10.7: Fourier transform

Definition: Fourier Transform

The Fourier transform of a function \(f(x)\) is defined by

\[\hat{f} (\omega) = \int_{-\infty}^{\infty} f(x) e^{-ix \omega}\ dx \nonumber \]

This is often read as '\(f\)-hat'.

Theorem \(\PageIndex{1}\): Fourier Inversion Formula

We can recover the original function \f(x)\) with the Fourier inversion formula

\[f(x) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f} (\omega) e^{ix \omega} \ d \omega. \nonumber \]

So, the Fourier transform converts a function of \(x\) to a function of \(\omega\) and the Fourier inversion converts it back. Of course, everything above is dependent on the convergence of the various integrals.

Proof

We will not give the proof here. (We may get to it later in the course.)

Example \(\PageIndex{1}\)

Let

\[f(t) = \begin{cases} e^{-at} & \text{for } t > 0 \\ 0 & \text{for } t < 0 \end{cases} \nonumber \]

where \(a > 0\). Compute \(\hat{f} (\omega)\) and verify the Fourier inversion formula in this case.

Solution

Computing \(\hat{f}\) is easy: For \(a > 0\)

\[\hat{f} (\omega) = \int_{-\infty}^{\infty} f(t) e^{-i\omega t} \ dt = \int_{0}^{\infty} e^{-at} e^{-i \omega t}\ dt = \dfrac{1}{a + i \omega} (\text{recall } a > 0). \nonumber \]

We should first note that the inversion integral converges. To avoid distraction we show this at the end of this example.

Now, let

\[g(z) = \dfrac{1}{a + iz} \nonumber \]

Note that \(\hat{f} (\omega) = g(\omega)\) and \(|g(z)| < \dfrac{M}{|z|}\) for large \(|z|\).

To verify the inversion formula we consider the cases \(t > 0\) and \(t < 0\) separately. For \(t > 0\) we use the standard contour (Figure \(\PageIndex{1}\)).

Figure \(\PageIndex{1}\): Standard contour. (CC BY-NC; Ümit Kaya)

Theorem 10.2.2(a) implies that

\[\lim_{x_1 \to \infty, x_2 \to \infty} \int_{C_1 + C_2 + C_3} g(z) e^{izt}\ dz = 0 \nonumber \]

Clearly

\[\lim_{x_1 \to \infty, x_2 \to \infty} \int_{C_4} g(z) e^{izt}\ dz = \int_{-\infty}^{\infty} \hat{f} (\omega) \ d \omega \nonumber \]

The only pole of \(g(z) e^{izt}\) is at \(z = ia\), which is in the upper half-plane. So, applying the residue theorem to the entire closed contour, we get for large \(x_1, x_2\):

\[\int_{C_1 + C_2 + C_3 + C_4} g(z) e^{izt}\ dz = 2 \pi i \text{Res} (\dfrac{e^{izt}}{a + iz}, ia) = \dfrac{e^{-at}}{i}. \nonumber \]

Combining the three equations 10.8.6, 10.8.7 and 10.8.8, we have

\[\int_{-\infty}^{\infty} \hat{f} (\omega) \ d \omega = 2\pi e^{-at} \ \ \ \ \text{for } t > 0 \nonumber \]

This shows the inversion formulas holds for \(t > 0\).

For \(t < 0\) we use the contour in Figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\): (CC BY-NC; Ümit Kaya)

Theorem 10.2.2(b) implies that

\[\lim_{x_1 \to \infty, x_2 \to \infty} \int_{C_1 + C_2 + C_3} g(z) e^{izt}\ dz = 0 \nonumber \]

Clearly

\[\lim_{x_1 \to \infty, x_2 \to \infty} \dfrac{1}{2\pi} \int_{C_4} g(z) e^{izt} \ dz = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f} (\omega) \ d \omega \nonumber \]

Since, there are no poles of \(g(z) e^{izt}\) in the lower half-plane, applying the residue theorem to the entire closed contour, we get for large \(x_1, x_2\):

\[\int_{C_1 + C_2 + C_3 + C_4} g(z) e^{izt}\ dz = -2\pi i \text{Res} (\dfrac{e^{izt}}{a + iz}, ia) = 0. \nonumber \]

Thus,

\[\dfrac{1}{2\pi} \int_{-\infty}^{\infty} \hat{f} (\omega) \ d \omega = 0 \ \ \ \ \text{for } t < 0 \nonumber \]

This shows the inversion formula holds for \(t < 0\).

Finally, we give the promised argument that the inversion integral converges. By definition

\[\begin{array} {rcl} {\int_{-\infty}^{\infty} \hat{f} (\omega) e^{i \omega t} \ d \omega} & = & {\int_{-\infty}^{\infty} \dfrac{e^{i \omega t}}{a + i \omega} d \omega} \\ {} & = & {\int_{-\infty}^{\infty} \dfrac{a \cos (\omega t) + \omega \sin (\omega t) - i \omega \cos (\omega t) + ia \sin (\omega t)}{a^2 + \omega^2} \ d \omega} \end{array} \nonumber \]

The terms without a factor of \(\omega\) in the numerator converge absolutely because of the \(\omega ^2\) in the denominator. The terms with a factor of \(\omega\) in the numerator do not converge absolutely. For example, since

\[\dfrac{\omega \sin (\omega t)}{a^2 + \omega ^2} \nonumber \]

decays like \(1/\omega\), its integral is not absolutely convergent. However, we claim that the integral does converge conditionally. That is, both limits exist and are finite.

\[\lim_{R_2 \to \infty} \int_{0}^{R_2} \dfrac{\omega \sin (\omega t)}{a^2 + \omega ^2} d \omega \ \ \text{and} \ \ \lim_{R_1 \to \infty} \int_{-R_1}^{0} \dfrac{\omega \sin (\omega t)}{a^2 + \omega ^2} d \omega \nonumber \]

The key is that, as \(\sin (\omega t)\) alternates between positive and negative arches, the function \(\dfrac{\omega}{a^2 + \omega ^2}\) is decaying monotonically. So, in the integral, the area under each arch adds or subtracts less than the arch before. This means that as \(R_1\) (or \(R_2\)) grows the total area under the curve oscillates with a decaying amplitude around some limiting value.

Figure \(\PageIndex{3}\): Total area oscillates with a decaying amplitude. (CC BY-NC; Ümit Kaya)