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The integral is the following : $$\int_{-\infty}^\infty \prod_{k=1}^\infty \cos\bigg(\frac{x}{2^k}\bigg)dx$$

I need first to evaluate this product, but how ?

My attempts :

Using : $$\sin(2x)=2\sin(x)\cos(x) \Leftrightarrow \cos(x)=\frac{\sin(2x)}{2\sin(x)}$$ So basically this can help us : $$\prod_{k=1}^\infty \frac{\sin\big(\frac{x}{2^k}\big)}{2\sin\big(\frac{x}{2^k}\big)} =\frac{\sin x}{2\sin(x/2)}\frac{\sin (x/2)}{2\sin(x/4)}\color{red}{\dots}$$ It's important to eveluate this product but I couldn't any hints ? I'll be thankfull !

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    $\begingroup$ Can you see how infinitely many terms can cancel in this telescoping product? As you have shown correctly$$\prod_{k=1}^n\cos{\left(\frac{x}{2^k}\right)}=\frac{\sin{(x)}}{2^n\sin{(x/2^n)}}\to\frac{\sin{(x)}}x$$so the integral is just the Dirichlet integral namely equal to $\pi$. $\endgroup$ – Peter Foreman Jun 9 at 17:12
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    $\begingroup$ See youtube.com/watch?v=qn7z9T-dezw $\endgroup$ – Math2718 Jun 9 at 17:20
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Note

$$\frac{\sin x}x =\frac{\sin \frac x2 }{\frac x2}\cos\frac x2 = \frac{\sin \frac x{2^2}}{ \frac x{2^2}} \cos\frac x{2^2}\cos\frac x2 \\ = \lim_{n\to \infty}\left(\frac{\sin \frac x{2^n}}{\frac x{2^n}}\right)\cos\frac{x}{2^n}...\cos\frac x{2^2}\cos\frac x2 = \prod_{k=1}^\infty \cos\frac{x}{2^k} $$ where $ \lim_{t\to 0}\frac{\sin t}{t} =1$ is used. Thus, $$\int_{-\infty}^\infty \prod_{k=1}^\infty \cos\frac{x}{2^k}dx= \int_{-\infty}^\infty \frac{\sin x}x dx = \pi $$

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