1. ## Difficult double integral

Hi!
I'm stuck at this problem:
Let D be the region given by |x| + |y| >= 1. Determine if the improper integral
$\int\limits_D {\int {\frac{{dxdy}}{{(x^2 + y^2 )^2 }}} }$
exists, and if it does, evaluate it.

My attempt:
Looking at the first quadrant and changing to polar coordinates I get $r\cos \theta + r\sin \theta \ge 1 \Leftrightarrow r \ge \frac{1}{{\cos \theta + \sin \theta }}$

This gives me the integral

$
4\int\limits_0^{\pi /2} {d\theta \int\limits_{\frac{1}{{\cos \theta + \sin \theta }}}^\infty {1/r^3 dr = } } 2\int\limits_0^{\pi /2} {\frac{{d\theta }}{{(\cos \theta + \sin \theta )^2 }}}
$

I don't know how to do that last iteration.

2. Hello,

Originally Posted by kloda

$
4\int\limits_0^{\pi /2} {d\theta \int\limits_{\frac{1}{{\cos \theta + \sin \theta }}}^\infty {1/r^3 dr }}
$

I don't know how to do that last iteration.
It's ok until here

An antiderivative of $\frac{1}{r^3}$ is indeed $\frac{-2}{r^2}$

But when substituting r by $\frac{1}{\cos \theta+\sin \theta}$, it yields :

$-2 (\cos \theta+ \sin \theta)^2$

So the latter integral should be $2 \int_0^{\frac{\pi}{2}} (\cos \theta+\sin \theta)^2 d \theta$

Note that $(\cos \theta+\sin \theta)^2=\cos^2 \theta + \sin^2 \theta+2 \cos \theta \sin \theta=1+\sin 2 \theta$

3. Originally Posted by Moo
But when substituting r by $\frac{1}{\cos \theta+\sin \theta}$, it yields :

$-2 (\cos \theta+ \sin \theta)^2$
Oops!