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Does anyone have any insight as to how I might find a closed-form expression for $\sum_{i=1}^N \sum_{j=1}^N\frac{1}{(i^2 + j^2)}$ ?

It feels like it should be straightforward because it's easy to do as an integral in polar coordinates if you include the axes. But here the summation is over a square lattice, and excludes the axis. Wolfram Alpha can't do it, and neither can I!

jms547
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1 Answers1

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An asymptotic formula for large $N$ may be derived as follows. First, $$ \sum\limits_{m = 1}^N {\frac{1}{{m^2 + n^2 }}} = - \frac{1}{{2n^2 }} + \frac{\pi }{{2n}}\coth (\pi n) - \frac{{\operatorname{Im} \psi (1 + N + \mathrm{i}n)}}{n} $$ where $\psi$ is the digamma function. From the known asymptotics of the digamma function, \begin{align*} \operatorname{Im} \psi (1 + N + \mathrm{i}n) &= \arg (N + \mathrm{i}n) + \frac{n}{{2(n^2 + N^2 )}} + \mathcal{O}\!\left( {\frac{1}{{N^2 }}} \right) \\ & = \arctan \left( {\frac{n}{N}} \right) + \frac{n}{{2(n^2 + N^2 )}} + \mathcal{O}\!\left( {\frac{1}{{N^2 }}} \right). \end{align*} Now \begin{align*} \sum\limits_{n = 1}^N {\frac{1}{n}\arctan \left( {\frac{n}{N}} \right)} & = \frac{1}{N}\sum\limits_{n = 1}^N {\frac{N}{n}\arctan \left( {\frac{n}{N}} \right)} \\ & = \int_0^1 {\frac{{\arctan x}}{x}\mathrm{d}x} + \mathcal{O}\!\left( {\frac{1}{N}} \right) = G + \mathcal{O}\!\left( {\frac{1}{N}} \right) \end{align*} where $G$ is Catalan's constant. Also $$ \sum\limits_{n = 1}^N {\frac{1}{{2(n^2 + N^2 )}}} = \mathcal{O}\!\left( {\frac{1}{N}} \right),\;\, \sum\limits_{n = 1}^N {\mathcal{O}\!\left( {\frac{1}{{N^2 }}} \right)} = \mathcal{O}\!\left( {\frac{1}{N}} \right),\;\, \sum\limits_{n = 1}^N { - \frac{1}{{2n^2 }}} = - \frac{{\pi ^2 }}{{12}} + \mathcal{O}\!\left( {\frac{1}{N}} \right) $$ and \begin{align*} \sum\limits_{n = 1}^N {\frac{\pi }{{2n}}\coth (\pi n)} & = \frac{\pi }{2}\sum\limits_{n = 1}^N {\frac{1}{n}} + \frac{\pi }{2}\sum\limits_{n = 1}^N {\frac{{\coth (\pi n) - 1}}{n}} \\ & = \frac{\pi }{2}\log N + \frac{\pi }{2}\gamma + \frac{\pi }{2}\sum\limits_{n = 1}^\infty {\frac{{\coth (\pi n) - 1}}{n}} + \mathcal{O}\!\left( {\frac{1}{N}} \right), \end{align*} using the asymptotics of the harmonic numbers ($\gamma$ is the Euler–Mascheroni constant). In summary $$ \sum\limits_{n = 1}^N {\sum\limits_{m = 1}^N {\frac{1}{{m^2 + n^2 }}} } = \frac{\pi }{2}\log N + C + \mathcal{O}\!\left( {\frac{1}{N}} \right) $$ with $$ C = \frac{\pi }{2}\gamma - G - \frac{{\pi ^2 }}{{12}} + \frac{\pi }{2}\sum\limits_{n = 1}^\infty {\frac{{\coth (\pi n) - 1}}{n}} =-0.825861175988396\ldots . $$

Addendum. Note that, using the Dedekind $\eta$-function, $$ \sum\limits_{n = 1}^\infty {\frac{{\coth (\pi n) - 1}}{n}} = - 2\log ({\rm e}^{\pi /12} \eta ({\rm i})) = 2\log \left( {\frac{{2\pi ^{3/4} }}{{\Gamma (1/4)}}} \right) - \frac{\pi }{6} $$ and hence $$ C = \frac{\pi }{2}\gamma - G - \frac{{\pi ^2 }}{6} + \pi \log \left( {\frac{{2\pi ^{3/4} }}{{\Gamma (1/4)}}} \right). $$

Gary
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  • Thanks Gary! Probably asymptotics are the best I can do. It's nice to see the log(N) come out as expected... – jms547 Aug 14 '21 at 17:23
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    @jms547 For your interest, the error term can be improved to $$ \frac{{\pi + 4}}{4}\frac{1}{N} + \mathcal{O}!\left( {\frac{1}{{N^2 }}} \right). $$ – Gary Aug 14 '21 at 19:27
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    Gary, your solution is a shortcut (+1). I also found the asymptotics, though in a bit longer way - https://www.quora.com/How-do-you-evaluate-L_1-lim_-n-to-infty-Big-sum_-k-1-n-sum_-l-1-n-frac-1-k-2-l-2-frac-pi-ln-n-2-Big-and-L_2-lim_-n-to-infty-n-Big-sum_-k-1-n-sum_-l-1-n-frac-1-k-2-l-2-frac-pi-ln-n-2-L_1-Big . It is interesting to note that the modular form has a closed form in this case, though its evaluation is beyond my capability :) I took a ready answer for it. – Svyatoslav Sep 02 '22 at 12:19