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I'm trying to get estimate on the following partial summation using Abel-Plana Summation formula:

$$\sum_{n=1}^x \frac{\sin^2(n)}{n}$$

I can handle the first integral in the formula but I'm stuck at the following functional:

$$T(x)=\int_0^\infty\frac{ (\frac{\sin^2(x+iy)}{(x+iy)}-\frac{\sin^2(x-iy)}{(x-iy)})}{(e^{2πy}-1)}dy$$

Questions:

(1) Sharp upper and lower bounds on $T(x)$.

(2) Graph of $T(x)$ (need an idea about it's growth).

Note: I tried to expand $\sin$ as complex variable in terms of hyperbolic functions ; but then I can't handle integral after expansion .

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  • $\begingroup$ Close votes !? What more clarity can I add? $\endgroup$ – Bambi Feb 24 at 10:03
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If I am not mistaken, the difference in the numerator is $$ \frac{{e^{2y} (2y\cos ^2 x + x\sin (2x) - y) + e^{ - 2y} (2y\cos ^2 x - x\sin (2x) - y) - 2y}}{{2(x^2 + y^2 )}} $$ (two times the imaginary part of the first term). Now for large positive $y$, this is $$ = \mathcal{O}(1)\frac{{e^{2y} (x + y)}}{{x^2 + y^2 }} = \mathcal{O}(1)\frac{{e^{2y} }}{{\sqrt {x^2 + y^2 } }} = \mathcal{O}(1)\frac{1}{x}\frac{{e^{2y} }}{y}, $$ and for small $y$, it is $\mathcal{O} (y/x)$. Hence, $T(x)=\mathcal{O}(1/x)$.

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Just looking at the sum, it is less than $\ln(x)+O(1)$ and, by looking at successive terms of $\sin^2(n)$, I'm pretty sure that I could show the s is greater than $c\ln(x)$ where $c > 1/2$ or so.

The key would be showing that if $\sin^2(n)$ is small then $\sin^2(n+1)$ is not small.

I'm on my phone, so I'll wait till later.

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    $\begingroup$ I believe that by the Euler--Maclaurin summation formula and properties of the cosine integral, the sum is $\frac{1}{2}\log x + \mathcal{O}(1) = \log \sqrt x + \mathcal{O}(1)$. $\endgroup$ – Gary Feb 24 at 7:12

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