Family of Products

Lemma (left as an exercise for the reader)

$\large \prod_{n=1}^N\sum_{k=0}^{r-1}x^{\frac{k}{r^n}}=\sum_{k=0}^{r^N-1}x^{\frac{k}{r^N}}$

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For odd $r$ :

$\large \begin{aligned} \displaystyle \prod_{n=1}^{\infty}\left(\frac{1}{r}\sum_{k=-\frac{r-1}{2}}^{\frac{r-1}{2}}\cos\left(\frac{kx}{r^n}\right)\right) &= \displaystyle \lim_{N \rightarrow \infty} \prod_{n=1}^N\frac{1}{r}\sum_{k=-\frac{r-1}{2}}^{\frac{r-1}{2}}\cos\left(\frac{kx}{r^n}\right) \\ \displaystyle &=\lim_{N \rightarrow \infty}\prod_{n=1}^N\frac{1}{r}\sum_{k=-\frac{r-1}{2}}^{\frac{r-1}{2}}\frac{e^{i\frac{kx}{r^n}}+e^{-i\frac{kx}{r^n}}}{2} \\ \displaystyle &=\lim_{N \rightarrow \infty}\prod_{n=1}^N\frac{1}{r} \sum_{k=-\frac{r-1}{2}}^{\frac{r-1}{2}}e^{i\frac{kx}{r^n}} \\ \displaystyle &=\lim_{N \rightarrow \infty}\prod_{n=1}^N\frac{1}{r}\exp\left(-i\frac{r-1}{2}\cdot\frac{x}{r^n}\right)\sum_{k=0}^{r-1}e^{i\frac{kx}{r^n}} \\ \displaystyle &=J \end{aligned}$

For even $r$ :

$\large \begin{aligned} \displaystyle \prod_{n=1}^{\infty}\left(\frac{1}{r}\sum_{k=-\frac{r}{2}+1}^{\frac{r}{2}}\cos\left(\Big(k-\frac{1}{2}\Big)\frac{x}{r^n}\right)\right) &=\displaystyle \lim_{N \rightarrow \infty} \prod_{n=1}^N\frac{1}{r}\sum_{k=-\frac{r}{2}+1}^{\frac{r}{2}}\cos\left(\left(k-\frac{1}{2}\right)\frac{x}{r^n}\right) \\ &=\displaystyle\lim_{N \rightarrow \infty} \prod_{n=1}^N\frac{1}{r}\sum_{k=-\frac{r}{2}+1}^{\frac{r}{2}}\frac{e^{i\left(k-\frac{1}{2}\right)\frac{x}{r^n}}+e^{-\left(k-\frac{1}{2}\right)\frac{x}{r^n}}}{2} \\ &=\displaystyle\lim_{N \rightarrow \infty} \prod_{n=1}^N\frac{1}{r}\sum_{k=-\frac{r}{2}+1}^{\frac{r}{2}}e^{i\left(k-\frac{1}{2}\right)\frac{x}{r^n}} \\ &=\displaystyle\lim_{N \rightarrow \infty} \prod_{n=1}^N\frac{1}{r}\exp\left(-i\frac{r-1}{2}\cdot\frac{x}{r^n}\right)\sum_{k=0}^{r-1}e^{ik\frac{x}{r^n}} \\ &=J \end{aligned}$

So:

$\large \begin{aligned} J &= \displaystyle \lim_{N \rightarrow \infty} \prod_{n=1}^N\frac{1}{r}\exp\left(-i\frac{r-1}{2}\cdot\frac{x}{r^n}\right)\sum_{k=0}^{r-1}e^{ik\frac{x}{r^n}} \\ &= \displaystyle \lim_{N \rightarrow \infty} \exp\left(-ix\frac{1-r^{-N}}{2}\right)\cdot\frac{1}{r^N}\prod_{n=1}^N\sum_{k=0}^{r-1}e^{i\frac{kx}{r^n}} & \text{ Via Lemma} \\ &= \displaystyle \lim_{N \rightarrow \infty} \exp\left(-ix\frac{1-r^{-N}}{2}\right)\cdot\frac{1}{r^N}\sum_{k=0}^{r^N-1}e^{i\frac{kx}{r^N}} & \text{ Via Riemann Sum} \\ &= \displaystyle \exp\left(\frac{-ix}{2}\right)\cdot\int_0^1e^{ixy}dy \\ &= \displaystyle e^{-\frac{ix}{2}}\cdot\frac{1}{ix}\left(e^{ix}-1\right) \\ &= \displaystyle \frac{1}{ix}\left(e^{\frac{ix}{2}}-e^{-\frac{ix}{2}}\right) \\ &= \displaystyle \frac{2}{x}\sin\left(\frac{x}{2}\right) \end{aligned}$

Finally:

$\large \int_0^{2\pi}xf\left(x\right)=\int_0^{2\pi}2\sin\left(\frac{x}{2}\right)=\boxed{8}$

(I'll solve the integral first, will figure out the identities later.)

Note that

$\begin{aligned} \sin \dfrac{x}{2^{i + 1}} \prod_{n = 1}^i \cos \dfrac{x}{2^{n + 1}} &= \sin \dfrac{x}{2^{i + 1}} \cos \dfrac{x}{2^{i + 1}} \prod_{n = 1}^i \cos \dfrac{x}{2^n} \\ &= \dfrac{1}{2} \sin \dfrac{x}{2^i} \cos \dfrac{x}{2^i} \prod_{n = 1}^{i - 1} \cos \dfrac{x}{2^n} \\ &= \dfrac{1}{2^2} \sin \dfrac{x}{2^{i - 1}} \cos \dfrac{x}{2^{i - 1}} \prod_{n = 1}^{i - 2} \cos \dfrac{x}{2^n} \\ &= \cdots \\ &= \dfrac{1}{2^k} \sin \dfrac{x}{2^{i + 1 - k}} \cos \dfrac{x}{2^{i + 1 - k}} \prod_{n = 1}^{i - k} \cos \dfrac{x}{2^n} \\ &= \cdots \\ &= \dfrac{1}{2^i} \sin \dfrac{x}{2}. \end{aligned}$

Thus,

$\begin{aligned} f(x) &= \lim_{i \rightarrow \infty} \prod_{n = 1}^i \cos \dfrac{x}{2^{n + 1}} \\ &= \lim_{i \rightarrow \infty} \dfrac{\sin \frac{x}{2}}{2^i \sin \frac{x}{2^{i + 1}}} \\ &= \sin \dfrac{x}{2} \lim_{2^i \rightarrow \infty} \dfrac{1}{2^i \sin \frac{x}{2^{i + 1}}} \\ &= \sin \dfrac{x}{2} \lim_{j \rightarrow \infty} \dfrac{1}{j \sin \frac{x}{2j}} \quad \small {\color{cyan} \text{(let } j = 2^i \text{)}} \\ &= \sin \dfrac{x}{2} \lim_{j \rightarrow \infty} \dfrac{\frac{1}{j}}{\sin \frac{x}{2j}} \\ &= \sin \dfrac{x}{2} \lim_{j \rightarrow \infty} \dfrac{-\frac{1}{j^2}}{ -\frac{x}{2j^2} \cos \frac{x}{2j}} \quad \small {\color{cyan} \text{(applying L'Hopital's rule)}} \\ &= \sin \dfrac{x}{2} \lim_{j \rightarrow \infty} \dfrac{2}{x \cos \frac{x}{2j}} \\ &= \dfrac{2}{x} \sin \dfrac{x}{2}, \end{aligned}$

and

$\begin{aligned} \int_0^{2\pi} x f(x) \, dx &= \int_0^{2\pi} x \left ( \dfrac{2}{x} \sin \dfrac{x}{2} \right ) \, dx \\ &= \int_0^{2\pi} 2 \sin \dfrac{x}{2} \, dx \\ &= \left [ -4 \cos \dfrac{x}{2} \right ]_0^{2\pi} \\ &= -4 \cos \pi + 4 \cos 0 \\ &= \boxed{8}. \end{aligned}$