For showing the double series $\sum_{i,j \in \mathbb{Z}^+}\frac{1}{i^2 + j^2}$ diverges, I can compare the partial sums with the double integral:
$$\sum_{i=1}^I\sum_{j=1}^J\frac{1}{i^2 + j^2} > \int_1^{I+1}\int_1^{J+1} \frac{dxdy}{x^2 + y^2}$$
and take limit of the integral as $I,J \to \infty$. I know that $dxdy/(x^2+y^2)$ behaves like $rdrd\theta/r^2$ in polar coordinates so the integral should diverge like $\log(r)$ as $r \to \infty$.
I'm having trouble seeing how the region $[1,I+1]\times[1,J+1]$ is transformed in polar coordinates and I'd like to see a rigorous proof that the integral diverges.
Thank you.
Best Answer
Assuming WLOG that $J \leqslant I$ and changing variables as $x = 1+s, \, y = 1+t$, we have
$$\begin{align} A_{I,J}=\int_1^{I+1}\int_1^{J+1} \frac{dx \, dy}{x^2 + y^2}\end{align} \geqslant \int_0^{J}\int_0^{J} \frac{ds \, dt}{s^2 + t^2 + 2(s+t) + 2}$$
Now switch to polar coordinates $s = r \cos \theta, \, t = r \sin \theta $.
The sector $S = \{(r,\theta): 0 \leqslant r \leqslant J, \,0 \leqslant \theta \leqslant \pi/2\}$ is a subset of the square $[0,J]^2$. Since the integrand is nonnegative and $1 \leqslant \cos \theta + \sin \theta \leqslant \sqrt{2}$ for $0 \leqslant \theta \leqslant \pi/2$ we have
$$\begin{align}A_{I,J} &\geqslant \int_0^{\pi/2}\int_0^J \frac{r\, dr\, d\theta}{r^2 + 2r(\cos \theta + \sin \theta)+2}\\ &\geqslant \int_0^{\pi/2}\int_0^J \frac{r\, dr\, d\theta}{r^2 + 2\sqrt{2}r+2} \\ &= \frac{\pi}{2}\int_0^J\frac{r}{(r+\sqrt{2})^2} \, dr \\ &=\frac{\pi}{2}\int_0^J\frac{1}{r+\sqrt{2}} \, dr - \frac{\pi}{2}\int_0^J\frac{\sqrt{2}}{(r+\sqrt{2})^2} \, dr \\ &= \frac{\pi}{2}\left(\log(J + \sqrt{2}) + \frac{\sqrt{2}}{J + \sqrt{2}} - (1 + \log \sqrt{2}) \right)\end{align}$$
The RHS tends to $+\infty$ as $I,J \to \infty$.