Fourier series - How to calculate f(x) in a specific value?

Question

I need to solve this problem related with Fourier series:

${f(x)=\left\{\begin{matrix} -6& x\in [-3,0)\\ -4& x= 0\\ -2& x\in (0,3] \end{matrix}\right.}$

I would like to determine $a_{0}$, $a_{n}$ and $b_{n}$. Since this function is not a 2$\pi$-periodic function, I have to use $2L$ formula:

$a_{0}= \frac{1}{L}\int_{-L}^{L} f(x)dx$

$a_{n}= \frac{1}{L}\int_{-L}^{L} f(x) cos(\frac{n\pi}{L}x)dx$

$b_{n}= \frac{1}{L}\int_{-L}^{L} f(x) sin(\frac{n\pi}{L}x)dx$

$a_{0}$ after substitution:

$a_{0}= \frac{1}{L}\int_{-3}^{0} -6 dx + ??? + \frac{1}{L}\int_{0}^{3} -2dx$

I suppose that my main problem is clear: I don't know how to prescribe the second integral (???), because its $Df$ is not an interval, it is a specific value.

Answer 1

Welcome to MSE!

You can safely ignore it. Changing the value of a function at one point does not change the value of the integral (unless you're integrating with respect to some weird measure, which you aren't in this case). So just integrating $\frac{1}{L} \left ( \int_{-3}^0 -6\ dx + \int_0^3 -2 \ dx \right )$ will do what you want.

If you've ever heard that the fourier series only lets you reconstruct your function "almost everywhere", this is part of the reason why! By changing your function on some set of points, the fourier coefficients don't actually change. So the function you've described here has the same fourier coefficients as the function with any other value at $0$. That said, you can still recover the value of $f$ almost everywhere.

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I hope this helps ^_^