The Fourier transform tells us that any signal can be represented in time or frequency. Imagine we have a pure tone of 1000 Hz sounding with a strength of 1 decibel. We can represent this tone by tabulating its values at all times (in an infinitely large table). This is the time representation, also known as the time-domain signal and written as \(x(t)\).
Alternatively, we can represent this signal by specifying the strength of all frequencies that make up the signal. In this case we have just 1 frequency at 1 decibel. So the frequency representation would be simply “1 decibel at 1000 Hz”. The frequency representation of the signal, also known as the frequency-domain signal is written as \(X(\omega)\).
Both representations are provide the same information about the signal. In other words, an alien can construct the signal from either from a table of \(x(t)\) or a specification of \(X(\omega)\). We write this equivalence as:
\[x(t) \leftrightarrow X(\omega)\]\(X(\omega)\) can be computed given \(x(t)\) and vice versa
\[\begin{align} X(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x(t)\, e^{-i\omega t}\, dt \\[0.1in] x(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}X(\omega)\, e^{i\omega t}\, d\omega \\ \end{align}\]Now lets talk about the total energy of the signal. In time representation, we can calculate the total energy of the signal by adding up (or integrating) its strength at each point in time. In the frequency representation, we would add up (integrate) the strength of each frequency component. Since the representations are identical, we should expect that the answers obtaied from the time and the frequency representations should match. Plancherel’s theorem is the statement that they do.
In the above notation, the Plancherel’s theorem states that
\[\begin{equation} \int_{-\infty}^{\infty} \left\vert x(t) \right\vert^2 dt = \int_{-\infty}^{\infty} \left\vert X(\omega) \right\vert^2 d\omega \end{equation}\]Main application of the Fourier Transform is in signal processing. In fact, it is not an overstatement to say that the Fourier Transform and its computational algorithm, the Fast Fourier Transform (FFT) underlie all modern communication systems. There are plenty of books on this subject.
The present post, however, has a different purpose. We will twist the mathematics of the Fourier transform to evaluate some definite integrals. What we primarily seek from this exercise is entertainment. I’d imagine that using Fourier Transforms to compute well-known definite integrals is of little practical use. With that disclaimer, let us evaluate a few integrals!
The general recipe is the following:
Choose a function \(x(t)\) and compute its Fourier Transform
Apply the Plancherel’s theorem
Evaluate either the left- or the right hand side explicitly
Write down the value of the integral on the other side
We start from the basic top-hat function.
\[\begin{align*} x(t) &= \begin{cases} 1 & -a \leq t \leq a\\[0.15in] 0 & \text{otherwise} \end{cases} \\[0.2in] X(\omega) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}x(t)\, e^{-i\omega t} dt = \int_{-a}^{a}e^{-i\omega t} dt\\[0.1in] &=\sqrt{\frac{2}{\pi}}\,\frac{\sin a\omega}{\omega} \end{align*}\]Applying the Plancherel’s theorem:
\[\begin{align*} \int_{-\infty}^{\infty} \left\vert x(t) \right\vert^2 dt &= \int_{-\infty}^{\infty} \left\vert X(\omega) \right\vert^2 d\omega \\[0.1in] \int_{-a}^{a} dt &= \frac{2}{\pi}\int_{-\infty}^{\infty} \frac{\sin^2 a\omega}{\omega^2} d\omega \\[0.1in] 2a &\stackrel{u\leftarrow a\omega}{=} \frac{2a}{\pi}\int_{-\infty}^{\infty} \frac{\sin^2 u}{u^2} du \\[0.1in] \end{align*}\]giving us
\[\int_{-\infty}^{\infty}\left(\frac{\sin u}{u}\right)^2\, du= \pi\]But we can go further.
We can prove that
\[\int_{-\infty}^{\infty}\left(\frac{\sin u}{u}\right)^2\, du = \int_{-\infty}^{\infty}\frac{\sin u}{u}\, du\]To do that integrate the left-hand side by parts, applying the limits:
\[\begin{align*} \int_{-\infty}^{\infty}\left(\frac{\sin u}{u}\right)^2\, du &= -\frac{\sin u}{u}\bigg\vert_{-\infty}^{\infty} + \int_{-\infty}^{\infty}\frac{1}{u} 2 \sin u\cos u\, du \\ &= \int_{-\infty}^{\infty}\frac{\sin 2u}{u} \, du \stackrel{v\leftarrow 2u}{=} \int_{-\infty}^{\infty}\frac{\sin v}{v} \, dv \end{align*}\]Thus we have a neat identity
where
\[\mathcal{J} = \int_{-\infty}^{\infty}e^{-t^2}dt\]Applying the Plancherel’s theorem:
\[\begin{align*} \int_{-\infty}^\infty \vert x(t)\vert^2dt &= \int_{-\infty}^\infty \vert X(\omega)\vert^2d\omega\\ \int_{-\infty}^\infty \left\vert e^{-t^2/2}\right\vert^2dt &= \int_{-\infty}^\infty \left\vert \frac{e^{-\omega^2/2}}{\sqrt{\pi}}\mathcal{J}\right\vert^2d\omega\\ \int_{-\infty}^\infty e^{-t^2}dt &= \frac{\mathcal{J}^2}{\pi}\int_{-\infty}^\infty e^{-\omega^2}d\omega\\ \mathcal{J} &= \frac{\mathcal{J}^3}{\pi}\\ \Rightarrow \mathcal{J} &= \sqrt{\pi} \end{align*}\]Writing out explicitly, we have the well-known identity:
\[\int_{-\infty}^{\infty}e^{-t^2}dt = \sqrt{\pi}\]Putting \(t^2 = \alpha y^2\) we get a slightly more general identity \(\int_{-\infty}^{\infty}e^{-\alpha y^2}dy = \sqrt{\pi/\alpha}\). Substituting \(\alpha = i\) (technically this step needs to be performed using contour integration, but here the answer will be correct even without it):
\[\begin{align*} \int_{-\infty}^{\infty}e^{-it^2}\,dt &= \sqrt{\frac{\pi}{i}} \\ \int_{-\infty}^{\infty}\left(\cos t^2 -i\sin t^2\right)\,dt &= \sqrt{\frac{\pi}{2}} (1 + i) \\ \end{align*}\]Equating the real and the imaginary parts we get the limiting values of the Fresnel integrals
Plancherel’s theorem can be used to evaluate a selected class of definite integrals. The integral must be a part of a Fourier transform pair. For a successful application it should be possible to simplify or explicitly evaluate the integral of the original function, or its Fourier Transform.
Fourier Transforms have a number of interesting properties that can prove useful for evaluating definite integrals. For example, can we exploit the derivative identity for attacking other integrals or proving interesting identities involving definite integrals?
Written on June 18th, 2021 by Rohan