Double integral $\int_0^1\int_0^1 \frac {x^2-y^2}{(x^2+y^2)^2} dy dx$

Question

I came across this one today.

$$\int_0^1\int_0^1 \frac {x^2-y^2}{(x^2+y^2)^2} dy dx$$

What I saw was that for any $(x,y)$ in $[0,1]\times[0,1], f(x,y) = -f(y,x)$

and since this region is symmetric across the line $y=x,$ this suggests to me that this integral will evaluate to $0$.

I applied a change in coordinates.

$u =x+y, v = x-y\\ 2 (\int_0^1\int_{-v}^{v} \frac{uv}{(u^2+v^2)^2} du dv + \int_1^2\int_{v-2}^{2-v} \frac{uv}{(u^2+v^2)^2} du dv)$

That appears to me to also evaluate to $0$.

But wolfram-$\alpha$ says $\frac {\pi}{4}$

http://www.wolframalpha.com/input/?i=int+%5B0+to+1%5D+int+%5B0+to+1%5D+(x%5E2-y%5E2)%2F(x%5E2%2By%5E2)%5E2+dy+dx

Where is the hole in my logic?

Answer 1

As Frpzzd noted in a comment, each of your "it's zero because of cancellation" symmetry arguments assumes $\infty-\infty=0$. But once you've noted $$\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dy=\int_0^1\bigg(\frac{1}{x^2+y^2}-\frac{2y^2}{(x^2+y^2)^2}\bigg)dy=\bigg[\frac{y}{x^2+y^2}\bigg]_0^1=\frac{1}{1+x^2},$$the rest is trivial.

Answer 2

This is my try, it may not be correct: $$I=\int\limits_0^1\int\limits_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dydx$$ now we use polar coordinates: $$I=\int_0^{\pi/2}\int_0^\sqrt{2}\frac{x^2-y^2}{r^4}rdrd\theta=\int_0^{\pi/2}\int_0^\sqrt{2}\frac{\cos^2(\theta)-\sin^2(\theta)}{r}drd\theta$$$$ =\int_0^{\pi/2}\int_0^\sqrt{2}\frac{\cos(2\theta)}{r}drd\theta$$$$ =\frac{1}{2}\ln(2)\int_0^{\pi/2}\cos(2\theta)d\theta=\frac{1}{4}\ln(2)\left[\sin(2\theta)\right]_0^{\pi/2}=0$$ this has come to the wrong answer but I do not know how

Answer 3

Let

$$I=\int_0^1\int_0^1\frac{x^2-y^2}{\left(x^2+y^2\right)^2}\,\mathrm{d}y\,\mathrm{d}x=-\int_0^1\mathrm{d}y\underbrace{\int_0^1\frac{x^2-y^2}{\left(x^2+y^2\right)^2}\,\mathrm{d}x}_{\equiv I_x}$$

So,

\begin{align} I_x&=\int_0^1\frac{x^2-y^2}{\left(x^2+y^2\right)^2}\,\mathrm{d}x \\ & = \int_0^1\left(\frac{x^2+y^2}{\left(x^2+y^2\right)^2}-\frac{2y^2}{\left(x^2+y^2\right)^2}\right)\mathrm{d}x \\ & = \int_0^1\frac{1}{x^2+y^2}\,\mathrm{d}x-2y^2\int_0^1\frac{1}{\left(x^2+y^2\right)^2}\,\mathrm{d}x. \end{align}

To evaluated these integrals you can use the following reduction formula:

$$\int\dfrac{1}{\left(\mathtt{a}x^2+\mathtt{b}\right)^{\mathtt{n}}}\,\mathrm{d}x={\frac{2\mathtt{n}-3}{2\mathtt{b}\left(\mathtt{n}-1\right)}}\int\frac{1}{\left(\mathtt{a}x^2+\mathtt{b}\right)^{\mathtt{n}-1}}\,\mathrm{d}x+\frac{x}{2\mathtt{b}\left(\mathtt{n}-1\right)\left(\mathtt{a}x^2+\mathtt{b}\right)^{\mathtt{n}-1}}$$

Such that

$$I_x= -\frac{1}{y^2+1}.$$

Thereby,

$$I=\int_0^1 \frac{1}{y^2+1} \, \mathrm{d}y.$$

Using again the reduction formula:

$$I=\frac{\pi}{4}.$$