A Sinful Sinfonia!

The key result needed to solve the problem is the following:

$\prod_{k=0}^{n-1}\sin \left(x+\frac{k\pi}{n}\right)=\frac{\sin nx}{2^{n-1}}$

Proof: Denoting $a=\pi/n$ , use Euler's formula to obtain: $\prod_{k=0}^{n-1}\sin \left(x+\frac{k\pi}{n}\right)=\prod_{k=0}^{n-1}\frac{e^{i(x+ka)}-e^{-i(x+ka)}}{2i}\\=\frac{e^{i(nx+(n-1)\pi/2)}}{(2i)^n}\prod_{k=0}^{n-1}(1-e^{-i(2x+2ka)})\\=\frac{e^{inx}}{2^n i}e^{-2inx}\prod_{k=0}^{n-1}(e^{i2x}-e^{-i2k\pi/n})\\=\frac{e^{-inx}}{2^n i}(e^{2inx}-1)=\frac{\sin nx}{2^{n-1}}$ where the penultimate step follows from the identity, $x^n-1=\prod_{k=0}^{n-1}(x-e^{-i2k\pi/n}),\ \forall x$ , and the last step followd form Euler's formula. $\blacksquare$

Thus, the desired integral becomes $I=\int_{0}^{\pi/3}\sum_{n=1}^{\infty}\frac{\sin nx}{2^{n-1}}dx$ Now, defining $f_N(x)=\sum_{n=1}^N \frac{\sin nx}{2^{n-1}}$ , it is easy to see that $f_N\to f$ uniformly where $f(x)=\sum_{n=1}^{\infty} \frac{\sin nx}{2^{n-1}}=2\Im\left(\frac{e^{ix}/2}{1-e^{ix}/2}\right)$ . We observe that $I=\int_{0}^{\pi/3}f(x) dx$ . Since $|f_n|\le \sum_{n\ge 1}\frac{1}{2^{n-1}}=2$ , using Bounded Convergence Theorem , we can write, $I=\lim_{N\to \infty }\int_{0}^{\pi/3} f_N(x) dx=\lim_{N\to \infty} \int_{0}^{\pi/3}\sum_{n=1}^N \frac{\sin nx}{2^{n-1}} dx\\=\lim_{N\to \infty} \sum_{n=1}^N\int_{0}^{\pi/3} \frac{\sin nx}{2^{n-1}}$ Now, $\int_{0}^{\pi/3}{\sin nx} dx=\frac{1-\cos(n\pi/3)}{n}$ Thus, $I=\sum_{n\ge 1}\frac{1-\cos(n\pi/3)}{n\cdot 2^{n-1}}=2\ln 2+2\Re\left(\ln(1-e^{i\pi/3}/2)\right)=\ln 4+\ln\left((1-1/4)^2+(\sqrt{3}/4)^2\right)=\ln(3)$ Thus giving $a=\boxed{3}$ .