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I'm studying the following product:

$$p(a,\omega)=\prod_{k=1}^{\infty}a\sin (k\omega\pi),\quad \omega \in \Bbb R,\quad a\in \Bbb R_+.$$ It's easy to see that for $a\in (0,1]$ this product diverges to zero for all values of $\omega$. I want to estimate the rate of this divergence and find what to do when $a>1$.

For rational $\omega$ the product is zero after a finite number of multiplications. Suppose $\omega$ irrational, for simplicity, $\omega=1/\pi$. Then we have $$|p(a,1/\pi)|=\prod_{k=1}^{\infty}a|\sin (k)|.$$ We write a logarithm of this product: $$\sum_{k=1}^{\infty} \ln |a\sin (k)|.$$ On the period of $\sin$ the set $\{\Bbb N\mod \pi\}$ is dense, looks like (not formally) samples of uniformly distributed random variable on $[0,\pi]$, therefore, this sum seems to approach to integral:

$$\frac \pi N\sum_{k=1}^{N} \ln |a\sin (k)|\approx\int_0^\pi \ln|a\sin x|dx=:I(a),$$ so $$\sum_{k=1}^{N} \ln |a\sin (k)|\approx \frac N\pi I(a),$$ and $$\prod_{k=1}^{N} |a\sin (k)|\approx \exp\left(\frac N\pi I(a)\right).$$

Now we want to conjecture that if $I(a)<0$, then the product still diverges to zero and when $I(a)>0$, then the product diverges to infinity.

I have several question regarding my reasoning.

First, apparently, is that our "samples" are not truly random, therefore we need a Monte Carlo method with non-independent samples, etc. Is there an entry-level source on this subject?

Second, if we can deal with the previous question, is it possible to formalise the described approach and prove the conjecture rigorously? Unless, of course, there's already a proof of this fact somewhere. Anyway, I still want to generalise this aproach to the product of arbitrary periodic continuous functions.

Third, what to do with the case when $I(a)=0$?

I'll be glad to hear all critique and suggestions.

mathematics2x2life
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TZakrevskiy
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    I presume you meant 'converges to zero'... – copper.hat Jan 01 '14 at 21:38
  • @copper.hat: No. An infinite product is said to diverge to $0$ (in as much as $\ln0=-\infty$). – Lucian Jan 01 '14 at 22:45
  • @Lucian: Thanks! – copper.hat Jan 01 '14 at 22:47
  • The sequence $n\bmod\pi$ is uniformly distirbuted on $[0,\pi]$, but that's not enough to imply that the sum approaches the integral. For that, you need the integrand to be bounded and continuous, which it isn't. There are theorems that bound the difference between the integral and the sum. They involve the discrepancy of the sequence $n\bmod\pi$, which in turn involves diophantine approximation properties of $\pi$, which in turn involves partial quotients in the continued fraction for $\pi$. – Gerry Myerson Sep 16 '19 at 02:52
  • Note also that $$\int_0^{\pi}\log|a\sin x|,dx=\int\log|a|+\int\log|\sin x|=\pi\log|a|-\pi\log2$$ See also https://math.stackexchange.com/questions/381254/the-infinite-product-of-the-sines-of-all-positive-integers-is-zero – Gerry Myerson Sep 16 '19 at 02:57

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