\(\text{Integral of the month: } \iint_S dt' dt f(t'-t)\)

Abstract

When the function depends only on the difference of the parameters, we can simplify the double integral using a clever change of variables. This technique is essential for understanding the Wiener-Khinchin theorem and appears frequently in Fourier analysis.

Keywords

Integral, Wiener-Khinchin theorem

\(\require{cancel}\) \(\def\oot{\frac{1}{2}}\)

We want to compute the integral \(I=\int_{\frac{-T}{2}}^{\frac{T}{2}}\int_{\frac{-T}{2}}^{\frac{T}{2}} dt' dt f(t'-t)\).

The argument of the function begs for a change of coordinates:

\[\begin{eqnarray} u=t'-t, \quad \text{and} \quad v=t+t' \label{eq:trans}, \end{eqnarray}\] and the associated inverse transform reads: \[\begin{eqnarray} t'=\frac{u+v}{2}, \quad \text{and} \quad t=\frac{v-u}{2}. \label{eq:transi} \end{eqnarray}\]

This transformation will rotate and scale the integration domain as shown in Figure 1.

Figure 1: The integration domain in the \(t-t'\) domain (left) and \(u-v\) domain(right). Since there is no \(v\) dependence, \(v\) integration gives the height of the green and blue slices.

The equation of the top boundary on the right can be written as \(v=T-u\), and on the left as $ v= T+u$. We can actually combine them as \(v=T-|u|\). We can do the same analysis for the lower boundaries to see that the height of the slices at a given \(u\) is \(2(T-|u|)\). This will help us easily integrate \(v\) out as follows: \[\begin{eqnarray} I&=&\int_{\frac{-T}{2}}^{\frac{T}{2}}\int_{\frac{-T}{2}}^{\frac{T}{2}} dt' dt f(t'-t) =\iint_{S_{u,v}}\left|\frac{\partial(t,t')}{\partial(u,v)}\right| dv du f(u)\nonumber\\ &=&\int_{-T}^T 2(T-|u|) \times\frac{1}{2} dv du f(u)=\int_{-T}^T du f(u)(T-|u|) \label{eq:transeq}, \end{eqnarray}\] where \(\left|\frac{\partial(t,t')}{\partial(u,v)}\right|=\frac{1}{2}\) is the determinant of the Jacobian matrix associated with the transformation in Eq. \(\ref{eq:transi}\).

In a typical problem, such as the proof of Wiener-Khinchin theorem, we need to evaluate the time average of the integral \[\begin{eqnarray} \bar{I}&=& \lim_{T\to \infty}\frac{1}{T}\int_{-T}^{T} du f(u)(T-|u|)=\int_{-\infty}^{\infty} du f(u), \label{eq:WautocFF2} \end{eqnarray}\] where we assume that \(uf(u)\) dies quickly enough so that the term drops out in the limit.