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Using Poisson Summation Formula, how do you evaluate the following infinite sum $\sum_{k=1}^{\infty}\left(\frac{\sin(tk)}{k}\right)^2$?

The Poisson Summation Formula states that: $\sum_{k=-\infty}^{\infty}f(2\pi k)=\frac{1}{2\pi}\sum_{k=-\infty}^{\infty}\hat{f}(k)$.

I tried playing around with the idea that if $f(x)=\mathbf{1}_{[-1,1]}$ then $\hat{f}(\xi)=2\frac{\sin(i\xi)}{i\xi}$. How should I go ahead from here to compute the summation?

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Compute the Fourier Transform $$ \begin{align} &\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,e^{-2\pi ix\xi}\,\mathrm{d}x\\ &=\int_{-\infty-i}^{\infty-i}\frac{e^{2iz}-2+e^{-2iz}}{-4z^2}\,e^{-2\pi iz\xi}\,\mathrm{d}z\tag1\\ &=\int_{-\infty-i}^{\infty-i}\frac{e^{2iz(1-\pi\xi)}-2e^{-2\pi iz\xi}+e^{-2iz(1+\pi\xi)}}{-4z^2}\,\mathrm{d}z\tag2\\[6pt] &=\pi(1-\pi\xi)[\pi\xi\le1]+2\pi(\pi\xi)[\pi\xi\lt0]-\pi(1+\pi\xi)[\pi\xi\lt-1]\tag3\\[12pt] &=\pi(1-\pi\xi)[0\le\pi\xi\le1]+\pi(1+\pi\xi)[-1\le\pi\xi\le0]\tag4\\[12pt] &=\pi(1-\pi|\xi|)\big[\,\pi|\xi|\le1\,\big]\tag5 \end{align} $$ Explanation:
$(1)$: write $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\phantom{\text{(1):}}$ shift the contour since there are no singularities
$(2)$: combine exponents
$(3)$: use the contour $[-R-i,R-i]\cup Re^{i\pi[0,1]}-i$
$\phantom{\text{(3):}}$ for exponentials with a positive coefficient of $iz$
$\phantom{\text{(3):}}$ use the contour $[-R-i,R-i]\cup Re^{-i\pi[0,1]}-i$
$\phantom{\text{(3):}}$ for exponentials with a negative coefficient of $iz$
$\phantom{\text{(3):}}$ we need only count the residues from
$\phantom{\text{(3):}}$ the exponentials with a positive coefficient of $iz$
$(4)$: simplify
$(5)$: simplify

For $t\gt0$, substitute $x\mapsto x/t$ and apply $(5)$: $$ \begin{align} \int_{-\infty}^\infty\frac{\sin^2(tx)}{x^2}\,e^{-2\pi ix\xi}\,\mathrm{d}x &=t\int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,e^{-2\pi ix\xi/t}\,\mathrm{d}x\tag6\\[6pt] &=\pi(t-\pi|\xi|)\big[\,\pi|\xi|\le t\,\big]\tag7 \end{align} $$


Apply Poisson Summation $$ \begin{align} t^2+2\sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2} &=\sum_{k\in\mathbb{Z}}\frac{\sin^2(tk)}{k^2}\tag8\\[6pt] &=\sum_{k\in\mathbb{Z}}\pi(t-\pi|k|)\big[\,\pi|k|\le t\,\big]\tag9\\ &=\pi t+2\sum_{k=1}^{\lfloor t/\pi\rfloor}\pi(t-\pi k)\tag{10}\\[9pt] &=\pi t+\left(2\pi t-\pi^2\right)\lfloor t/\pi\rfloor-\pi^2\lfloor t/\pi\rfloor^2\tag{11} \end{align} $$ Explanation:
$\phantom{1}(8)$: make a sum over $\mathbb{Z}$
$\phantom{1}(9)$: Poisson summation
$(10)$: make a sum over $\mathbb{N}$
$(11)$: sum in $k$

Solve for the sum: $$ \bbox[5px,border:2px solid #C0A000]{\sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2}=\frac12\left((\pi-t)t+\left(2\pi t-\pi^2\right)\lfloor t/\pi\rfloor-\pi^2\lfloor t/\pi\rfloor^2\right)}\tag{12} $$


A Dilogarithmic Identity

As Claude Leibovici shows using $\sin(tk)=\frac{e^{itk}-e^{-itk}}{2i}$, $$\newcommand{\Li}{\operatorname{Li}} \sum_{k=1}^\infty\frac{\sin^2(tk)}{k^2}=\frac{\pi^2}{12}-\frac14\left(\Li_2\left(e^{2it}\right)+\Li_2\left(e^{-2it}\right)\right)\tag{13} $$ which gives a nice identity: $$ \hspace{-18pt}\bbox[5px,border:2px solid #C0A000]{\Li_2\left(e^{it}\right)+\Li_2\left(e^{-it}\right)=\frac{\pi^2}3-\frac12\left((2\pi-t)t+4\!\left(\pi t-\pi^2\right)\left\lfloor\frac{t}{2\pi}\right\rfloor-4\pi^2\left\lfloor\frac{t}{2\pi}\right\rfloor^2\right)}\tag{14} $$

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Without Poisson summation formula $$S=\sum_{k=1}^{\infty}\left(\frac{\sin(tk)}{k}\right)^2=\sum_{k=1}^{\infty}\frac{\sin^2(tk)}{k^2}=\frac 12\sum_{k=1}^{\infty}\frac{1-\cos(2tk)}{k^2}$$ make the cosine exponentials to get $$S=\frac{\pi^2}{12}-\frac{1}{4} \left(\text{Li}_2\left(e^{-2 i t}\right)+\text{Li}_2\left(e^{2 i t}\right)\right)$$

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  • $\begingroup$ (+1) This is quite interesting as a comparison $\endgroup$ – robjohn Apr 6 at 7:23

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