A very challenging question integral of an infinite product.

Question

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.

Answer 1

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?

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}