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I need to calculate the coefficients $$c_n=\frac{1}{2L}\int_{-L}^Lf(x)e^{-\frac{in\pi x}{L}}dx,\qquad n=0,\pm1,\pm2,\cdots$$ of the complex Fourier series for the function: $$f(x)=\begin{cases}-x\ &0<x<1,\\ x\ &1<x<2,\\ \end{cases}$$ but clearly this function is not defined over an interval of the form $(-L,L)$, but $(0,2)$.

I already know that, for the trigonometric real Fourier series, I can fix this by projecting the function in a certain way over $(-2,0)$, making $f(x)$ an ever, odd or periodic function. I thought I could use this for my case, but how do I apply this for the complex series? Do I need to "re-transform" $$e^{-\frac{in\pi x}{L}}=\cos\left(\frac{n\pi x}{L}\right)-i\sin\left(\frac{n\pi x}{L}\right)?$$

Xander Henderson
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Armando Rosas
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    The function is periodic. You can always integrate over a symmetric interval if you like. That being said, $2L$ is just the period of the function, so you can replace it with the period. – Xander Henderson Mar 20 '20 at 20:06
  • @XanderHenderson So I suppose that my $f$ is periodic over the interval $(-2,2)$, to make the period and the limit $2L=2,\ L=1$? – Armando Rosas Mar 20 '20 at 20:14
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    What do you mean "So I suppose that my function is periodic over the interval $(-2,2)$"? The function you describe is defined on the interval $(0,2)$, is periodic on $\mathbb{R}$, and has period $2$. Integrate over $[0,2]$, $[-1,1]$, or any other interval of length $2$. In this context, yes, $L=1$. – Xander Henderson Mar 20 '20 at 20:18
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    It may also be worth noting that $$c_n = \frac{1}{P} \int_{a}^{a+P} f(x) \mathrm{e}^{2\pi n x / P} ,\mathrm{d}x, $$ where $P$ is the period of the function, and $a$ is any real number. That is, you can find the Fourier coefficients by integrating over any interval of length $P$; because the function is periodic, it doesn't matter which interval you integrate over. – Xander Henderson Mar 20 '20 at 20:21
  • I would take $L=1$ and extend $f(x) = f(x+2)$ for $x \in [-1,0]$. – copper.hat Mar 20 '20 at 20:56

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