$\int_{0}^{\infty}\int_{1}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$ diverges?

I want someone to review my proof that $$\int_{1}^{\infty}\int_{0}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$$ does not converge.

To make things easier, I said let's look at the entire first quadrant and then subtract the integral over the small rectangle that we added.

move to polar coordinates:

$$\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}r^2\frac{(\cos^2\theta-\sin^2\theta)}{r^3}drd\theta=\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{\cos^2\theta-\sin^2\theta}{r}drd\theta=\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1-2\sin^2\theta}{r}drd\theta\leq\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1}{r}drd\theta$$

The inequality part is true since $-2\sin^2 x$ is always negative.

So:

$$\int_{0}^{\frac{\pi}{2}}\int_{o}^{\infty}\frac{1}{r}drd\theta=\frac{\pi}{2}\ln(r)|_0^{\infty}=\frac{\pi}{2}\ln(\frac{\infty}{0})=\frac{\pi}{2}\ln({\infty})=\infty$$

So over the entire first quadrant it diverges.

Is there a point to checking the small rectangle that we added? I mean, even if it diverges, the answer would still diverge.

• Just realizedthat the inequality, while true, is meaningless since i need to find a lower bound, not upper. sigh... – Oria Gruber May 24 '14 at 19:12
• \sin \theta and \cos \theta ... you never put latex for trig functions. – Santosh Linkha May 24 '14 at 19:13
• According to wolf, the integral converges to $\frac \pi 2$ – Santosh Linkha May 24 '14 at 19:17
• Is there a typo in the original problem? Do you intend to have $0\le x,y<\infty$? – Ted Shifrin May 24 '14 at 22:07
• no. $1 \leq y < \infty$ – Oria Gruber May 25 '14 at 9:18

As an iterated integral, this integral converges. Indeed

$$\int_{0}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dx = \left[ - \frac{x}{x^{2} + y^{2}} \right]_{0}^{\infty} = 0.$$

So the whole integral also converges to 0. But it does not converge absolutely. Indeed,

$$\int_{0}^{\infty} \left|\frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}\right| \, dx = \int_{0}^{y} \frac{y^{2} - x^{2}}{(x^{2} + y^{2})^{2}} \, dx + \int_{y}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dx = \frac{1}{y}$$

and therefore

$$\int_{1}^{\infty} \int_{0}^{\infty} \left|\frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}}\right| \, dxdy = \infty.$$

Much frustrating fact is that if you switch the order of integration, you get

$$\int_{1}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dy = \left[ \frac{y}{x^{2} + y^{2}} \right]_{1}^{\infty} = - \frac{1}{x^{2}+1}$$

and thus the whole inegral becomes

$$\int_{0}^{\infty}\int_{1}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dydx = -\frac{\pi}{2}.$$

Since we are considering the region in the first quadrant above $y=1$, in polar coordinates we may write it as: \begin{align*} \int_{1}^{\infty}\int_{0}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}\, dx\, dy &= \lim_{N\to\infty} \int_{\arcsin(1/N)}^{\pi/2} \, \int_{1/\sin\theta}^{N} \, \frac{\cos\left(2\, \theta\right)}{r}\, dr\, d\theta\\ &= \lim_{N\to\infty}\, \frac{1}{2N}\sqrt{1-\frac{1}{N^2}}+\frac{1}{2}\arcsin\frac{1}{N} - \frac{\pi}{4}\\ &= -\frac{\pi}{4} \end{align*}

If we calculate without transformation:

\begin{align*} \lim_{N\to\infty} \int_{1}^{N} \, \int_{0}^{N} \, \frac{x^2-y^2}{(x^2+y^2)^2}\, dx\, dy &= \lim_{N\to\infty} \arctan\left(\frac{1}{N}\right)- \frac{\pi}{4}\\ &= -\frac{\pi}{4} \end{align*}

Using mpmath's gauss-legendre quadrature with $\mathrm{maxdegree} > 6$ indicates $\displaystyle -\frac{\pi}{2}$ as answer, and when $\mathrm{maxdegree} \le 6$ it stays close to $\displaystyle -\frac{\pi}{4}$. 'tanh-sinh' is close to neither.

Also note that if we choose \begin{align*} \lim_{N_1\to\infty} \lim_{N_2\to\infty} \int_{1}^{N_2} \, \int_{0}^{N_1} \, \frac{x^2-y^2}{(x^2+y^2)^2}\, dx\, dy &= \lim_{N_1\to\infty}\lim_{N_2\to\infty} \arctan\left(\frac{1}{N_1}\right) - \arctan\left(\frac{N_2}{N_1}\right) \end{align*} So, the actual answer depends on the ratio $\displaystyle \frac{N_2}{N_1}$. If we consider "$\infty$" in both directions to be the same, the answer is $\displaystyle -\frac{\pi}{4}$