# Convergence of $1/(x^2+y^2)^a$.

I'm curious about the convergence of the series: \begin{align} \sum_{x,y=1}^\infty \frac{1}{(x^2+y^2)^\alpha}\ ,\ \alpha \in \mathbb{N} \end{align} I'm wondering for what values of $\alpha$ this converges, and if so is there an analytic way to find the limit? I'm not really sure where to start, it's been a long time since I studied sequences and series.

Let $N,M \in \Bbb N$. On the square $[N,N+1]\times [M,M+1]$,

$$\frac{1}{[(N+1)^2 + (M + 1)^2]^\alpha} \le \frac{1}{(x^2 + y^2)^\alpha} \le \frac{1}{(N^2 + M^2)^\alpha}.$$

Integrating the inequality over the square, we find

$$\frac{1}{[(N+1)^2 + (M+1)^2]^\alpha} \le \int_N^{N+1}\int_M^{M+1} \frac{dx\, dy}{(x^2 + y^2)^{\alpha}} \le \frac{1}{(N^2 + M^2)^\alpha}.$$

Summing from over all $N,M \ge 1$, we have

$$\sum_{N,M\ge 2} \frac{1}{(N^2 + M^2)^\alpha} \le \int_1^\infty \int_1^\infty \frac{dx\, dy}{(x^2 + y^2)^\alpha} \le \sum_{N,M\ge 1} \frac{1}{(N^2 + M^2)^\alpha}$$

Hence, the series $\sum_{x,y\ge 1} (x^2 + y^2)^{-\alpha}$ converges if and only if $\int_1^\infty \int_1^\infty (x^2 + y^2)^{-\alpha}dx\, dy$ converges. Using the transformation $x = u\tan \theta$, $y = u$, we obtain

$$\int_1^\infty \int_1^\infty \frac{dx\, dy}{(x^2 + y^2)^\alpha} = C_\alpha\int_1^\infty u^{1 - 2\alpha}\, du$$

where $C_\alpha = \int_{\pi/4}^{\pi/2} \cos^{2\alpha - 2}\theta\, d\theta$. Since $C_\alpha$ is finite (since $\alpha \ge 1$) and $\int_1^\infty u^{1 - 2\alpha}\, du$ converges if and only if $\alpha > 1$, it follows that $\sum_{x,y \ge 1} (x^2 + y^2)^{-\alpha}$ converges if and only if $\alpha > 1$.

• This is a very nice proof technique! Mar 28 '15 at 20:22

By the Gauss circle problem the number of lattice points inside $\{(x,y)\in\mathbb{R}^2: x^2+y^2\leq R^2\}$ is given by $\pi R^2+O(R)$, hence the series is convergent for any $\alpha > 1$. In order to compute the sum for a particular $\alpha$, we may exploit identities like: $$\int_{0}^{+\infty}\frac{\sin(ax)}{a} x \frac{e^{-bx}}{b}\,dx = \frac{2}{(a^2+b^2)^2}\tag{1}$$ that leads to: $$\sum_{x,y\geq 1}\frac{1}{(x^2+y^2)^2}=-\frac{1}{2}\int_{0}^{+\infty}W(x)\, x\log(1 - e^{-x})\,dx \tag{2}$$ where $W(x)$ is the sawtooth wave with period $2\pi$ such that $W(0)=\frac{\pi}{2}$.

For different $\alpha$s, we just need to differentiate $(1)$ with respect to $b$ the right number of times. That gives:

$$S_\alpha = \sum_{x,y\geq 1}\frac{1}{(x^2+y^2)^\alpha}=C_\alpha\int_{0}^{+\infty}W(x)\, x^{\alpha-1}\operatorname{Li}_{\alpha-1}(e^{-x})\,dx.$$

By using Dirichlet series and the fact that the representation function for the sum of two squares is well-known we also have:

$$S_2 = K\cdot\zeta(2)-\zeta(4)$$

where $K$ is the Catalan constant, as well as:

$$S_\alpha = -\zeta(2\alpha)+\zeta(\alpha)\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{\alpha}}.$$

• Thanks, that's a lot of interesting functions to learn about. Mar 28 '15 at 20:24
The convergence of this series equivalent to the convergence of following series : $$S_{\alpha}=\sum_{n=1}^{+\infty} \frac{r_2(n)}{n^{\alpha}}$$
where $r_2(n)$ denotes the number of ways of writing the integer $n$ as sum of positive two squares we have the following formula: $$r_2(n)=\sum_{d|n}\sin(\frac{\pi}{2}d)$$
see here for other formulas. from this we know that :$r_2(n) \leq d(n)$ and $\sum_{i=1}^n \frac{d(n)}{n^{\alpha}}$ converges if and only if $\alpha>1$, this analysis suffices to show that $S_{\alpha}$ converges if and only if $\alpha>1$ because it diverges when $\alpha=1$