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$$\sum_{n=1}^∞ \frac{(-1)^n\sin(nx)}{n^3}$$

I'm brushing up on Fourier series. I was looking at deriving the Fourier series for:

$$ f(x) = x^3 - \pi^2 x, x\in [-\pi,\pi] $$

I have an undergrad in maths but my skills are rusty, and I don't recall how to approach sums like this. I tried googling the result, but couldn't find anything. Wolfram Alpha didn't offer a solution either.. I'm not sure if this is a well known series, or if it's even possible to simplify. Any guidance would be much appreciated!

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    $\begingroup$ If $f(y)=\sum_n \frac{y^n}{n^3}$ then your function is `the imaginary part of $$f\left(-e^{ix}\right)$$ which can be written as $$\frac{f(-e^{ix})-f(-e^{-ix})}{2i}$$ I don't see a reason for it to have a name. $f$ might have a name $\endgroup$ Commented Sep 21 at 1:19
  • $\begingroup$ I don't have much time to do the full calculations, but here's something that most likely works: Note that $(-1)^n\sin(nx)=\sin(n(x+\pi))$ and $\sum_{n\ge1}\sin(ny)/n$ has a closed formula valid for $0 < y < 2 \pi$, so when you integrate twice you get the formula for your sum, where $y = x + \pi$. $\endgroup$ Commented Sep 21 at 1:26
  • $\begingroup$ Have a look at en.wikipedia.org/wiki/Polylogarithm and use what @Thomas Andrews suggested. Simple expression and nice function. $\endgroup$ Commented Sep 21 at 2:21

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As I wrote in comments, the definition of the polylogarithm function is $$\text{Li}_s(z)=\sum_{n=1}^\infty \frac {z^n}{n^s}$$

Using Euler representation of the trigonometric function, you then have $$\sum_{n=1}^∞ \frac{(-1)^n\sin(nx)}{n^3}=\frac{i}{2} \Big(\text{Li}_3\left(-e^{-i x}\right)-\text{Li}_3\left(-e^{i x}\right)\Big)$$ which is very close to $-\sin(x)$.

It is also related to the Clausen function as you already wrote.

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  • $\begingroup$ Thank you for the response! This was the path I ended up going down $\endgroup$ Commented Sep 21 at 4:49
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After Googling, I learned that the function is definitionally just a phase shift of the Clausen function of order three.

$$ f(x) = Cl_3(x+\pi) $$

Pretty cool that taking a Fourier series of a cubic equation can lead to the world of Dirichlet L-series, polylogarithms, and even Bernoulli polynomials.

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    $\begingroup$ What cubic equation? You never mentioned cubic equations in the title. It might have helped readers find the rights wa if you told them how you got the function. $\endgroup$ Commented Sep 21 at 1:37
  • $\begingroup$ @ThomasAndrews Good point. It was $$ f(x) = x^3 - \pi^2 x $$ $\endgroup$ Commented Sep 21 at 2:57
  • $\begingroup$ Put it in te question. Help people help you. $\endgroup$ Commented Sep 21 at 3:15
  • $\begingroup$ @ThomasAndrews will do! $\endgroup$ Commented Sep 21 at 4:44
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Actually, there is a very simple answer to your question.

For representing a non-periodic function on the interval $[-L,L]$, we use \begin{equation*} f(x)=k_1+\frac{k_2x}{L}+\sum_{n=1}^{\infty} \left(A_n\cos\frac{n\pi x}{L}+B_n\sin\frac{n\pi x}{L}\right). \end{equation*} By evaluating the above series at $x=-L,L$, we get \begin{equation*} k_2=\frac{f(L)-f(-L)}{2}. \end{equation*} Let \begin{equation*} \tilde{f}(x)=f(x)-\frac{k_2x}{L}. \end{equation*} Then using the orthogonality of the sine and cosine functions, we get \begin{align*} k_1&=\frac{1}{2L}\int_{-L}^L \tilde{f}(\hat{x})\,d\hat{x}, \\ A_n&=\frac{1}{L}\int_{-L}^L \tilde{f}(\hat{x})\cos\frac{n\pi\hat{x}}{L}\,d\hat{x}, \\ B_n&=\frac{1}{L}\int_{-L}^L \tilde{f}(\hat{x})\sin\frac{n\pi\hat{x}}{L}\,d\hat{x}. \end{align*} For your function with $L=\pi$, we get \begin{equation*} x^3-\pi^2x=12\sum_{n=1}^{\infty} \frac{(-1)^n\sin(nx)}{n^3}, \end{equation*} from which you get your desired summation.

For why the $x$ term should be present while representing non-periodic functions, see https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2022-03/mr_3_2022_fourier.pdf

Besides eliminating the Gibbs phenomenon, you will get a much faster rate of convergence with the $x$ term present compared to when it is absent.

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