calculusdefinite integralsintegrationlimitsreal-analysis
Compute the following limit:
$$\lim_{n \to \infty}\int_{0}^{2\pi}\cos\left(x\right)\cos\left(2x\right)\ldots
\cos\left(nx\right)\,{\rm d}x$$
Today I was working on a W. L. Putnam competition's problem containing this
integral and wondered how I may compute its value when $n$ goes to $\infty$. So far I've found no answer.
Could you give me some suggestions about the way I should go?
First note that $$\int_{-\infty}^{\infty} \frac{1-\cos ax}{x^2} \; dx = \left[ -\frac{1-\cos ax}{x}\right]_{-\infty}^{\infty} + a \int_{-\infty}^{\infty} \frac{\sin ax}{x} \; dx = \pi \, |a|,$$ by the Dirichlet integral. Also, by mathematical induction we can easily prove that $$ \prod_{k=1}^{n} \cos \theta_k = \frac{1}{2^n} \sum_{\mathrm{e}\in S} \cos\left( e_1 \theta_1 + \cdots + e_n \theta_n \right),$$ where the summation runs over the set $S = \{ -1, 1\}^n$. Thus we have $$ \begin{align*} I_n = \int_{-\infty}^{\infty} \frac{1-\cos x \cdots \cos nx}{x^2} \; dx &= \frac{1}{2^n} \sum_{\mathrm{e}\in S} \int_{-\infty}^{\infty} \frac{1-\cos(e_1 x + \cdots + e_n nx)}{x^2} \; dx \\ &= \frac{\pi}{2^n} \sum_{\mathrm{e}\in S} \left|e_1 + \cdots + e_n n\right|. \end{align*}$$ For example, if $n = 3$, we have $\left|\pm 1 \pm 2 \pm 3\right| = 0, 0, 2, 2, 4, 4, 6, 6$ and hence $$I_3 = \frac{\pi}{8}(0 + 0 + 2 + 2 + 4 + 4 + 6 + 6) = 3\pi.$$ Let the summation part as $$ A_n = \sum_{\mathrm{e}\in S} |e_1 + \cdots + e_n n|.$$ The first 10 terms of $(A_n)$ are given by $$ \left(A_n\right) = (2, 8, 24, 72, 196, 500, 1232, 2968, 7016, 16280, \cdots ), $$ and thus the corresponding $(I_n)$ are given by $$ \left(I_n\right) = \left( \pi ,2 \pi ,3 \pi ,\frac{9 \pi }{2},\frac{49 \pi }{8},\frac{125 \pi }{16},\frac{77 \pi }{8},\frac{371 \pi }{32},\frac{877 \pi }{64},\frac{2035 \pi }{128} \right).$$ So far, I was unable to find a simple formula for $(A_n)$, and I guess that it is not easy to find such one.
p.s. The probability distribution of $S_n = e_1 + \cdots + e_n n$ is bell-shaped, and fits quite well with the corresponding normal distribution $X_n \sim N(0, \mathbb{V}(S_n))$. Thus it is not bad to conjecture that $$ \frac{A_n}{2^n} = \mathbb{E}|S_n| \approx \mathbb{E}|X_n| = \sqrt{\frac{n(n+1)(2n+1)}{3\pi}},$$ and hence $$ I_n \approx \sqrt{\frac{\pi \, n(n+1)(2n+1)}{3}}.$$ Indeed, numerical experiment shows that
I was able to prove a much weaker statement: $$ \lim_{n\to\infty} \frac{I_n}{n^{3/2}} = \sqrt{\frac{2\pi}{3}}. $$ First, we observe that for $|x| \leq 1$ we have $$ \log \cos x = -\frac{x^2}{2} + O\left(x^4\right).$$ Thus in particular, $$ \sum_{k=1}^{n} \log\cos\left(\frac{kx}{n}\right) = \sum_{k=1}^{n}\left[-\frac{k^2 x^2}{2n^2} + O\left(\frac{k^4x^4}{n^4}\right)\right] = -\frac{nx^2}{6} + O\left(x^2 \vee nx^4\right).$$ Now let $$ \begin{align*}\frac{1}{n^{3/2}} \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n}\cos (kx)}{x^2} \; dx &= \frac{1}{\sqrt{n}} \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n}\cos \left(\frac{kx}{n}\right)}{x^2} \; dx \qquad (nx \mapsto x) \\ &= \frac{1}{\sqrt{n}} \int_{|x|\leq 1} + \frac{1}{\sqrt{n}} \int_{|x| > 1} =: J_n + K_n. \end{align*}$$ For $K_n$, we have $$ \left|K_n\right| \leq \frac{1}{\sqrt{n}} \int_{1}^{\infty} \frac{2}{x^2}\;dx = O\left(\frac{1}{\sqrt{n}}\right).$$ For $J_n$, the substitution $\sqrt{n} x \mapsto y$ gives $$ \begin{align*} J_n &= \frac{1}{\sqrt{n}} \int_{|x|\leq 1} \left( 1 - \exp\left( -\frac{nx^2}{6} + O\left(x^2 \vee nx^4\right) \right) \right) \; \frac{dx}{x^2} \\ &= \int_{|y|\leq\sqrt{n}} \left( 1 - \exp\left( -\frac{y^2}{6} + O\left(\frac{y^2}{n}\right) \right) \right) \; \frac{dy}{y^2} \\ &\xrightarrow[]{n\to\infty} \int_{-\infty}^{\infty} \frac{1 - e^{-y^2/6}}{y^2} \; dy \\ &= \left[-\frac{1-e^{-y^2/6}}{y}\right]_{-\infty}^{\infty} + \frac{1}{3} \int_{-\infty}^{\infty} e^{-y^2/6} \; dy = \sqrt{\frac{2\pi}{3}}. \end{align*}$$ This completes the proof.
By letting $t=x/2$, we have that $$\int_{1/n}^{n}\cos(x/2)\frac{\ln (x)}{x}dx=\int_{1/(2n)}^{n/2}\cos(t)\frac{\ln(t) +\ln(2)}{t}dt$$ Hence $$\begin{align}\int_{1/n}^{n}\left(\cos x-\cos(x/2)\right)\frac{\ln(x)}{x}\,dx&= \int_{n/2}^{n}\frac{\cos(x)}{x}\ln(x)dx-\int_{1/(2n)}^{1/n}\frac{\cos(x)}{x}\ln(x)\,dx \\ &\qquad-\ln(2)\int_{1/(2n)}^{n/2}\frac{\cos(x)}{x}\,dx.\end{align}$$ Now show that $$\begin{align} &\lim_{n\to \infty}\int_{n/2}^{n}\frac{\cos(x)}{x}\ln(x)\,dx=0,\\ &\lim_{n\to \infty}\int_{1/(2n)}^{1/n}\frac{\cos(x)-1}{x}\ln(x)\,dx=0. \end{align}$$ Morever $$\lim_{n\to \infty}\left(\int_{1/(2n)}^{n/2}\frac{\cos(x)}{x}\,dx-\ln(n)\right) =\lim_{n\to \infty} (-\text{Ci}(\frac{1}{2n})-\ln(n))=\ln(2)-\gamma.$$ where $\text{Ci}(x)$ is the cosine integral (recall that $\text{Ci}(x)=\ln(x)+\gamma+o(1)$, as $x\to 0$).
Therefore $$\int_{1/(2n)}^{n/2}\frac{\cos(x)}{x}dx=\ln(n)+\ln(2)-\gamma+o(1),$$ and $$\int_{1/(2n)}^{1/n}\frac{\cos(x)}{x}\ln(x)\,dx =-\ln(2)\ln(n)-\frac{\ln^2(2)}{2}+o(1).$$ Thus we may conclude that $$\lim_{n \to \infty}\int_{1/n}^{n}\left(\cos x-\cos(x/2)\right)\frac{\ln (x)}{x} \, dx=\boxed{\gamma \ln (2)-\frac{\ln^2(2)}{2}}.$$
Best Answer
Let $u = e^{ix}$. Then $\cos x \cos 2x ... \cos nx = \frac{1}{2^n}\prod_{k=1}^n (u^k + u^{-k})$.
Write $$\prod_{k=1}^n (z^k + z^{-k}) = \sum_{j=-N}^N a_j z^j$$
where $N=\frac{n(n+1)}2$.
Then the integral above is just $\frac{2\pi}{2^n}a_0$.
Combinatorially, we can see that $a_0$ is the number of ways of picking a subset of $\{1,2,...,n\}$ which adds up to $\frac{n(n+1)}{4}$.
Now, the set of subsets of $\{1,...,n\}$ that add up to a particular value is an antichain, so we have a bound on the size of $a_0$, namely, $$0\leq a_0 \leq \binom {n}{\lfloor n/2\rfloor}$$.
And we know that $$\lim_{n\to\infty}\frac{1}{2^n}\binom {n}{\lfloor n/2\rfloor} =0$$
So we know your limit is zero.
(The maximum size of an antichain in the Boolean poset is Sperner's theorem.)
Note that when $n\equiv 1,2\pmod 4$, $\frac{n(n+1)}4$ is not an integer, so in those cases, $a_0=0$. Anybody have an "interesting" lower bound on $a_0$ for the other cases?
[In comments below, Robert Israel shows that when $n\equiv 0,3\pmod 4$, $a_0\geq 2^{\lfloor (n+1)/4\rfloor}$. I suspect that there is a polynomial, $q$, such that $a_0\geq \frac{2^n}{q(n)}$.]
This proof generalizes. Given any sequence of positive integers, $\{b_1,...,b_n\}$, we have the same result:
$$\int_{0}^{2\pi} \cos b_1 x \cos b_2 x ... \cos b_n x dx= \frac{2\pi}{2^n}A$$
where $A$ is the count of an antichain, so $0\leq A\leq \binom {n}{\lfloor n/2\rfloor}$
If $b_i=1$ for all $i$, then we get exact equality when $n$ is even, which is to say that:
$$\int_{0}^{2\pi} \cos^{2m} x dx = \frac{2\pi}{2^{2m}} \binom {2m}{m}$$