A very challenging question integral of an infinite product. Evaluate:
$$\int_{0}^{\infty}\sqrt{\prod_{k=0}^{\infty}\frac{\cos(θ/2^k)+1}{2}}dθ.$$
This was a problem from the book 'S.P.Patterson's Rectreational Problems in Advanced Mathematics'.
The problem was included in the 'Additional problems' section and did not have the solution.
I don't even know where to start.
I thought of first decomposing the infinte product into an integrable function of  $θ$, but I don't know how to .
Even wolfram alpha was not able to comprehend it..
If you have a solution, please do share it.
 A: Hint. Note that
$$\sqrt{\prod_{k=0}^{\infty}\frac{\cos(θ/2^k)+1}{2}}=\left|\prod_{k=0}^{\infty}\cos(θ/2^{k+1})\right|$$
then use Finding the limit $\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) $ and see Improper integral $\sin(x)/x $ converges absolutely, conditionaly or diverges?
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\sin\pars{\theta} & = 2\sin\pars{\theta \over 2}\cos\pars{\theta \over 2}
\\
\sin\pars{\theta} & = 2^{2}\sin\pars{\theta \over 4}
\cos\pars{\theta \over 4}\cos\pars{\theta \over 2}
\\
\sin\pars{\theta} & = 2^{3}\sin\pars{\theta \over 8}
\cos\pars{\theta \over 8}\cos\pars{\theta \over 4}\cos\pars{\theta \over 2}
\\
\vdots\phantom{AA} & \,\,\vdots\phantom{AAAAAAAAAAAAAA}\vdots
\\
\sin\pars{\theta} & = 2^{n}\sin\pars{\theta \over 2^{n}}
\cos\pars{\theta \over 2^{n}}
\cos\pars{\theta \over 2^{n - 1}}\cdots\cos\pars{\theta \over 2}
\\[5mm] \implies & 
\bbx{\prod_{k = 0}^{n - 1}\cos\pars{\theta \over 2^{k + 1}} =
{\sin\pars{\theta} \over 2^{n}\sin\pars{\theta/2^{n}}}}
\\[5mm] \implies &
\bbx{\prod_{k = 0}^{\infty}\cos\pars{\theta \over 2^{k + 1}} = {\sin\pars{\theta} \over \theta}}
\end{align}
