# Showing $\int^\infty_{-\infty} \frac{\sin(x/n)}{x} \prod_{i=1}^n \cos(x/i)dx>c/n$, for some $c,N>0$ and all positive integers $n>N$

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 $0$?

• Do you mean $\prod_{i=1}^{n}\cos(x/i)$, not $\prod_{i=1}^{n}\cos(x/\color{red}{n})$? Jan 2 at 7:52
• Yes, that’s fixed.
– Eric
Jan 2 at 8:00
• 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 $|D'|\frac \pi{2^n}$. Jan 18 at 19:38
• I am a bit confused, since Eric's answer to the linked question already shows this combinatorial statement? @AlexRavsky Jan 19 at 2:18
• 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 $0$. Jan 19 at 2:27

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.

• Why not show something like : $$\frac{\sin(x/n)}{x} \prod_{i=1}^n \cos\left(\frac xi\right)=f(x),\int_{0}^{\infty}f(x)dx>\int_{0}^{\infty}n(f(x))^2dx$$ Jan 20 at 9:41
• @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 :-) Jan 20 at 9:57
• I think we can use Euler product in the proof of the Vieta's formula and something like $$\cos(\pi+x/n)\simeq -\cos(x/2^m)cos(x/n)-sin(x/n)/n$$ Jan 20 at 11:14
• @DesmosTutu "I think we can". Then just do it ;-) Jan 20 at 13:21
• If so I was right with Vieta's see math.stackexchange.com/questions/151997/… Jan 20 at 14:35

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 .

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

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

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

• 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 . Jan 23 at 9:49