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

Question

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]

Answer 1

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:

$$e^{\frac{-a x}{L}} = \frac{(1-e^{-a})}{a} + \sum_{n=1}^{\infty}\bigg[\frac{2a(1-e^{-a})}{(a)^2 + (2n\pi)^2}\cos\bigg(\frac{2n\pi x}{L}\bigg) + \frac{4n\pi(1-e^{-a})}{(a)^2 + (2n\pi)^2}\sin\bigg(\frac{2n\pi x}{L}\bigg)\bigg]$$

Answer 2

For representing a non-periodic function on the interval $[0,L]$, we use \begin{equation*} f(x)=k_1+\frac{k_2x}{L}+\sum_{n=1}^{\infty} \left(A_n\cos\frac{2n\pi x}{L}+B_n\sin\frac{2n\pi x}{L}\right). \end{equation*} By evaluating the above series at $x=0,L$, we get \begin{equation*} k_2=f(L)-f(0). \end{equation*} Let \begin{equation*} \tilde{f}(x)=f(x)-\frac{k_2x}{L}. \end{equation*} Then using the orthogonality of the sine and cosine functions, we get \begin{align*} k_1&=\frac{1}{L}\int_0^L \tilde{f}(\hat{x})\,d\hat{x}, \\ A_n&=\frac{2}{L}\int_0^L \tilde{f}(\hat{x})\cos\frac{2n\pi\hat{x}}{L}\,d\hat{x}, \\ B_n&=\frac{2}{L}\int_0^L \tilde{f}(\hat{x})\sin\frac{2n\pi\hat{x}}{L}\,d\hat{x}. \end{align*} For the function $e^{-ax/L}$, we get \begin{equation*} f(x)=\frac{(a+2)(1-e^{-a})}{2a}+\frac{(e^{-a}-1)x}{L}+a(1-e^{-a})\sum_{n=1}^{\infty} \frac{2n\pi\cos\frac{2n\pi x}{L}-a\sin\frac{2n\pi x}{L}}{n\pi[a^2+(2n\pi)^2]}. \end{equation*} For why the $x$ term should be present while representing non-periodic functions, see https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2022-03/mr_3_2022_fourier.pdf

With the $x$ term present, you will get a much faster rate of convergence compared to when it is not present.