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I’d like to show for some $c,N>0$ and all positive integer $n>N$ that $$\int^\infty_{-\infty} \frac{\sin(x/n)}{x} \prod_{i=1}^n \cos\left(\frac xi\right)dx>\frac cn$$

If I just look at the integral near $0$, I can get that for $x$ much less than $1$, the cosine terms are positive and approximately quadratic and the sin term is odd and approximately linear, so the integrand is positive and around $(1-\pi^2 x^2/6)/n$ which when integrating from $-1$ to $1$ gives a positive value proportionate to $1/n$, but I’m having trouble bounding the potentially large oscillations of the tails.

I have noticed that $x$ and $2\pi k n! -x$ give the same value for the cosine product and opposite signs for the sine, so this could lead to a significant amount of cancellation, but making this rigorous seems tricky.

This problem came up while I was looking at the characteristic function of the partial sums of a random variant of the harmonic series while digging into this problem: How often do we expect a random walk with decreasing step size to cross $0$?

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    $\begingroup$ Do you mean $\prod_{i=1}^{n}\cos(x/i)$, not $\prod_{i=1}^{n}\cos(x/\color{red}{n})$? $\endgroup$ Jan 2 at 7:52
  • $\begingroup$ Yes, that’s fixed. $\endgroup$
    – Eric
    Jan 2 at 8:00
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    $\begingroup$ I expect that the integral has the following combinatorial interpretation, but I guess that you already know it. Let $n\ge 3$ and $D=\{-1,1\}^{n-1}$. For each sequence $x=(x_1,\dots,x_{n-1})\in D$ let $\Sigma x=1+\sum_{i=2}^n \frac {x_{i-1}}i$ and $D'=\left\{x\in D:|\Sigma x|<\frac 1n\right\}$. Then the integral equals $|D'|\frac \pi{2^n}$. $\endgroup$ Jan 18 at 19:38
  • $\begingroup$ I am a bit confused, since Eric's answer to the linked question already shows this combinatorial statement? @AlexRavsky $\endgroup$
    – abacaba
    Jan 19 at 2:18
  • $\begingroup$ Well that answer only shows that you can get within $2 / n$ distance $0$ with $\Theta(n^{-1})$ probability, but you can "reserve" more elements into $W_n$(e.g. reserve a pair $(x_i / i, x_j / j)$ such that $1/i - 1/j \in [1.5/n, 1/n]$ and another pair $(x_u / u, x_v / v)$ with $1/u - 1/v \in [0.75/n, 0.5/n]$) and get within $1/n$ distance of $0$. $\endgroup$
    – abacaba
    Jan 19 at 2:27

3 Answers 3

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Equivalently, this means $n$ times the proportion of $(e_1,…,e_n)\in \{−1,1\}^n$ such that $\sum_{k=1}^n e_k/k$ belongs to $[−1/n,1/n]$ is bounded below.

That is not too hard. We'll just use the cheap observation that if $2^m\le n<2^{m+1}$, then $\sum_{j=0}^m \frac{e_{2^j}}{2^j}$ has uniform distribution on the arithmetic progression of $2^{m+1}$ terms with step $2^{1-m}$ running essentially from $-2$ to $2$.

Now, if the requested interval were $[-2/n,2/n]$, that would be the end of the story because the expectation of the square of the rest of the series is less than $\sum_{k\ge 3}\frac 1{k^2}\le \sum_{k\ge 3}\frac 1{(k-1)k}=\frac 12$, so with probability $\ge \frac 12$, the rest adds up to a number in $[-1,1]$ and the shift of our arithmetic progression by that number would have at least one term in the desired interval (because $2^{1-m}<\frac 4n$), yielding the lower bound $2^{-m-2}\ge\frac 1{4n}$.

In our case the interval is a bit shorter, so for large $m$ and $n$ we will also consider $e_p,e_q$ with $\frac{2^{m-1}}p\approx \frac 54, \frac{2^{m-1}}q\approx \frac 98$. Then considering the contribution of $\frac{e_p}{p}+\frac{e_q}q$ as well, we see that we have a small perturbation of an arithmetic progression with step $2^{-m-1}<\frac 1n$, so we are still fine but with the bound $\frac 1{16n}$.

That takes care of large $n$, so all that remains is to show that the probability is never $0$. That can be achieved by the greedy choice of signs going in the direction of $0$ every time. The trivial induction shows that we'll be under or at $1/j$ in absolute value after adding $\pm 1/j$.

It is also worth mentioning that the integral is not this probability, but this probability minus one half of the probability that the sum is exactly $\pm 1/n$ (the Fourier integral for a function with nice jump discontinuity converges to the value in the middle of the jump). That doesn't change much though: the final bound merely drops at most twice more.

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  • $\begingroup$ Why not show something like : $$\frac{\sin(x/n)}{x} \prod_{i=1}^n \cos\left(\frac xi\right)=f(x),\int_{0}^{\infty}f(x)dx>\int_{0}^{\infty}n(f(x))^2dx$$ $\endgroup$ Jan 20 at 9:41
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    $\begingroup$ @DesmosTutu There is absolutely no reason why. I cannot do it this way, but you are welcome to try. Just keep in mind that $f$ changes sign :-) $\endgroup$
    – fedja
    Jan 20 at 9:57
  • $\begingroup$ I think we can use Euler product in the proof of the Vieta's formula and something like $$\cos(\pi+x/n)\simeq -\cos(x/2^m)cos(x/n)-sin(x/n)/n$$ $\endgroup$ Jan 20 at 11:14
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    $\begingroup$ @DesmosTutu "I think we can". Then just do it ;-) $\endgroup$
    – fedja
    Jan 20 at 13:21
  • $\begingroup$ If so I was right with Vieta's see math.stackexchange.com/questions/151997/… $\endgroup$ Jan 20 at 14:35
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Partial answer.

Let $(\epsilon_k)_{k \ge 1}$ be a sequence of independent Rademacher random variables and $U$ be a uniform random variable on $[-1,1]$, independent of $(\epsilon_k)_{k \ge 1}$. The characteristic function of the random variable $$Z_n := \frac{U}{n} + \sum_{k=1}^n \frac{\epsilon_k}{k}$$ is given by $$\phi_{z_n}(t) = \frac{\sin(t/n)}{t/n} \prod_{k=1}^n \cos\left(\frac{t}{k}\right).$$ If it was Lebesgue integrable on $\mathbb{R}$, then the random variable $Z_n$ would have a continuous density on $\mathbb{R}$, and the integral would be $2\pi$ times the density at $0$.

Actually, the random variable $$S_n := \sum_{k=1}^n \frac{\epsilon_k}{k}$$ is discrete, and `the' density of $Z_n$ is the step function given by $$f_{Z_n}(z) = \frac{1}{2^n}\sum_s f_{U/n}(z-s),$$ where $f_{U/n}$ is the density of $U/n$.

I still expect that we have
$$\int^\infty_{-\infty} \frac{\sin(t/n)}{t/n} \prod_{k=1}^n \cos\left(\frac {t}{k}\right)dt = 2\pi f_{Z_n}(0),$$ where we choose the density of $U/n$ as $f_{U/n} := (n/2)\mathbb{1}_{]-1/n,1/n[} + (n/4)\mathbb{1}_{\{-1/n,1/n\}}$ to have $2f_{Z_n}(z) = f_{Z_n}(z+) + f_{Z_n}(z-)$ at every point $z$.

Actually, the random variable $S_n$ cannot take the value $0$, since the $2$-adic valuation of the rational number $S_n$ is $2^{-\lfloor \log_2 n \rfloor}$, so $f_{Z_n}$ is continuous at $0$, and to compute we may take the density of $U/n$ equal to $(n/2)\mathbb{1}_{[-1/n,1/n]}$.

We have to prove that $f_{Z_n}(0)$ is bounded away from $0$. Equivalently, this means $n$ times the proportion of $(e_1,\ldots,e_k) \in \{-1,1\}^k$ such that $\sum_{k=1}^n e_k/k$ belongs to $[-1/n,1/n]$ is bounded below.

It is already known that the random harmonic series $\sum_k \frac{\epsilon_k}{k}$ converges almost surely, in $L^2$ and that its sum has a continuous and everywhere positive density. View Random Harmonic Series, The American Mathematical Monthly Volume 110, 2003 - Issue 5 https://doi.org/10.1080/00029890.2003.11919978 What we need here is a `local version' of this convergence in distribution .

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  • $\begingroup$ I don’t think this actually solves the problem? It does however do a good job of explaining the combinatorics which inspired the problem. $\endgroup$
    – Eric
    Jan 20 at 1:03
  • $\begingroup$ @Eric If that is where the problem came from, the direct proof without going to the Fourier side is quite simple: see my answer. $\endgroup$
    – fedja
    Jan 20 at 4:58
  • $\begingroup$ Why was my partial answer down voted? $\endgroup$ Jan 21 at 15:52
  • $\begingroup$ Sorry I believe it was you but see my answer I solve the problem in the big line (+1) $\endgroup$ Jan 22 at 8:37
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Some thoughts :

Integration over $(-1,1)$ :

First step : As commented we can use Polya-Szego inequality to give a bound in term of square integrable function as the function is positive and bounded it gives a $17/100<c<40/100$

Second step : as the oscillating remainders is not shown to be positive or negative we use Cauchy-Schwarz to give for $n$ sufficiently large:

$$\sqrt{\int_{1}^{\infty}\left(\prod_{i=1}^{\infty}\cos\left( x/i\right)\right)^2dx}\simeq C$$

The other integral called Dirichlet integral are well know

For some literature see https://arxiv.org/pdf/math/0411380

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  • $\begingroup$ In fact the integral above is lower bounded 1 with generalized holder inequality and if we use $x+1\leq e^x$ or modified Wirtinger inequality (reversed) we have a sufficient upper bound cutting at $\pi/2$ instead of one . $\endgroup$ Jan 23 at 9:49

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