<s>[INST] The fourier series $\sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a}$ A Fourier series arising in perturbation theory in quantum mechanics is
$$\sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} \, .$$
where $n$ is an odd positive integer and $n$ runs through all odd positive integers other than $n$. (The numbers are odd so that the Fourier terms are zero at $\pm a$.)
I have no idea what kind of function produces this series. Is it familiar to anyone?
 [/INST] Taking $n=1$ and $a= \frac{1}{2}$, wolframalpha gives me $$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{1-m^2} = \frac{1}{4}(\pi(1-2x)\sin(\pi x) - \cos(\pi x))$$
Taking $n=3$ and $a= \frac{1}{2}$, wolframalpha gives me $$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{9-m^2} = \frac{1}{36}(3\pi(1-2x)\sin(3\pi x) - \cos(3\pi x))$$
Following this my hunch would be for $a= \frac{1}{2}$
$$\displaystyle \sum_{m = \text{odd}}^{\infty} \frac{\cos(m \pi x)}{n^2-m^2} = \frac{1}{4n^2}(n\pi(1-2x)\sin(n\pi x) - \cos(n\pi x))$$
Playing around a bit more with wolfram alpha, my new guess is 
For $x \in [0,a]$,
$$\displaystyle \sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} = \frac{1}{4n^2} \left(n\pi \left(1-\frac{x}{a} \right) \sin \left(\frac{n\pi x}{2a} \right) - \cos \left(\frac{n\pi x}{2a} \right) \right)$$
and For $x \in (-a,0]$,
$$\displaystyle \sum_{m\neq n} \frac{1}{n^2 - m^2} \cos \frac{m\pi x}{2a} = \frac{1}{4n^2} \left(n\pi \left(-1-\frac{x}{a} \right) \sin \left(\frac{n\pi x}{2a} \right) - \cos \left(\frac{n\pi x}{2a} \right) \right)$$
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