Fourier series Even vs. Odd and effect of integral bounds? I understand that when you express a function in fourier series there are 3 coefficients you need to calculate ( a0, an, bn) and I have in the past made use of the symmetry of the function in  my integration to make one of the coefficients 0.( eg. integral bounds -L to L of a sin() function) but can I still take this approach if my bounds are not symmetrical about the Y axis? (eg. If I integrated from 0 to 2*pi?) Thanks!! (also, bonus points if you can tell me if its valid from 0 to pi)
 A: There are several issues embedded in this question. Let me know if I am hitting precisely where your question lies.
When $f(x)$ is given on $0<x<\ell$:
You need to extend $f$ to $-\ell<x<0$ before you can do a full Fourier series. If you do an even extension, 
$$f_e(x):=\begin{cases}f(-x), &-\ell<x<0,\\ f(x), &0<x<\ell,\end{cases}$$
then the full Fourier series will end up being a Fourier cosine series (all the sine coefficients will end up being zero). If you do an odd extension, 
$$f_o(x):=\begin{cases}-f(-x), &-\ell<x<0,\\ f(x), &0<x<\ell,\end{cases}$$
then the full Fourier series will end up being a Fourier sine series (all the cosine coefficients will end up being zero). If you do some arbitrary extension, then both sets of coefficients (on the cosines and sines) can survive and you get a bona fide full Fourier series.
This is the approach you would use in what you call the "nonsymmetrical case".
When $f(x)$ is given on $-\ell<x<\ell$:
In this case, the given function is either even, odd, neither, or both depending on the $f$. If $f$ is even, the full Fourier series ends up as a Fourier cosine series. If $f$ is odd, it ends up as a Fourier sine series. If neither, you just get a full Fourier series. If both, you get a series full of zeros. ;-)
Finally, I don't understand your very last question.
