Integration of an Infinite Product

This key to this problem is the identity $\frac{\sin x}{x} = \prod_{k=1}^{\infty}\cos\left(\frac{x}{2^k}\right)$ . Once this identity is applied, the integral becomes routine. This formula, I think, is credited to Euler and you may have come across it if you've studied Fourier transforms, but if you haven't seen it before, here's a quick proof:

By applying the double angle identity over and over we get: $\begin{aligned} \sin x &= 2 \sin\frac{x}{2} \cos\frac{x}{2} \\ &= 4 \sin \frac{x}{4} \cos \frac{x}{4} \cos \frac{x}{2} \\ & \space\space \vdots & \\ &= 2^n \sin \frac{x}{2^n} \prod_{k = 1}^{n} \cos \frac{x}{2^k} \end{aligned}$

Taking the limit as $n \to \infty$ we get $\sin x = \prod_{k = 1}^{\infty} \cos \frac{x}{2^k} \left( \lim_{n \to \infty} 2^n \sin \frac{x}{2^n} \right)$

Now, as $n \to \infty$ , $\frac{x}{2^n} \to 0$ , so by the small angle approximation $\sin \frac{x}{2^n} \to \frac{x}{2^n}$ . Hence, $\lim_{n \to \infty} 2^n \sin \frac{x}{2^n} = x$ .