오늘의 계산 47 - A Simple Integral Limit

Problem. Find the limit \begin{equation}\label{prob:lim} \lim_{n\to\infty} \frac{1}{n^{3/2}} \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n} \cos (kx)}{x^2} \; dx. \end{equation}

To find the value of \eqref{prob:lim}, we need some preliminaries.

Lemma 1. For any $\alpha \in \mathbb{R}$, we have \begin{equation*} \int_{-\infty}^{\infty} \frac{1 - \cos (\alpha x)}{x^2} \; dx = \pi |\alpha|. \end{equation*}

Proof. This is an immediate consequence of the Dirichlet integral. \begin{equation*} \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|. \end{equation*}

Lemma 2. For any $n \in \mathbb{N}$ and $\theta_1, \cdots, \theta_n$, we have \begin{equation*} \prod_{k=1}^{n} \cos \theta_k = \frac{1}{2^n} \sum_{\epsilon \in S} \cos\left( \epsilon_1 \theta_1 + \cdots + \epsilon_n \theta_n \right), \end{equation*} where $S = \{ -1, 1\}^{n}$.

Proof. This follows from the repeated application of the famous trigonometric formula \begin{equation*} 2 \cos A \cos B = \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right). \end{equation*}

Although it is possible to calculate \eqref{prob:lim} by brutal force^[1] of analysis, we are going to utilize the following famous version of the central limit theorem(CNT) in probability theory.

Theorem 3. (Lyapunov CNT)^[2] Let $\{ X_i \}$ be a sequence of independent random variables, each having a finite expected value $\mu_i$ and variance $\sigma_i^2$. Define $s_n^2 = \sigma_1^2 + \cdots + \sigma_n^2$. If for some $\delta > 0$, the Lyapunov’s condition \begin{equation}\label{thm3:lyap_cond} \lim_{n\to\infty} \frac{1}{s_n^{2+\delta}} \sum_{k=1}^{n} \mathbb{E}[|X_i - \mu_i|^{2+\delta}] = 0 \end{equation} is satisfied, then \begin{equation*} \frac{1}{s_n}\sum_{k=1}^{n} (X_k - \mu_k) \xrightarrow[ ]{d} N(0, 1) \end{equation*} as $n \to \infty$.

Then the solution goes as follows:

Solution of the Problem. Let $I_n$ be \begin{equation*} I_n := \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n} \cos (kx)}{x^2} \; dx, \end{equation*} and $S$ be as in Lemma 2. By Lemma 1 and 2, we have \begin{align*} I_n &= \frac{1}{2^n} \sum_{\epsilon \in S} \int_{-\infty}^{\infty} \frac{1-\cos(\epsilon_1 x + \cdots + n \epsilon_n x)}{x^2} \; dx \\ &= \frac{\pi}{2^n} \sum_{\epsilon \in S} \left|\epsilon_1 + \cdots + n \epsilon_n\right|. \end{align*} Consider $(X_k) = (k \epsilon_k)$ as a sequence of independent random variables satisfying \begin{equation*} \mathbb{P}(X_k = k) = \mathbb{P}(X_k = -k) = \frac{1}{2}. \end{equation*} Then the previous identity is rewritten, in probability language, as \begin{equation*} I_n = \pi \, \mathbb{E}\left| X_1 + \cdots + X_n \right|. \end{equation*} Since \begin{equation*} s_n^2 := \mathbb{V}(X_1) + \cdots + \mathbb{V}(X_n) = 1^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}, \end{equation*} the limit \eqref{prob:lim} is equivalent to \begin{equation*} \frac{\pi}{\sqrt{3}} \lim_{n\to\infty} \frac{1}{s_n} \, \mathbb{E}\left| X_1 + \cdots + X_n \right|. \end{equation*} Now, it is easy to verify that the Lyapunov's condition \eqref{thm3:lyap_cond} is satisfied for every $\delta > 0$. Thus by Theorem 3, we obtain the convergence in distribution \begin{equation*} \frac{X_1 + \cdots + X_n}{s_n} \xrightarrow[ ]{d} N(0, 1). \end{equation*} In particular, if $Z \sim N(0, 1)$, then \begin{equation*} \lim_{n\to\infty} \frac{1}{s_n} \, \mathbb{E}\left| X_1 + \cdots + X_n \right| = \mathbb{E}|Z| = \sqrt{\frac{2}{\pi}}. \end{equation*} Therefore we obtain \begin{equation*} \lim_{n\to\infty} \frac{1}{n^{3/2}} \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n} \cos (kx)}{x^2} \; dx = \frac{\pi}{\sqrt{3}} \lim_{n\to\infty} \frac{1}{s_n} \, \mathbb{E}\left| X_1 + \cdots + X_n \right| = \sqrt{\frac{2 \pi}{3}}. \end{equation*}

References

1. Compute $ I_{n}=\int_{-\infty}^\infty \frac{1-\cos x \cos 2x \cdots \cos nx}{x^2}\,dx$ in "Math StackExchange"
2. Central Limit Theorem - Lyapunov CLT in "Wikipedia"