Finding the Fourier series of the function f(x) shown graphically below shouldn’t be a problem. Since the function is defined on , we can easily apply the formulas.

Unless all the functions in the world are defined on , we will be satisfied with ourselves. Sadly, this is not the case and rightly so. How do we find the Fourier series of a function on [-L, L] graphically shown below.

The method is rather simple. We will use integration by substitution with a change in variable to alter the original definition to suit this problem.

We first need to relate two independent variables which define the x axis of the two graphs. Let them be t and x, and that

 and so

Notice here the respective domains for both independent variables. x is defined on [-L, L] and t is defined on . Next, we write a function in terms of t which translates the function f(x) such that

Now, we can use the previous definitions to find the Fourier coefficients of function g(t) as the domain of t is , and from there, from the respective Fourier coefficients of f(x).

We have employed the method of integration by substitution using the relationship . Doing the same for the other coefficients,

and

That wasn’t too hard. So our formal definition of the Fourier series and coefficients on [-L, L] is as follows.

Let  be integrable on .
1. The Fourier Coefficients of  on  are

,
, and

for n = 1, 2, 3, …

2. The Fourier Series of  on  is

,

in which the numbers  are the Fourier coefficients of  on .

For me, the most noticeable change is the variable inside the trigonometry functions. Once you can get used to the change from nx to , you should be fine. Now we can find the Fourier Series for another set of functions.