Determination of Fourier coefficient (Euler's Formulae) Is the integral from -π to π the same with integral from 0 to 2π?
 A: We generally use Fourier series decompositions for functions that are periodic, or that are defined on a bounded interval such that you can extend them into a periodic function by making copies of that interval over and over again.
The formula for finding Fourier coefficients always assumes you've scaled the sine and cosine functions to match the period of the function (or the width of the interval), but otherwise it doesn't matter how you take the integral because it's periodic - as long as the integral covers a whole period, you can chop it up and shift it in every direction.
For the case where the period is $2\pi$, then our function $f$ will always follow $f(x) = f(x + 2k\pi)$ for any integer $k$. Thus, for example,
$\begin{eqnarray}a_1 & = & \int_{-\pi}^\pi f(x) \sin x\ dx \\
& = & \int_{-\pi}^0 f(x) \sin x\ dx + \int_0^\pi f(x) \sin x\ dx \\
& = & \int_\pi^{2\pi} f(u - 2\pi) \sin (u - 2\pi)\ du + \int_0^\pi f(x) \sin x\ dx & \mbox{(where } u = x + 2\pi \mbox{)} \\
& = & \int_\pi^{2\pi} f(u) \sin u \ du + \int_0^\pi f(x) \sin x \ dx \\
& = & \int_\pi^{2\pi} f(x) \sin x \ dx + \int_0^\pi f(x) \sin x \ dx \\
& = & \int_0^\pi f(x) \sin x \ dx + \int_\pi^{2\pi} f(x) \sin x \ dx \\
& = & \int_0^{2\pi} f(x) \sin x \ dx
\end{eqnarray}$
And similarly for the other coefficients, so it doesn't actually matter where the interval is located. That means that the choice of whether to work on $[-\pi, \pi]$ or $[0, 2\pi]$ is generally just a matter of convention - sometimes it's nice to have the symmetry, sometimes it's easier to work with things that start at 0 (especially if you're using it for differential equations with boundary conditions, since it's quite common for 0 to be one of those boundaries).
