# A trigonometric series calculation

Let $a$ be a real number. I'm asked to find the sum of the series: $$\sum_{n = 1}^{\infty} (-1)^n\frac{\sin(nx)}{n(n^2 + a^2)}$$

I've tried to represent the series as sum of two series: $$\sum_{n = 1}^{\infty} (-1)^n\frac{\sin(nx)}{a^2n} - \sum_{n = 1}^{\infty} (-1)^n\frac{\sin(nx)n}{a^2(n^2 + a^2)}$$

Where the first series in $[-\pi, \pi]$ is $-\frac{x}{2a^2}$. I've tried to go to the complex form for the second series and convert it to the Taylor series of $\ln(1 + x)$ but did not succeed.

Is it some well-known Fourier series? Can anybody give me a hint?

• I forgot that I started to wirte "\alpha" at some point and began to use "a" instead. Thank you, I fixed it. – Webber Nov 27 '17 at 21:32

## 2 Answers

Let $$y(x)=\sum_{n=1}^{\infty}(-1)^n\frac{\sin nx}{n(n^2+a^2)}$$ Since the denominator goes as $n^3$, $y^{\prime}(x)$ is uniformly continuous so we can differentiate twice to get $$a^2y-y^{\prime\prime}=\sum_{n=1}^{\infty}(-1)^n\frac{\sin nx}{n}$$ This looks familiar and indeed if $$x=\sum_{n=1}^{\infty}b_n\sin nx$$ Then $$b_n=\frac2{\pi}\int_0^{\infty}x\sin nxdx=\left.\frac2{\pi}\left(-\frac xn\cos nx+\frac1{n^2}\sin nx\right)\right|_0^{\pi}=-\frac2n(-1)^n$$ So we have $$y^{\prime\prime}-a^2y=\frac12x$$ Solution is $$y=-\frac x{2a^2}+C_1\cosh ax+C_2\sinh ax$$ Now, $y(0)=C_1=0$, and the denominator of the original Fourier series goes as $n^3$, so the series must be continuous, even at its wraparound points: $$y(-\pi)=-y(\pi)=y(\pi)=0=-\frac{\pi}{2a^2}+C_2\sinh a\pi$$ So our result is $$\sum_{n=1}^{\infty}(-1)^n\frac{\sin nx}{n(n^2+a^2)}=-\frac x{2a^2}+\frac{\pi\sinh ax}{2a^2\sinh \pi a}$$

Considering the function $$f(z)=\frac{\pi}{\sin \pi z}\frac{z\sin xz}{a^2}\left( z^2 +a^2\right)$$ Using Jordan lemma, one can show that its integral along the large circle centered at the origin, with a radius $R$, vanishes when $R\to\infty$ \begin{equation} I=lim_{R\to\infty}\int_{C_R}f(z)\,dz=0 \end{equation} The function has poles at $z=\pm ia$ with identical residues $\rho_{\pm}=\frac{\pi\sinh ax}{2a^2\sinh \pi a}$ and poles at $z=n$, where $n$ is an integer. The corresponding residues are $\rho_n=(-1)^n\frac{n\sin nx}{a^2\left( n^2+a^2 \right)}$ which are exactly the terms of the series. In the contour, the series is counted twice (for positive and negative $n$, as the residues are even function of $n$. Then, from the residue theorem \begin{equation} I=0=2i\pi\left[2\sum_{n=0}^\infty\rho_n+2\rho_\pm\right] \end{equation} Finally \begin{equation} \sum_{n=0}^\infty(-1)^n\frac{n\sin nx}{a^2\left( n^2+a^2 \right)}=-\frac{\pi\sinh ax}{2a^2\sinh \pi a} \end{equation} This expression is valid for $-\pi<x<\pi$ and extended periodically to all reals.