Fourier series of non-periodic function $f(x)=e^{-\frac{ax}{L}}$ The definition of Fourier series states that 

It decomposes any periodic function or periodic signal into the weighted sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equivalently, complex exponentials)

I was a little confused as to how then, non-periodic functions like $f(x) = e^{\frac{-ax}{L}}$ defined over an interval $[0,L]$ can have a Fourier expansion ?
We know that
$$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty}a_n\cos(nx) + \sum_{n=1}^{\infty}b_n\sin(nx)$$
Note $\rightarrow$ $a,L$ are constants $>0$
What I fail to understand is, how is this possible for a non-periodic function like an exponential ? Somewhere on another question in MSE, I learned that The Fourier series is described for the periodic extension of a non-periodic function, which failed to clarify my doubts.
The motivation to ask this question comes from my attempts to solve 
Two fluids flowing perpendicular in thermal contact with a Wall [Help to mathematically model] and 
Evaluating Coefficients for a Fourier Series when Exponential terms are present [Approach needed]
 A: Thanks to Ian and Xander Henderson for their suggestions.
We can consider any function defined on a finite interval $(a,b)$ or $[a,b]$ as a periodic function defined on $R$ by thinking that the function is extended to $R$ by repeating the values in $[a,b]$ to the remaining part of $R$.
Thus for $f(x)$ defined on $[a,b]$ where $(\frac{b-a}{2}) = l$, we have 
$$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \bigg[ a_n \cos\bigg(\frac{n\pi x}{l}\bigg) + b_n \sin\bigg(\frac{m\pi x}{l}\bigg)\bigg]$$
where,
$$a_0 = \frac{1}{l} \int_a^b f(x) \mathrm{d}x$$
$$a_n = \frac{1}{l} \int_a^b f(x)\cos\bigg(\frac{n\pi x}{l}\bigg)\mathrm{d}x$$
$$b_n = \frac{1}{l} \int_a^b f(x)\sin\bigg(\frac{n\pi x}{l}\bigg)\mathrm{d}x$$
Applying these for our function $f(x) = e^{-\frac{ax}{L}}$ defined on $x \in [0,L]$.
Hence, $a = 0$,$b = L$ and $l=\frac{L}{2}$, leads us to: 
$$e^{\frac{-a x}{L}} = \frac{(1-e^{-a})}{a} + \sum_{n=1}^{\infty}\bigg[\frac{2a(1-e^{-a})}{(a)^2 + (2n\pi)^2}\cos\bigg(\frac{2n\pi x}{L}\bigg) + \frac{4n\pi(1-e^{-a})}{(a)^2 + (2n\pi)^2}\sin\bigg(\frac{2n\pi x}{L}\bigg)\bigg]$$
