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Question: $$\sum_{n=1}^{\infty}\frac{\sin(n^2)}{n^2}=\,?$$

Previously I calculated a similar summation but it was more luck than wisdom, and insight led me to believe my methods were super incorrect (but gave some clue of how it should be done correctly because the answer was correct after all). $$\sum_{x=1}^{\infty} c^{-x^2}=\frac{\sqrt{\pi}}{2 \sqrt{\ln(c)}}-\frac{1}{2}+\frac{ \sqrt{\pi}}{\sqrt{\ln(c)}}\sum_{x=1}^{\infty}e^{-\frac{\pi^2 x^2}{\ln(c)}}.$$

So I wanted to try out a similar sum. I thought it might be common, however, I couldn't find it, and the quick estimated guess, gave a surprisingly large error, I still got to give it a try with my old incorrect method, but I wondered how to "alternatively/formally" derive the answer.

The estimate would be: $\sqrt{\pi/2}-1/2$ which is way to far off, yet the answer is surprisingly close to $\sqrt{\pi/8}$ with a small error, which confused me.

Question: How to calculate the summation, and what's the answer including the error term?

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  • $\begingroup$ Perhaps the Poisson summation formula could help. $\endgroup$
    – Gary
    Commented Dec 5, 2021 at 0:13
  • $\begingroup$ Possibly Related: math.stackexchange.com/questions/3472703/… $\endgroup$
    – Chris Grossack
    Commented Dec 5, 2021 at 0:15
  • $\begingroup$ Also worth noting, we can use sage to compute this to 8 places: $\approx 0.62674514...$. Then we can ask an inverse symoblic calculator to try and find a symbolic representation of this number, but it looks like there isn't one that it knows about. All this to say there probably isn't a known way to sum this exactly. I would love to be proven wrong, though ^_^ $\endgroup$
    – Chris Grossack
    Commented Dec 5, 2021 at 0:40
  • $\begingroup$ I think I want to apply Poisson summation but I don't know how and would love an example, I hope I will figure the answer with some trial and error of course, I also wondered indeed if there are more general answers known, so thanks for the link! All I figured is that you got to expand it's power series r(n) of a function f(x), e.g. $$f(x^d)=\sum_{n \in \mathbb{Z} }^{\infty}x^{n} r(n/d) \sum_{k=0}^{d-1} \frac{e^{\frac{2i\pi*km}{d}}}{d} $$ So here we'd get $\sum_ {n \in \mathbb{Z} }\frac{(sin(n \pi/4)-sin(n \pi 3/4)+sin(n \pi 5/4)-sin(n \pi 7/4))}{4} \frac{\zeta(n-2)}{(n/2)!}$ $\endgroup$
    – Gerben
    Commented Dec 5, 2021 at 1:00
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    $\begingroup$ @HallaSurvivor I expect something along the lines of $\sqrt{\pi/8}+c_1\sum erf(xc_2)/x$ to be the answer, with the sum being small but I been a bit sloppy with the derivation, and all and I am sure there's a more rigorous general method for such sums. $\endgroup$
    – Gerben
    Commented Dec 5, 2021 at 2:33

1 Answer 1

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Knowing that:

$$\frac 12\left(\vartheta_3\left(e^{ix}\right)-1\right)=\sum_{n=1}^\infty e^{i xn^2}$$

we try an elliptic theta function of the third kind $\vartheta_3(x)$

$$\frac 12\int \left(\vartheta_3\left(e^{ix}\right)-1\right)dx\mathop\iff^?\sum_{n=1}^\infty\frac{e^{i xn^2}}{n^2}$$

Therefore:

$$\sum_{n=1}^\infty\frac{\sin\left(n^2\right)}{n^2}=\frac12\lim_{k\to1^-}\text{Re}\int_{i\infty}^1\vartheta_3\left(ke^{it}\right)-1\ dt$$

and

$$\sum_{n=1}^\infty\frac{\cos\left(n^2\right)}{n^2}=\frac12\text{Im}\lim_{k\to0}\int_{i\infty}^1\vartheta_3\left(ke^{i(t+k i)}\right)-1\ dt $$

which works

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