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fixed bug pointed out by Will.
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Igor Rivin
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I assume that $x$ and $y$ are integral, in which case your sum is one quarter of the sum

$$\sum_{r=1}^\infty \exp(-\sqrt{r}) s(r),$$ where $s(r)$ is the number of representations of $r$ as the sum of two squares. Clearly, $\sum_{r=1}^R s(r) \sim \pi r^2,$$\sum_{r=1}^R s(r) \sim \pi r,$ which means you can estimate the original sum via summation by parts, followed by approximating the sum by the integral.

I assume that $x$ and $y$ are integral, in which case your sum is one quarter of the sum

$$\sum_{r=1}^\infty \exp(-\sqrt{r}) s(r),$$ where $s(r)$ is the number of representations of $r$ as the sum of two squares. Clearly, $\sum_{r=1}^R s(r) \sim \pi r^2,$ which means you can estimate the original sum via summation by parts, followed by approximating the sum by the integral.

I assume that $x$ and $y$ are integral, in which case your sum is one quarter of the sum

$$\sum_{r=1}^\infty \exp(-\sqrt{r}) s(r),$$ where $s(r)$ is the number of representations of $r$ as the sum of two squares. Clearly, $\sum_{r=1}^R s(r) \sim \pi r,$ which means you can estimate the original sum via summation by parts, followed by approximating the sum by the integral.

Source Link
Igor Rivin
  • 96.4k
  • 11
  • 153
  • 366

I assume that $x$ and $y$ are integral, in which case your sum is one quarter of the sum

$$\sum_{r=1}^\infty \exp(-\sqrt{r}) s(r),$$ where $s(r)$ is the number of representations of $r$ as the sum of two squares. Clearly, $\sum_{r=1}^R s(r) \sim \pi r^2,$ which means you can estimate the original sum via summation by parts, followed by approximating the sum by the integral.