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I was wondering how to evaluate $\sum_{p \ prime}\frac{1}{1+p^2}$. For the integer analog, here is the closed-form solution.

Edit: If there may not exist a closed-form solution, how to calculate the value with high precision, say rounding to 30 significant digits?

Hang Wu
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1 Answers1

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I would not bet on the existence of a nice closed form, but since $$ \frac{1}{1+p^2} \approx -\log\left(1-\frac{1}{p^2}\right)+\frac{3}{2}\log\left(1-\frac{1}{p^4}\right) $$ is a pretty accurate approximation for any $p\geq 3$, by Euler's product we have $$ \sum_{p}\frac{1}{1+p^2}\approx \frac{1}{5}+\log\prod_{p\geq 3}\frac{(1-1/p^4)^{3/2}}{1-1/p^2}=\frac{1}{5}+\log\frac{\zeta(2)\left(1-\frac{1}{4}\right)}{\zeta(4)^{3/2}\left(1-\frac{1}{16}\right)^{3/2}} $$ or $$ \sum_{p}\frac{1}{1+p^2}\approx \frac{1}{5}+\log\frac{48\sqrt{6}}{\pi^4} = \color{green}{0.38}8161\ldots $$ In order to compute the LHS with more accuracy, it is enough to find the coefficients $a_1,a_2,\ldots,a_n$ such that the Maclaurin series of $\frac{x^2}{x^2+1}$ and $\sum_{k=1}^{n}-a_k \log\left(1-x^{2k}\right)$ agree up to the $x^{2n}$ term - this is equivalent to the inversion of a structured matrix$^{(*)}$, and it leads to an approximate identity of the $$ \sum_{p}\frac{1}{1+p^2} \approx \log\prod_{k=1}^{n}\zeta(2k)^{a_k}$$ kind, which can be further improved by computing $\sum_{k=1}^{n}\frac{1}{1+p_k^2}$ in a explicit way, then exploiting the previous manipulations in terms of "delayed" Euler products $\prod_{p>p_n}\left(1-\frac{1}{p^{2k}}\right)^{-1}$.

$(*)$ In terms of the algebra of the arithmetic functions with respect to the Dirichlet convolution, this inversion problem can be written as $(-1)^{m+1}=(a*\text{Id}^{-1})(m)$ for any $m\geq 1$. It leads to the recursion $a_1=1$ and $$a_m = (-1)^{m+1}-\frac{1}{m}\sum_{\substack{d\mid m\\ d<m}}da_d.$$

For instance, the improved approximation $$ \frac{1}{1+p^2}\approx -\log\left(1-\tfrac{1}{p^2}\right)+\tfrac{3}{2}\log\left(1-\tfrac{1}{p^4}\right)-\tfrac{2}{3}\log\left(1-\tfrac{1}{p^6}\right)$$ leads to $$ \sum_{p}\frac{1}{1+p^2}\approx \frac{3}{10}+\log\frac{\zeta(2)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\zeta(6)^{2/3}\left(1-\frac{1}{64}\right)^{2/3}\left(1-\frac{1}{729}\right)^{2/3}}{\zeta(4)^{3/2}\left(1-\frac{1}{16}\right)^{3/2}\left(1-\frac{1}{81}\right)^{3/2}}$$ or $$ \sum_{p}\frac{1}{1+p^2}\approx \frac{3}{10}+\frac{1}{6}\log\frac{16663715523}{9765625000}=\color{green}{0.3890}61\ldots$$ Up to $30$ figures we have $$ \sum_{p}\frac{1}{p^2+1}=\sum_{m\geq 1}a_m\log\zeta(2m) \approx \color{green}{0.3890595553169683717104143896924963581384} $$ This is a practical acceleration formula since the coefficients $a_m$ are simple to compute and bounded, while $\log\zeta(2m)$ has an exponential decay as $m\to +\infty$.

Jack D'Aurizio
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  • $\sum_{p \ prime}\frac{1}{1+p^2}$ equals to $0.0.3890595553...$. – Hang Wu Sep 04 '18 at 19:23
  • @HangWu: agreed. I outlined a technique for a computation with arbitrary accuracy. – Jack D'Aurizio Sep 04 '18 at 20:26
  • I've somehow obtained a formula $\sum_{p \ prime}\frac{1}{p^2+1}=\sum_{k=1}^{\infty}\frac{\mu(k)}{k}\sum_{i=1}^{\infty} (-1)^{i+1}log(\zeta(2ik))$, where $\mu(\cdot)$ is the mobius function and $\zeta(\cdot)$ is the Riemann zeta function. I'm not sure whether it's correct, but I've checked up to 10 digits and it works. Moreover, it seems that $\sum_{p \ prime}\frac{1}{p^s+1}=\sum_{k=1}^{\infty}\frac{\mu(k)}{k}\sum_{i=1}^{\infty} (-1)^{i+1}log(\zeta(sik))$ – Hang Wu Sep 04 '18 at 20:59
  • @HangWu: indeed my approach gives $\sum_{p}\frac{1}{p^2+1}=\sum_{m\geq 1}a_m \log\zeta(2m)$, so you just have to check that your arithmetic function fulfills the wanted constraint. – Jack D'Aurizio Sep 04 '18 at 21:05
  • Yes. They're equivalent. – Hang Wu Sep 06 '18 at 06:38