Challenge 19: Infinite products

[micromass]

Infinite Products

This weeks challenge is a short one:

Calculate
\int_0^\infty \prod_{n=1}^{+\infty} \cos\left(\frac{x}{2^n}\right)dx

Please list any sources that you have used to solve this question. Using google or other search engines is forbidden. Wikipedia is allowed but not its search engine. Wolfram alpha and other mathematical software are not allowed.

Points will be given as follows:
1) First person to post a correct solution to one of the above points will receive 3 points (So if you solved 3 and 4, you will receive 6 points).

2) Anybody to post a (original) correct solution to one of the above points will receive 1 point.

3) Anybody posting a (nontrivial) generalization of some of the above questions will receive 2 points

4) The person who posts a solution with the least advanced mathematical machinery will receive 1 extra point for his solution (for example, somebody solving this with basic calculus will have an "easier" solution than somebody using singular homology), if two people use the same mathematical machinery, then we will look at how complicated the proof is.

5) The person with the most elegant solution will receive 1 extra point for his solution (I decide whose solution is most elegant)

Private messages with questions, problem suggestions, feedback, etc. are always welcome!

 

[D H]

Too easy. I'll post the final result first, then add posts to derive this. The answer is:

Spoiler

$\pi / 2$

 

[D H]

Answer:

Spoiler

$\pi / 2$

Attacking the infinite product first, I'll write that product more formally as $$\lim_{N\to\infty} \Pi_{n=1}^N \cos\left( \frac x {2^n} \right)$$
I'll denote the N^th partial product $\Pi_{n=1}^N \cos(x/{2^n})$ as $P_N(x)$. Is there a simple way to express $P_N(x)$?

Note that for N=1, we have $P_1(x) = \cos \frac x 2$. Multiplying both sides by $2\sin\frac x 2$ yields $2 \sin\frac x 2 P_1(x) = 2 \sin\frac x 2 \cos \frac x 2 = \sin x$. Thus $P_1(x) = \frac {\sin x}{2 \sin\frac x 2}$. Multiplying $P_1(x)$ by $\cos \frac x 4$ yields $P_2(x)$: $P_1(x) \cos \frac x 4 = \cos \frac x 2 \cos \frac x 4 = P_2(x) = \frac {\sin x \cos \frac x 4}{2 \sin\frac x 2}$. The denominator $2 \sin \frac x 2$ can be rewritten as $2 (2 \sin \frac x 4 \cos \frac x 4)$ using the fact that $\sin(2t) = 2\sin t \cos t$. Thus $P_2(x) = \frac {\sin x}{4 \sin\frac x 4}$. This suggests that $$P_N(x) = \frac {\sin x}{2^N \sin \frac x {2^N}}$$.

The base case has already been established (twice). All that remains to prove this is to show that the above relation holds for $P_{N+1}(x)$ assuming that it holds for $P_N(x)$. Multiplying both sides of the above expression for $P_N(x)$ by $\cos \frac x {2^{N+1}}$ yields $$P_N(x) \cos \frac x {2^{N+1}} = \frac {\sin x \cos \frac x {2^{N+1}}}{2^N \sin \frac x {2^N}}$$ Via the double angle formula, $\sin \frac x {2^N} = 2\sin \frac x {2^{N+1}} \cos x \frac x {2^{N+1}}$ Thus $$P_N(x) \cos \frac x {2^{N+1}} = P_{N+1}(x) = \frac {\sin x}{2^{N+1} \sin \frac x {2^{N+1}}}$$

The desired infinite product is the limit $\lim_{N\to\infty} P_N(x)$. Since $\lim_{x\to a} \frac 1 f(x) = \frac 1 {\lim_{x \to a} f(x)}$ if the second limit exists and is non-zero, we have $$\lim_{N\to\infty} P_N(x) = \frac {\sin x}{\lim_{N\to\infty} 2^N \sin\frac x {2^N}} = \frac {\sin x} x$$

Thus we have $\int_0^{\infty} \frac {\sin x}{x} dx = \lim_{t\to\infty} \int_0^t \frac {\sin x}{x} dx$. This is the sine integral, $\operatorname{Si}(t)$. Integrating $\frac {\sin x} x$ from 0 to infinity is a well-known result. It's $\frac {\pi} 2$.There's one problem with the above: What if $\sin \frac x {2^N}$ is zero for some N?

 

[D H]

That $\sin \frac x{2^N} = 0$ for some positive integer N means that $x$ is an integral multiple of $2\pi$. There are two cases to worry about: $x=0$ and $x=2k\pi$ where $k$ is a positive integer. When $x=0$, $\cos \frac x{2^N}$ is identically equal to one for all N, and thus the infinite product is one in this case. Since $\lim x\to 0 \frac {\sin x } x = 1$, there's no problem at x=0.

The other problematic cases involve x being a positive integral multiple of $2\pi$. With these, there is always going to be one term in the infinite product that is zero, thus making the infinite product zero. At these points, $\frac {\sin x}x$ is zero, so there's no problem at $x=2k\pi$.Resources:
- Double angle formula $\sin(2t) = 2\sin t \cos t$
- Common knowledge that $\int_0^{\infty} \frac {\sin x} x dx = \frac{\pi} 2$.
- Simple handing of special cases $x=2k\pi$.

 

[Saitama]

DH already did most of the stuff. Only thing left to prove is the integral.

$$ \begin{aligned}
\int_0^{\infty} \frac{\sin x}{x}\,dx &= \int_0^{\infty} \int_0^{\infty} e^{-xt}\sin(x)\,dx\,dt\\
&=\int_0^{\infty} \Im\left(\int_0^{\infty} e^{-(i+t)x}\,dx\right)\,dt\\
&=\int_0^{\infty} \Im\left(\frac{1}{i+t}\right)\,dt\\
&=\int_0^{\infty} \frac{1}{1+t^2}\,dt \\
&=\frac{\pi}{2}
\end{aligned}$$

And yes, this problem is definitely very easy.