Simple Prob

$\begin{aligned} I & = \int_0^\frac \pi 4 x {\color{#3D99F6} \prod_{k=1}^\infty \cos \left(\frac x{2^k} \right)} dx & \small \color{#3D99F6} \text{Note that }\prod_{k=1}^\infty \cos \left(\frac x{2^k} \right) = \text{sinc }(x) \text{ (see proof)} \\ & = \int_0^\frac \pi 4 x {\color{#3D99F6} \text{ sinc }(x)} \ dx & \small \color{#3D99F6} \text{where }\text{sinc }(x) = \begin{cases} 1 & \text{for }x = 0 \\ \frac {\sin x}x & \text{otherwise} \end{cases} \\ & = \int_0^\frac \pi 4 \sin x \ dx = - \cos x \ \bigg|_0^\frac \pi 4 \\ & = - \frac 1{\sqrt 2} + 1 \approx \boxed{0.293} \end{aligned}$ .

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Proof:

$\begin{aligned} \prod_{k=1}^n \cos \left(\frac x{2^k} \right) & = \cos \frac x{2^n} \cos \frac x{2^{n-1}} \cdots \cos \frac x4 \cos \frac x2 \\ & = \frac {{\color{#3D99F6}\sin \frac x{2^n}} \cos \frac x{2^n} \cos \frac x{2^{n-1}} \cdots \cos \frac x4 \cos \frac x2}{\color{#3D99F6} \sin \frac x{2^n}} \\ & = \frac {{\color{#3D99F6}\sin \frac x{2^{n-1}}} \cos \frac x{2^{n-1}} \cos \frac x{2^{n-2}} \cdots \cos \frac x4 \cos \frac x2}{{\color{#3D99F6}2} \sin \frac x{2^n}} \\ & = \frac {{\color{#3D99F6}\sin \frac x{2^{n-2}}} \cos \frac x{2^{n-2}} \cos \frac x{2^{n-3}} \cdots \cos \frac x4 \cos \frac x2}{{\color{#3D99F6}4} \sin \frac x{2^n}} \\ & = \cdots \quad \cdots \quad \cdots \\ & = \frac {\sin \frac x2 \cos \frac x2}{2^{n-1}\sin \frac x{2^n}} = \frac {\sin x}{2^n \sin \frac x{2^n}} \end{aligned}$

Therefore, $\displaystyle \lim_{n \to \infty} \prod_{k=1}^n \cos \left(\frac x{2^k} \right) = \lim_{n \to \infty} \frac {\sin x}{2^n \sin \frac x{2^n}} = \lim_{n \to \infty} \frac {\sin x}{\frac {x\sin \frac x{2^n}}{\frac x{2^n}}} = \frac {\sin x}x$ . Note that for $x=0$ , $\displaystyle \prod_{k=1}^\infty \cos \left(\frac x{2^k} \right) = 1$ .

$\displaystyle \implies \prod_{k=1}^\infty \cos \left(\frac x{2^k} \right) = \text{sinc }(x) = \begin{cases} 1 & \text{for }x=0 \\ \dfrac {\sin x}x & \text{otherwise} \end{cases}$