Estimate on partial sum $\sum_{n=1}^x \frac{\sin^2(n)}{n}$ using Abel Plana Summation Formula :

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 .

• Close votes !? What more clarity can I add? – Bambi Feb 24 at 10:03

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