Integral related to sine series

이번 포스팅에서는 푸리에 사인 급수와 관련된 적분을 다뤄보겠습니다.

 

Theorem. Let $f:(0,\infty)\to\mathbb{C}$ be a locally bounded Lebesgue-measurable function such that the improper integral $$
\int_{0}^{\infty} \frac{f(x)}{x} \, \mathrm{d}x = \lim_{\substack{a\to 0^+ \\ b\to\infty}} \int_{a}^{b} \frac{f(x)}{x} \, \mathrm{d}x
$$ exists in $\mathbb{C}$. Then, for any $(a_n)_{n\geq 1}\subset\mathbb{C}$ satisfying $\sum_{n\geq1} |a_n| < \infty$, we have $$
\int_{0}^{\infty} \frac{1}{x}\left[ \sum_{n=1}^{\infty} a_n f(nx) \right] \, \mathrm{d}x = \biggl( \int_{0}^{\infty} \frac{f(x)}{x} \, \mathrm{d}x \biggr) \biggl( \sum_{n=1}^{\infty} a_n \biggr).
$$ Here, the convergence of the left-hand side is part of the statement to be established.

Proof. The assumption ensures the existence of an antiderivative $G(x)$ of $f(x)/x$ on $(0,\infty)$, in the sense that

$$
\int_{a}^{b} \frac{f(x)}{x} \, \mathrm{d}x = G(b) - G(a)
$$

for any $0 < a < b < \infty$. Moreover, both $G(0):=\lim_{x\to0^+}G(x)$ and $G(\infty)=\lim_{x\to\infty}G(x)$ exist in $\mathbb{C}$ by the assumption. So, thls allows us to regard $G$ as a continuous function on $[0,\infty]$. Now, by the Weierstrass $M$-test, the series $x^{-1} \sum_{n=1}^{\infty} a_n f(nx) $ converges uniformly on any compact interval $[a, b]$ of $(0, \infty)$. So,

\begin{align*}
\int_{a}^{b} \frac{1}{x}\left[ \sum_{n=1}^{\infty} a_n f(nx) \right] \, \mathrm{d}x
&= \sum_{n=1}^{\infty} a_n \int_{a}^{b} \frac{f(nx)}{x} \, \mathrm{d}x \\
&= \sum_{n=1}^{\infty} a_n [G(nb) - G(na)].
\end{align*}

Since the last sum converges unifornly on $[0, \infty]$ by the Weierstrass $M$-test, the desired conclusion follows by letting $a\to0^+$ and $b\to\infty$. $\square$

 

아래는 위 정리의 예제입니다.

 

Example 1. We have
\begin{align*}
\int_{0}^{\infty} \frac{\operatorname{Im}\exp(e^{ix})}{x} \, \mathrm{d}x
&= \int_{0}^{\infty} \frac{1}{x} \left[ \sum_{n=1}^{\infty} \frac{\sin(nx)}{n!} \right] \, \mathrm{d}x \\
&= \biggl( \sum_{n=1}^{\infty} \frac{1}{n!} \biggr) \biggl( \int_{0}^{\infty} \frac{\sin x}{x} \, \mathrm{d}x \biggr) \\
&= \frac{\pi}{2}(e - 1).
\end{align*}

 

Example 2. We compute the integral $$
I = \int_{0}^{\infty} \frac{ \left|\cos\left(x-\frac{\pi}{4}\right)\right| - \left| \cos\left(x+\frac{\pi}{4}\right) \right| }{x} \, \mathrm{d}x.
$$ Using the Fourier series
\begin{align*}
\left|\cos\left(x-\frac{\pi}{4}\right)\right| - \left| \cos\left(x+\frac{\pi}{4}\right) \right|
& = \frac{8}{\pi} \sum_{k=0}^{\infty} \frac{(-1)^k}{(4k+2)^2 - 1} \sin((4k+2)x),
\end{align*} it follows that
\begin{align*}
I
&= \frac{\pi}{2} \biggl( \frac{8}{\pi} \sum_{k=0}^{\infty} \frac{(-1)^k}{(4k+2)^2 - 1} \biggr) \\
&= 2 \left( 1 - \frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - \frac{1}{11} - \frac{1}{13} + \frac{1}{15} + \cdots \right). \end{align*} The last sum can be evaluated by invoking the Fourier series for $\log \left| \tan x \right| $: $$
\log\left|\tan x\right| = -2 \sum_{k=0}^{\infty} \frac{\cos(2(2k+1)x)}{2k+1}.
$$ Indeed, plugging $x = \frac{\pi}{8}$ into the formula above, we get
\begin{align*} \log\left(\tan \frac{\pi}{8}\right) &= -2 \sum_{k=0}^{\infty} \frac{\cos((2k+1)\frac{\pi}{4})}{2k+1} \\ &= -\sqrt{2} \left( 1 - \frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - \frac{1}{11} - \frac{1}{13} + \frac{1}{15} + \cdots \right) \\ &= -\frac{I}{\sqrt{2}}. \end{align*} Therefore $$
I = -\sqrt{2}\log\left(\tan \frac{\pi}{8}\right) = \sqrt{2}\log(1+\sqrt{2}). $$