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?
 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.
A: 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
$$\sum_{n=1}^{\infty}(-1)^n\frac{\sin nx}{n(n^2+a^2)}=-\frac x{2a^2}+\frac{\pi\sinh ax}{2a^2\sinh \pi a}$$
