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I'm trying to tackle the following question:

Use Fourier series of the function $f(x)=x(\pi+|x|)$ in $[-\pi,\pi]$ to compute the infinite sum $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{(2n-1)^3}$$

So, I found that the coefficients of the series are $\displaystyle c_n=\begin{cases}0,& n\text{ even} \\ \frac{8}{in^3},& n\text{ odd}\end{cases}$.

How should I continue? how do the coefficients help me compute the sum?

Please help, thank you!

Galc127
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2 Answers2

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That is a special case of a general identity proved here.

With a Fourier-analytic approach, we may notice that: $$ \sum_{n\geq 1}\frac{(-1)^{n+1} \sin((2n-1)x)}{(2n-1)^2} $$ is the Fourier sine series of a triangle wave, hence: $$ \sum_{n\geq 1}\frac{(-1)^n}{(2n-1)^3} = -\int_{0}^{\pi/2}\left(\frac{\pi x}{4}\right)\,dx = \color{red}{-\frac{\pi^3}{32}}.$$

Community
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Jack D'Aurizio
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  • First, thank you for your answer. Second, how finding the coefficients help me? – Galc127 Nov 05 '15 at 18:56
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    @Galc127: my answer gives another approach, but if you know a function $f(x)$ such that $f(x)=\sum_{k\geq 0}\frac{8\sin((2k-1)x)}{(2k-1)^3}$, you just need to evaluate it at $x=\frac{\pi}{2}$ to recover the value of the given series. – Jack D'Aurizio Nov 05 '15 at 18:59
  • Perfect, thanks a lot! – Galc127 Nov 05 '15 at 19:55
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You can use Poisson summation formula on $F$ which extends $f$ to be zero outside $[-\pi,\pi]$.

Ben
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  • Unfortunately, I'm not familiar with Poisson summation and I don't think we need to use it here. – Galc127 Nov 05 '15 at 18:50