Prove an integral formula for ∑ (sin (nx)) / n2

Prove that the series

$\sum_{n=1}^{\infty} \frac{\sin (nx)}{n^2}$

converges for all $x \in \mathbb{R}$ and let $f(x)$ denote the value of this sum for each $x$ . Prove that $f(x)$ is continuous for $x \in [0, \pi]$ and prove that

$\int_0^{\pi} f(x) \, dx = 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3}.$

Proof. First, the series converges for all real $x$ by the comparison test since

$|\sin (nx)| \leq 1 \quad \implies \quad \left|\frac{\sin (nx)}{n^2}\right| \leq \frac{1}{n^2}$

for all $n$ . Therefore, the convergence of $\sum \frac{1}{n^2}$ implies the convergence of $\sum \frac{\sin (nx)}{n^2}$ . Furthermore, this convergence is uniform by the Weierstrass $M$ -test with $M_n$ given by $\frac{1}{n^2}$ , and again $\sum \frac{1}{n^2}$ converges. Thus, by Theorem 11.2 (page 425 of Apostol),

$f(x) = \sum_{n=1}^{\infty} \frac{\sin (nx)}{n^2}$

is continuous on the interval $[0, \pi]$ . Therefore, we may apply Theorem 11.4 (page 426 of Apostol):

$\begin{align*} && \sum_{k=1}^{\infty} \int_0^{\pi} \frac{\sin (kt)}{k^2} \, dt &= \int_0^{\pi} \sum_{k=1}^{\infty} \frac{ \sin (kt)}{t^2} \, dt \\[9pt] \implies && \sum_{k=1}^{\infty} \left( \frac{-\cos (kt)}{k^3} \Bigr \rvert_0^{\pi} \right) &= \int_0^{\pi} f(x) \, dx \\[9pt] \implies && \int_0^{\pi} f(x) \, dx &= 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3} \end{align*}$

since $\cos 0 - \cos (k \pi) = 0$ if $k = 2n$ and equals 2 if $k = 2n-1. \qquad \blacksquare$

Stumbling Robot

Prove an integral formula for ∑ (sin (nx)) / n²

Related

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply