The Fourier Transform is an extremely important tool in applied mathematics. It appears in electro-dynamics, the study of waves, and quantum mechanics, and has strong links with Green's functions. It is also extremely useful in pure mathematics, as it can be used to prove the existence of solutions to certain classes of differential equations.

The Fourier Transform is closely linked to the Fourier Series. The Fourier Series allows us to express periodic functions as discrete sums of sine waves, while the Fourier Transform allows us to express any function a continuous integral of sine waves.


For a complex function f(x) which satisfies the condition that

integral(|f(x)|dx, x=-inf...inf)

takes a finite value, the Fourier Transform F(k) is given by

F(k)= 1/sqrt(2π) integral(f(x)exp(-ikx)dx, x=-inf...inf).


The function F is often denoted as f with a tilde ("~"). We think of F(k) as a representation of f in Fourier space, which is dual to normal space. In applied mathematics the factor of 1/sqrt(2π) is omitted: since we never directly compare f(x) and F(k) it is seldom needed. Because the integral is from negative infinity to positive infinity, the techniques of contour integration, particularly Jordan's lemma, are often useful.

There is a corresponding vector result: if f(x) is a function on n dimensions, then it has Fourier Transform

F(ξ) = (2π)n/2 integral(f(x)exp(-i x.ξ)dx, x in Rn).

Basic properties

From the definition, we can prove the following basic results, which are useful when manipulating Fourier expressions.

  1. Linearity. Let h(x) = f(x) + g(x). Then H(k) = F(k) + G(k).
  2. Translation. Let g(x) = f(x-a). Then G(k) = exp(-ika) F(k).
  3. Frequency shift. Let g(x) = exp(ilx) f(x). Then G(k) = F(k-l).
  4. Scaling. Let g(x) = f(ax). Then G(k) = F(k/a)/|a|.
  5. Derivative. Let g(x) = f'(x). Then G(k) = ik F(k).
  6. Factors of x. Let g(x) = xf(x). Then G(k) = i dF(k)/dk.

These results are easily proved using the properties of integration. Note the duality of the results 2 and 3, and of the results 5 and 6. The fifth result is critical in many applications of Fourier Transforms since it allows us to rewrite differential equations in a form without derivatives. For example, consider the differential equation

f''(x) - 5f'(x) + 6f(x) = g(x)

where we wish to find f(x) for an arbitrary input function g(x). Then in the Fourier representation we have

-k2 F(k) - 5 ik F(k) + 6 F(k) = G(k)
F(k) = G(k) / (6 - 5ik - k2).

Using Fourier Transforms we now have an expression linking f and g not involving any derivatives.


Suppose F(k) is the Fourier Transform of f(x). Then if f is continuous we have the Inverse Fourier Transform

f(x) = 1/sqrt(2π) integral(exp(ikx)F(k)dk, k=-inf...inf).

If f is discontinuous at x, then we have the more general result

(f(x+) + f(x-))/2 = 1/sqrt(2π) integral(exp(ikx)F(k)dk, k=-inf...inf),

where f(x+) and f(x-) represent the values of f on either side of the discontinuity.