Check $\sum\limits_{p = 1}^\infty\frac{\sin(Qpa)}p\sin\left(\frac{qpa}2\right)^2=\frac\pi4\theta(Q-q)$

Question

I was going through a solid-state textbook when the following result appeared in the text. The author states that

$$\sum_{p = 1}^{\infty}\dfrac{\sin(Qpa)}{p} \sin\left(\dfrac{qpa}{2}\right)^2 =\frac{\pi}{4} \theta(Q-q)$$

where $\theta$ denotes the Heaviside step function. I have no idea why this should be true as I was not able to tackle this with my (limited) ability in Fourier series. I would appreciate if someone could provide a proof of this or some intuition as to why this should be the case.

Edit: As a way of convincing myself of the truth of the formula i have plotted the above sum as a function of q, with Q = 2 and the sum running from 1 to pmax = 1, 10, 50 in blue green and orange respectively and the results seem to make the formula plausible. This is of course far from a proof.

Answer 1

I think the OP has it backward. The sum is easily addressed by using a half-angle formula, and is equal to

$$\frac12 \sum_{p=1}^{\infty} \frac{\sin{(Q a p)}}{p} - \frac12 \sum_{p=1}^{\infty} \frac{\sin{(Q a p)}}{p} \cos{(q a p)} $$

Consider the Fourier series

$$f(x) = \sum_{p=-\infty}^{\infty} \frac{\sin{(k p)}}{p} \cos{(x p)} = \cases{\pi \quad |x| \lt k \\ 0 \quad |x| \gt k} $$

Thus,

$$\frac12 \sum_{p=1}^{\infty} \frac{\sin{(Q a p)}}{p} = \frac{\pi}{4} - \frac14 $$

$$\frac12 \sum_{p=1}^{\infty} \frac{\sin{(Q a p)}}{p} \cos{(q a p)} = \frac{\pi}{4} \theta(Q a-q a) - \frac14 $$

We can ignore the factor $a$ inside the Heaviside. Thus, the sum in question is equal to

$$\sum_{p=1}^{\infty} \frac{\sin{(Q a p)}}{p} \sin^2{\left (\frac{q a p}{2} \right )} = \frac{\pi}{4} \left ( 1 - \theta(Q-q) \right ) = \frac{\pi}{4} \theta(q-Q)$$

Answer 2

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With the Abel-Plana Formula:

\begin{align} &\bbox[10px,#ffd]{\ds{% \sum_{p = 1}^{\infty}{\sin\pars{Qpa} \over p}\,\sin^{2}\pars{qpa \over 2}}} = Qa\sum_{p = 1}^{\infty}\mrm{sinc}\pars{Qap}\sin^{2}\pars{{qa \over 2}\,p} \\[5mm] = &\ Qa\sum_{p = 0}^{\infty}\mrm{sinc}\pars{\verts{Qa}p}\sin^{2}\pars{{\verts{qa} \over 2}\,p} \\[5mm] = &\ Qa\int_{0}^{\infty}\mrm{sinc}\pars{\verts{Qa}x}\sin^{2}\pars{{\verts{qa} \over 2}\,x}\,\dd x \qquad\pars{~Abel\mbox{-}Plana\ Formula~} \\[5mm] \stackrel{\mu\ \equiv\ \verts{q/Q}}{=} &\ \,\mrm{sgn}\pars{Qa}\int_{0}^{\infty}{\sin\pars{x} \over x}\,\sin^{2}\pars{{1 \over 2}\,\mu x} \,\dd x = \,\mrm{sgn}\pars{Qa}\int_{0}^{\infty}{\sin\pars{x} \bracks{1 - \cos\pars{\mu x}}/2\over x}\,\dd x \\[5mm] = &\ {1 \over 2}\,\mrm{sgn}\pars{Qa}\braces{% \int_{0}^{\infty}{\sin\pars{x}\over x}\,\dd x - \int_{0}^{\infty}{\sin\pars{\bracks{1 + \mu}x}\over x}\,\dd x - \int_{0}^{\infty}{\sin\pars{\bracks{1 - \mu}x}\over x}\,\dd x} \\[5mm] = &\ -{1 \over 2}\,\mrm{sgn}\pars{Qa} \int_{0}^{\infty}{\sin\pars{\bracks{1 - \mu}x}\over x}\,\dd x \\[5mm] = &\ -{1 \over 2}\,\mrm{sgn}\pars{Qa} \bracks{\Theta\pars{1 - \mu}\int_{0}^{\infty}{\sin\pars{x}\over x}\,\dd x - \Theta\pars{\mu - 1}\int_{0}^{\infty}{\sin\pars{x}\over x}\,\dd x} \\[5mm] = &\ {\pi \over 4}\,\mrm{sgn}\pars{Qa}\,\mrm{sgn}\pars{\mu - 1} = \bbx{{\pi \over 4}\,\mrm{sgn}\pars{Qa}\,\mrm{sgn}\pars{\verts{q} - \verts{Q}}} \end{align}