Length of the arc of a Fourier series

I'm working modeling the behavior of periodic variable stars and I have a question about reducing the expression of a parameter involved in this analysis.

Let $$f(t)$$ be a Fourier series define as:

$$f(t)= \langle m \rangle + \displaystyle{\sum_{k=1}^{N} A_{k}\sin \left(\frac{2\pi kt}{P} + \phi_{k}\right)}.$$

The variable $$t$$ is real and defined in the so-called phase space, where $$0\leq t \leq 1$$. These series describe a periodic process with period $$P$$, $$A_i$$ and $$\phi_i$$ the individual Amplitude and phases for all the $$N$$ harmonics; and $$\langle m \rangle$$ the average value of the time-series that contains the process. All the shown variables are known. I want to figure out the length of this Fourier series using the definition:

$${\displaystyle s=\int _{a}^{b}{\sqrt {1+\left(f'(x)\right)^{2}}} dx.}$$

If we define $$x_k:= \frac{2\pi k}{P}$$, the derivative of the Fourier series $$f'(t)$$ will be:

$$f'(t) = \displaystyle{ \sum_{k=1}^{N} A_{k} x_{k} \cos(x_{k} t + \phi_{k}).}$$

Then: $$\begin{split} (f'(t))^{2}=\displaystyle{\left(\sum_{k=1}^{N} A_{k} x_{k} \cos(x_{k} t + \phi_{k})\right)^{2}}\\ = \left(\sum_{i=1}^{N} A_{i} x_{i} \cos(x_{i} t + \phi_{i})\right) \times \left(\sum_{j=1}^{N} A_{j} x_{j} \cos(x_{j} t + \phi_{j})\right) \\ = \sum_{i,j=1}^{N} a_i a_j \cos(x_{i} t + \phi_{i}) \cos(x_{j} t + \phi_{j}), \end{split}$$

where we also define $$a_k := A_k x_k$$. Finally our lenght will be:

$${\displaystyle s=\int _{a}^{b}{\sqrt {1+ \sum_{i,j=1}^{N} a_i a_j \cos(x_{i} t + \phi_{i}) \cos(x_{j} t + \phi_{j}) }}\ \ dt.}\ \ (1)$$

The integration limits are $$a=0$$ and $$b=1$$, given we live in this convenient phase space. Now comes the big question:

Did you have some idea or trick to attack and solve this integral? I tried with some trigonometric manipulations, but I got not great progress. I'm trying to obtain the less-expensive function to compute and so far, is the expression (1).

Thanks for reading this and every kind of feedback will be awesome :)

• by "solve" you mean a closed form expression? that is not likely to be forthcoming, but you can of course evaluate it numerically. – Carlo Beenakker Sep 22 at 6:02
• In the case a Fourier series approximating a jump function (i.e. when Gibbs phenomenon kicks in), there is this paper "Gibbs' phenomenon and arclength" (link.springer.com/article/10.1007/BF02511544) by Strichartz but the paper does not seem to contain a technique to calculate these integrals… – Dirk Sep 22 at 14:05
• @CarloBeenakker By solve I mean I want to obtain the more compact expression to compute numerically. – Nicolás Medina Sep 22 at 19:31
• Thanks @Dirk for the paper, I will look it carefully ;) – Nicolás Medina Sep 22 at 19:32