Can any continous,bounded function have a fourier series? In particular, can an oscillatory function with some decay term ( i.e $e^{-t} \cos(kt)$) have a fourier series representation? All the articles I read said that the function has to be periodic,but this one is not. 
 A: No. Since $$
f(x) = \sum_{n=0}^\infty \left(a_n\cos\left(2\pi n\frac{x}{T}\right) + b_n\sin\left(2\pi n\frac{x}{T}\right)\right)
$$
is always $T$-periodic (due to $\sin$ and $\cos$ being $2\pi$-periodic), a fourier series can only faithfully represent periodic functions.
You can still apply the fourier transform to non-periodic functions, but instead of coefficients $a_n$,$b_n$, you get continous function $\mathbb{R}\to\mathbb{C}$.
A: To actually get a Fourier Series, you must have a periodic function.  However, a non-periodic function can be considered to haven an infinite period.  Then your sum over sin/cos terms with integer frequencies becomes an integral over sin/cos terms with continuous frequencies, and your coefficients become a function.
ie.
$ \sum_{-\infty}^{\infty }c_{n}e^{\frac{in\pi x}{L}} \rightarrow  \int_{-\infty}^{\infty}F(w)e^{iwx}dw $
Here, n*pi/L is the frequency in the periodic case, and w is the frequency in the non-periodic case.  In both cases, we sum/integrate over all frequencies.
The function that would give you your coefficients in the periodic case ($c_{n}$) and which sits under the integral as the continuous analog of the coefficients in the non-periodic case (F(w)) is called the Fourier Transform.
So for a periodic function, the Fourier Transform returns discrete amplitudes, with which you can build a series.  For a non-periodic (infinite periodic) function, the Fourier Transform returns a complex-valued function, which is used under the integral to weight the different frequency (w) contributions in an exactly analogous way to the coefficients in the periodic case.  However, you can no longer really call it a Fourier series.
One last point: if you are only interested in a particular region of a non-periodic function, then you can construct a new function that is periodic and looks like your region of interest over a single period. Then you can find a Fourier series, but it will only be applicable to the region of the original function that you singled out.
A: It has to be periodic - suppose you may represent a function $f:\Bbb{R}\rightarrow\Bbb{R}$ as a Fourier series:
$$f(x)=\sum_{-\infty}^{\infty}{c_{n}e^{inx}}.$$
Then, as $e^{inx}$ is $2\pi$-periodic, so is $f$.
