Inspired by the previous example and ignoring any technical issues (such as convergence), we have been led to define the following:
Definition of the Fourier transform and inverse transform
The Fourier transform of a function is the function defined by the integral
if the integral exists. The function is sometimes denoted or .
The inverse Fourier transform of a function is the function defined by the integral
if the integral exists. The function is sometimes denoted or .
Convention warning
Although our definition of the Fourier transform seems like the most natural one, as an evolution of the Fourier series, in practice there are many competing definitions. Some sources switch the roles of what we have called the Fourier transform and the inverse Fourier transform. Some use the integral
or even simply the integral
In these notes will stick with the definitions we made above, but be wary when reading other resources.
Examples
The rectangle function
Based on our work in the previous example, the Fourier transform of the rectangle function is the function :
If we believe the rectangle function is a "fundamental" function, then its Fourier transform probably deserves to be considered fundamental, as well. Indeed, it is often given a name:
Definition of the function
Define the function (pronounced "sink") by
Convention warning
As with the definition of the Fourier transform, not all sources agree on this definition of the function. Some instead use the function .
So, we can now say . Also, for reference, here's a bit of the graph of the function:
We should note that in our build-up to the discovery of the Fourier transform, we also "proved" that the inverse Fourier transform of the function is the rectangle function , i.e.,
This would be fairly difficult to verify directly, since it would amount to calculating the integral
In fact, the above integral doesn't quite exactly agree with our rectangle function. It turns out that the above integral takes the value outside , takes the value inside , and takes the value at . In other words, it agrees with except at the two jump discontinuities (where the inverse transform takes the average of the values across the jump).
We won't worry about this for now.
A triangle function
Consider the "triangle function" defined piecewise by
The graph of this function looks like a isosceles right triangle of height and base centered at the origin:
We have chosen to use the capital lambda symbol for this function's name because that Greek character also happens to look like a triangle.
In any case, we can compute the Fourier transform of this function:
We could leave the answer like this, but this expression can be simplified even further, as follows:
So the Fourier transform of the triangle function is exactly the square of the Fourier transform of the rectangle function . We'll soon see that there is a secret reason for this!
One-sided exponential decay
Consider the one-sided exponential decay function
The graph of looks like
(We have not chosen a fun character to name this function, since there aren't any characters that look like the above graph.)
We compute the Fourier transform of this function:
(Note that here we are assuming only takes real values, in which case the exponential function is always a complex number on the unit circle. This is how we concluded that the term went to as .)
A Gaussian function
Consider the classic Gaussian function , whose graph is the usual bell curve:
Note that the constant was chosen so that the area under the bell curve is exactly , i.e.,
The Fourier transform of this function is difficult to compute directly:
We can determine without computing the above integral in a very sneaky way. Let's temporarily denote the above integral by . First we consider differentiating the mystery function , finding
If we live life reckless and pass the differentiation inside the integral, we find that
This new integral can be computed using integration-by-parts. Specifically, if we let and , then and , and so we have
This is the antiderivative formula, so what happens to the definite (improper) integral? We'll be a bit fast and loose here, but the main point is that as , the boundary term rapidly tends to . Indeed, if you converted it the usual real-and-imaginary parts, you would see that
The net effect is that
But this new integral is just the integral we started with, which defined . So we've discovered that
The general solution to this ordinary differential equation is
We can even determine the mystery constant , since and
So, we can finally conclude that . In other words, the Fourier transform of the Gaussian is the identical Gaussian .
If you think of the Fourier transform as some kind of "operator" that transforms functions into other functions, then you can also start to think of this particular Gaussian function as something like an "eigenfunction" with eigenvalue .