3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ I forgot that I started to wirte "\alpha" at some point and began to use "a" instead. Thank you, I fixed it. $\endgroup$ – Webber Nov 27 '17 at 21:32
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.