# Math Help - double integral

1. ## double integral

I'm trying to show that

$\int\int_{0\leq x,y\leq 1}\Bigg\vert\frac{x^2-y^2}{(x^2+y^2)^2}\Bigg\vert\hspace{1mm}dxdy = 2\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$

so that I can use Fubini's theorem to deduce that the iterated integrals of the integrand(without the absolute value function) are not the same.

But I got

$\int\int_{0\leq x,y\leq 1}\Bigg\vert\frac{x^2-y^2}{(x^2+y^2)^2}\Bigg\vert\hspace{1mm}dxdy = 2\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy + \frac{\pi}{4}_{.}$

I guess it does not really make any difference in the end because the right hand sides of both of the equalities are $\infty$.

But which one is right??

2. Originally Posted by willy0625
I'm trying to show that

$\int\int_{0\leq x,y\leq 1}\Bigg\vert\frac{x^2-y^2}{(x^2+y^2)^2}\Bigg\vert\hspace{1mm}dxdy = 2\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$

so that I can use Fubini's theorem to deduce that the iterated integrals of the integrand(without the absolute value function) are not the same.

But I got

$\int\int_{0\leq x,y\leq 1}\Bigg\vert\frac{x^2-y^2}{(x^2+y^2)^2}\Bigg\vert\hspace{1mm}dxdy = 2\int_{0}^{1}\int_{0}^{1}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy + \frac{\pi}{4}_{.}$

I guess it does not really make any difference in the end because the right hand sides of both of the equalities are $\infty$.

But which one is right??
Have you tried converting to polars?

3. First, because of the absolute value, the obvious thing to do is to write this as two integrals. As long as 0< y< x, the fraction is postive:
$\int_{x=0}^1\int_{y= 0}^x \frac{x^2- y^2}{(x^2+ y^2)^2}dy dx$
and as long as 0< x< y, the fraction is negative:
$-\int_{x=0}^1\int_{y= x}^1 \frac{x^2- y^2}{(x^2+ y^2)^2}dy dx$

Then do as Prove It suggested, convert to polar coordinates:
The line x= 1 is given by $r cos(\theta)= 1$ so $r= \frac{1}{cos(\theta)}= sec(\theta)$

$\int_{\theta= 0}^{\pi/4}\int_{r= 0}^{sec(\theta)} \frac{r^2(cos^2(\theta)- sin^2(\theta))}{r^4} r drd\theta$ $=\int_{\theta= 0}^{\pi/4}\int_{r= 0}^{sec(\theta)} \frac{cos(2\theta)}{r} drd\theta$

$-\int_{\theta= \pi/4}^{\pi/2}\int_{r= 0}^{sec(\theta)} \frac{r^2(cos^2(\theta)- sin^2(\theta))}{r^4} r drd\theta$ $=-\int_{\theta= \pi/4}^{\pi/2}\int_{r= 0}^{sec(\theta)} \frac{cos(2\theta)}{r} drd\theta$

The original integral is the sum of those, of course.