My first question regarding the Fourier Series is: What are the "certain conditions" necessary for a function to be expressible as a Fourier Series? Most treatments I've come across are either very involved, or omit any further discussion of these necessary conditions.
Here is my understanding of what a Fourier Series is:
(I'm following Georg Joos's development in http://store.doverpublications.com/0486652270.html his Theoretical Physics.)
A periodic function $f:\mathbb{R\to\mathbb{R}}$ is one that may be written as
$$ f\left[t\right]=f\left[\frac{1}{\nu}+t\right]=f\left[t+T\right]=f\left[t+\frac{2\pi}{\omega}\right]; $$
where $\nu$ is the periodic frequency, $T=\frac{1}{\nu}$ is the period of one cycle, and $\omega=2\pi\nu$ is the angular frequency.
Under certain conditions, which are not usually hard to satisfy the periodic function $f$ may be written as an infinite sum known as a Fourier Series
$$ f\left[t\right]=\sum_{n=-\infty}^{+\infty}a_{n}e^{in{\omega}t}. $$
The origin of the Fourier Series lies in the orthogonality properties of $e^{in\omega t}$, where $n$ is an integer.
$$ n\ne0\implies\int_{0}^{\frac{2\pi}{\omega}}a_{n}e^{in\omega t}dt=\frac{a_{n}}{in\omega}\left(e^{in2\pi}-e^{0}\right)=0. $$
$$ n=0\implies\int_{0}^{\frac{2\pi}{\omega}}a_{0}e^{i0\omega t}dt=\frac{2\pi}{\omega}a_{0}. $$
So
$$ \int_{0}^{\frac{2\pi}{\omega}}f\left[t\right]dt=\sum_{n=-\infty}^{+\infty}\int_{0}^{\frac{2\pi}{\omega}}a_{n}e^{ni{\omega}t}dt=\frac{2\pi}{\omega}a_{0}. $$
Thus $a_{0}$ becomes
$$ a_{0}=\frac{\omega}{2\pi}\int_{0}^{\frac{2\pi}{\omega}}f\left[t\right]dt=\overline{f\left[t\right]}. $$
Multiplying $f\left[t\right]$ by $e^{in\omega t}$ ($n\ne0$) and changing the name of the summation index leads to
$$ \int_{0}^{\frac{2\pi}{\omega}}f\left[t\right]e^{-i\color{blue}{n}\omega t}dt $$
$$ =\sum_{\color{red}{m}=-\infty}^{+\infty}\int_{0}^{\frac{2\pi}{\omega}}a_{\color{red}{m}}e^{i\omega(\color{red}{m}-\color{blue}{n})t}dt $$
$$ =\int_{0}^{\frac{2\pi}{\omega}}a_{\color{blue}{n}}dt=\frac{2\pi}{\omega}a_{\color{blue}{n}}. $$
Thus
$$ \frac{\omega}{2\pi}\int_{0}^{\frac{2\pi}{\omega}}a_{n}dt=a_{n}. $$ Replacing the symbols $a_{n}$ with these values produces
$$ f\left[\color{blue}{t}\right]=\sum_{n=-\infty}^{+\infty}\frac{\omega}{2\pi}e^{in\omega\color{blue}{t}}\int_{0}^{\frac{2\pi}{\omega}}f\left[\color{red}{\alpha}\right]e^{-in\omega\color{red}{\alpha}}d\color{red}{\alpha}. $$
Since the integrals do not involve the variable $\color{blue}{t}$, the expression $e^{in\omega\color{blue}{t}}$ may be moved under the integral
$$ f\left[\color{blue}{t}\right]=\sum_{n=-\infty}^{+\infty}\frac{\omega}{2\pi}\int_{0}^{\frac{2\pi}{\omega}}f\left[\color{red}{\alpha}\right]e^{in\omega\left(\color{blue}{t}-\color{red}{\alpha}\right)}d\color{red}{\alpha}. $$
Using $T=\frac{2\pi}{\omega}$ this may be written as
$$ f\left[t\right]=\sum_{n=-\infty}^{+\infty}\frac{1}{T}\int_{0}^{T}f\left[\alpha\right]e^{in\omega\left(t-\alpha\right)}d\alpha. $$
