$ \int_{-u}^{u}\frac{dv}{(u^2+v^2)^2} = \frac{\pi+2}{4u^3}. $$
Step 3 · Evaluate the outer integral
$$ I = \int_1^\infty 2\arctan u\cdot\frac{\pi+2}{4u^3}\,du = \frac{\pi+2}2\int_1^\infty\frac{\arctan u}{u^3}\,du. $$
Integrating by parts with $w=\arctan u$, $dz=u^{-3}du$ (so $dw=\frac{du}{1+u^2}$, $z=-\frac1{2u^2}$):
$$ \int_1^\infty\frac{\arctan u}{u^3}\,du = \left[-\frac{\arctan u}{2u^2}\right]_1^\infty + \int_1^\infty\frac{du}{2u^2(1+u^2)} = \frac{\pi}8 + \frac12\int_1^\infty\left(\frac1{u^2}-\frac1{1+u^2}\right)du. $$
$$ = \frac\pi8+\frac12\left[\left(1\right)-\left(\frac\pi2-\frac\pi4\right)\right] = \frac\pi8+\frac12\left(1-\frac\pi4\right) = \frac12. $$
Step 4 · Combine
$$ I = \frac{\pi+2}2\cdot\frac12 = \frac{\pi+2}4. $$
Additional Notes
- Verified numerically: direct double integration over the original region (truncated at a large cutoff and extended) gives $I\approx1.28539$, matching $\frac{\pi+2}4\approx1.28540$.
Result
The solution to the integral is: