Proving $ \sum_{n \in \mathbb{Z} } \left[\frac{\sin (n \alpha + \theta) }{ n \alpha + \theta} \right]^2 = \frac{\pi}{\alpha} $

Question

Proving $$ \sum_{n \in \mathbb{Z} } \left[\frac{\sin (n \alpha + \theta) }{ n \alpha + \theta} \right]^2 = \frac{\pi}{\alpha} \,\, \forall \alpha , \theta \in \mathbb{R} $$

My attempt

It is known that $$\sum_{n\in\mathbb{Z}}\frac{1}{(n\pi+x)^2}=\csc^2(x)\tag{1}$$ which can be yielded from $$\sin(\pi x)=\pi x\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2}\right)$$

Thus for the situation $\alpha = \pi$ we have $$ \sum_{n \in \mathbb{Z} } \left[\frac{\sin (n \pi + \theta) }{ n \pi + \theta} \right]^2 = \sum_{n \in \mathbb{Z} } \left(\frac{\sin\theta }{ n \pi + \theta} \right)^2 = 1 $$

But how to deal with the situation that $\alpha \ne \pi$?

Since $$ \displaystyle\sum_{n \in \mathbb{Z} } \frac{1}{ (n \alpha + \theta)^2} = \frac{\pi^2 \csc^2 \frac{\pi\theta}{\alpha} }{\alpha^2} $$ is a corollary of $(1)$, we only need to evaluate $ \displaystyle\sum_{n \in \mathbb{Z} } \frac{\cos (2n \alpha + 2\theta) }{ (n \alpha + \theta)^2} $ , which means to evaluate $$ \displaystyle\sum_{n \in \mathbb{Z} } \frac{\cos (2n \alpha ) }{ (n \alpha + \theta)^2} \tag{2} $$ and $$ \displaystyle\sum_{n \in \mathbb{Z} } \frac{\sin(2n \alpha ) }{ (n \alpha + \theta)^2} \tag{3} $$

In the post linked, we get a nice equation $$ \frac{\pi \cos(xy)}{\sin(\pi y)} =\sum_{n=-\infty}^\infty \frac{(-1)^n\cos(nx)}{y-n} $$

If there doesn't exist $(-1)^n$, we can evaluate $(2)$ and $(3)$ by differentiating and selecting appropriate $x$.

But I got stuck on that.

Any hints? Thanks in advance!

Answer 1

I will prove the stated equality for the case $|\alpha|\leq\pi$. (I believe it is false when $|\alpha|>\pi$.)

The Poisson summation formula tells us that if $f(x)$ is a function and $$ F(q) = \int_{-\infty}^{+\infty}dx\, f(x)\, e^{-i q x} $$ is its continuous Fourier transform (FT), then $$ \sum_{n = -\infty}^{+\infty} f(n)\;=\; \sum_{n = -\infty}^{+\infty} F(2\pi n) $$

For this problem, let $$ f(x) = {\mathrm{sinc}(\alpha x + \theta)}^2\, , $$ where $\mathrm{sinc}(x) = \sin(x)/x$. Then the FT of $f(x)$ is \begin{align*} F(q) &= \int_{-\infty}^{+\infty}dx\; {\mathrm{sinc}(\alpha x + \theta)}^2\; e^{-i q x}\\ &= e^{+i \, q\, \theta/\alpha} \int_{-\infty}^{+\infty}dx\; {\mathrm{sinc}(\alpha x)}^2\; e^{-i q x} \qquad \text{Change of variables: $x\rightarrow x - \theta/\alpha$}\\ &= e^{+i \, q\, \theta/\alpha} \frac{1}{|\alpha|}\int_{-\infty}^{+\infty}dz\; {\mathrm{sinc}(z)}^2\; e^{-i u z} \qquad z = |\alpha|x\, , \; u = q/|\alpha| \end{align*}

The FT of the square of a sinc function is a "tent" function: \begin{equation*} \int_{-\infty}^{+\infty}dz\; {\mathrm{sinc}(z)}^2\; e^{-i u z}\;=\; \begin{cases} \pi \left(1 - \frac{|u|}{2}\right) & |u| < 2\\ 0 & \text{else} \end{cases} \end{equation*} I'm not going to prove this here so as to not clutter up the derivation, but it can be found in various sources, and follows from the convolution theorem for Fourier transforms and the fact that the FT of a sinc function is a "top hat" function.

Putting this all together, we arrive at: \begin{equation} F(q) \;=\; e^{+i \, q\, \theta/\alpha} \; \frac{\pi}{|\alpha|} \; \begin{cases} 1 - \frac{|q|}{2|\alpha|} & |q| < 2|\alpha|\\ 0 & \text{else} \end{cases} \qquad\qquad (1) \end{equation}

The Poisson summation formula tells us that \begin{equation*} \sum_{n = -\infty}^{+\infty} {\left[\frac{\sin(n\alpha + \theta)}{n\alpha + \theta}\right]}^2\;=\; \sum_{n = -\infty}^{+\infty} f(n)\;=\; \sum_{n = -\infty}^{+\infty} F(2\pi n) \end{equation*}

In performing the latter sum over $F(2\pi n)$ using Eq. (1), we see that since $F(q)$ has compact support on $|q| < 2|\alpha|$, we need only sum $n$ over $|n| < |\alpha|/\pi$. When $|\alpha| \leq \pi$, only the $n = 0$ term contributes to the sum, and we are left with: \begin{equation*} \sum_{n = -\infty}^{+\infty} {\left[\frac{\sin(n\alpha + \theta)}{n\alpha + \theta}\right]}^2\;=\; F(0) \;=\; \frac{\pi}{|\alpha|} \qquad\qquad (|\alpha| \leq \pi) \end{equation*}

When $|\alpha| > \pi$, more terms with $n \neq 0$ enter the sum, and the expression becomes more complicated, the $\theta$ dependence does not go away, and I'm not sure there's a nice, closed-form expression. For instance, evaluating the sum in Mathematica for $\alpha = 4$ and $\theta = 0$ yields: \begin{equation*} \sum_{n = -\infty}^{+\infty} {\left[\frac{\sin(4 n)}{4 n}\right]}^2\;=\; \frac{6\pi - \pi^2}{8} \;\neq \; \frac{\pi}{4}\, . \end{equation*}

Edited to add:

The general sum is given by \begin{equation*} \sum_{n = -\infty}^{+\infty} {\left[\frac{\sin(n\alpha + \theta)}{n\alpha + \theta}\right]}^2\;=\; \frac{\pi}{|\alpha|}\left[ 1 \;+\; 2\sum_{n = 1}^{\left\lfloor |\alpha|/\pi \right\rfloor} \cos(2\pi \,n\, \theta\, /\, \alpha) \left(1 - \frac{\pi\, n}{|\alpha|}\right) \right]\, . \end{equation*} If this has a "nice" closed form, I'm unaware of it.

Answer 2

With a slight modification, we can use the residue theorem and the cotangent trick to evaluate the series.

Assume that $\frac{\theta}{\alpha} \notin \mathbb{Z}.$

Let's integrate the function $$f(z) = \frac{\pi \left(\cot(\pi z)\color{red}{-i} \right)\left(1- e^{i(2 \alpha z+ 2\theta)}\right) }{2(\alpha z +\theta)^{2}}$$ around a square contour with vertices at $\pm \left(N + \frac{1}{2} \right) \pm i \left(N + \frac{1}{2} \right)$, where $N$ is some positive integer large enough that $ z = - \frac{\theta}{\alpha}$ falls inside the contour.

If $0 < \alpha \le \pi$, the integral will vanish as $ N \to \infty$ through the positive integers.

This is due in part to the fact that the magnitude of $\left(\cot(\pi z)-i \right) = e^{-i \pi z} \csc(\pi z)$ decays like $2 e^{2 \pi \Im(z)}$ as $\Im(z) \to - \infty$.

Therefore, if $0 < \alpha \le \pi$, we have

$$ \begin{align} 0 &= \sum_{n \in \mathbb{Z}} \operatorname{Res}[f(z), n] + \operatorname{Res} \left[f(z), - \frac{\theta}{\alpha} \right] \\ &= \sum_{n \in \mathbb{Z}} \frac{1-e^{i(2n \alpha + 2\theta) } }{2(n \alpha+ \theta)^{2}} + \operatorname{Res} \left[f(z), - \frac{\theta}{\alpha} \right] \, , \end{align}$$

from which it follows that $$ \begin{align} \sum_{n \in \mathbb{Z}}\frac{1-e^{i(2 n \alpha + 2 \theta) }}{2(n \alpha+ \theta)^{2}} &= -\lim_{z \to - \frac{ \theta}{\alpha}} \left(z+ \frac{\theta}{\alpha} \right) f(z) \\&= -\frac{\pi}{2 \alpha^{2}}\lim_{ z \to -\frac{\theta}{\alpha}} \left(\cot(\pi z) -i \right) \, \frac{1-e^{i(2az+2 \theta)} }{z+ \frac{\theta}{\alpha}} \\ &= -\frac{\pi}{2\alpha^{2}} \left(-\cot \left(\frac{\pi \theta}{\alpha} \right) -i\right) \left(-2i \alpha \right) \\ &= \frac{\pi}{\alpha } \left(1- i\cot \left(\frac{\pi \theta}{\alpha} \right) \right). \end{align}$$

Equating the real parts on both sides of the above equation and using the trig identity $\sin^{2}(x) = \frac{1-\cos(2x)}{2}$, we get $$\sum_{n \in \mathbb{Z}} \frac{\sin^{2}\left(n \alpha + \theta \right)}{\left(n \alpha + \theta\right)^{2}} = \frac{\pi}{\alpha} \, , \quad 0 < \alpha \le \pi . $$

As a bonus, we also get $$\sum_{n \in \mathbb{Z}} \frac{\sin \left(2n \alpha + 2 \theta\right)}{\left(n \alpha + \theta\right)^{2}} = \frac{2 \pi}{\alpha} \, \cot \left(\frac{\pi \theta}{\alpha} \right) \, , \quad 0 < \alpha \le \pi. $$