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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]

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  • $\begingroup$ What are the "standard coefficient finding formulae"? The formula of which I am aware involves integrating over a single period of a periodic formula. How do you define the $n$-th Fourier coefficient of a non-periodic function? $\endgroup$ – Xander Henderson Feb 27 at 5:03
  • $\begingroup$ If you write down the formula for the Fourier coefficients of a non-periodic function, you get a representation of the Fourier series of its periodic extension (which is a discontinuous function even if the original function was continuous). $\endgroup$ – Ian Feb 27 at 5:31
  • $\begingroup$ @Ian My recollection is that the periodic extension of a function refers to the extension of a function which is defined on a finite interval to a function defined on $\mathbb{R}$. For example, the function $$ f : (-\frac{1}{2},\frac{1}{2})\to\mathbb{R} : x \mapsto x $$ is non-periodic, but has a sawtoolh function as its periodic extension. This is neither here nor there with respect to the question of a function defined on all of $\mathbb{R}$. $\endgroup$ – Xander Henderson Feb 27 at 5:41
  • $\begingroup$ @XanderHenderson I have added the relations I used to find the coefficients $\endgroup$ – Indrasis Mitra Feb 27 at 5:41
  • $\begingroup$ @IndrasisMitra In those formulae, $L$ the the period of the function. How are you choosing $L$ when $f$ is a non-periodic function defined on $\mathbb{R}$? $\endgroup$ – Xander Henderson Feb 27 at 5:42
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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:

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

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