Fourier transform

The Fourier transform (FT) is an integral transform which converts a function $$f(x)$$ into its frequency representation $$\tilde{f}(k)$$. Great volumes have already been written about this subject, so let us focus on the aspects that are useful to physicists.

The forward FT is defined as follows, where $$A$$, $$B$$, and $$s$$ are unspecified constants (for now):

\begin{aligned} \boxed{ \tilde{f}(k) = \hat{\mathcal{F}}\{f(x)\} = A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} } \end{aligned}

The inverse Fourier transform (iFT) undoes the forward FT operation:

\begin{aligned} \boxed{ f(x) = \hat{\mathcal{F}}^{-1}\{\tilde{f}(k)\} = B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k x) \dd{k} } \end{aligned}

Clearly, the inverse FT of the forward FT of $$f(x)$$ must equal $$f(x)$$ again. Let us verify this, by rearranging the integrals to get the Dirac delta function $$\delta(x)$$:

\begin{aligned} \hat{\mathcal{F}}^{-1}\{\hat{\mathcal{F}}\{f(x)\}\} &= A B \int_{-\infty}^\infty \exp(-i s k x) \int_{-\infty}^\infty f(x') \exp(i s k x') \dd{x'} \dd{k} \\ &= 2 \pi A B \int_{-\infty}^\infty f(x') \Big(\frac{1}{2\pi} \int_{-\infty}^\infty \exp(i s k (x' - x)) \dd{k} \Big) \dd{x'} \\ &= 2 \pi A B \int_{-\infty}^\infty f(x') \: \delta(s(x' - x)) \dd{x'} = \frac{2 \pi A B}{|s|} f(x) \end{aligned}

Therefore, the constants $$A$$, $$B$$, and $$s$$ are subject to the following constraint:

\begin{aligned} \boxed{\frac{2\pi A B}{|s|} = 1} \end{aligned}

But that still gives a lot of freedom. The exact choices of $$A$$ and $$B$$ are generally motivated by the convolution theorem and Parseval’s theorem.

The choice of $$|s|$$ depends on whether the frequency variable $$k$$ represents the angular ($$|s| = 1$$) or the physical ($$|s| = 2\pi$$) frequency. The sign of $$s$$ is not so important, but is generally based on whether the analysis is for forward ($$s > 0$$) or backward-propagating ($$s < 0$$) waves.

Derivatives

The FT of a derivative has a very interesting property. Below, after integrating by parts, we remove the boundary term by assuming that $$f(x)$$ is localized, i.e. $$f(x) \to 0$$ for $$x \to \pm \infty$$:

\begin{aligned} \hat{\mathcal{F}}\{f'(x)\} &= A \int_{-\infty}^\infty f'(x) \exp(i s k x) \dd{x} \\ &= A \big[ f(x) \exp(i s k x) \big]_{-\infty}^\infty - i s k A \int_{-\infty}^\infty f(x) \exp(i s k x) \dd{x} \\ &= (- i s k) \tilde{f}(k) \end{aligned}

Therefore, as long as $$f(x)$$ is localized, the FT eliminates derivatives of the transformed variable, which makes it useful against PDEs:

\begin{aligned} \boxed{ \hat{\mathcal{F}}\{f'(x)\} = (- i s k) \tilde{f}(k) } \end{aligned}

This generalizes to higher-order derivatives, as long as these derivatives are also localized in the $$x$$-domain, which is practically guaranteed if $$f(x)$$ itself is localized:

\begin{aligned} \boxed{ \hat{\mathcal{F}} \Big\{ \dv[n]{f}{x} \Big\} = (- i s k)^n \tilde{f}(k) } \end{aligned}

Derivatives in the frequency domain have an analogous property:

\begin{aligned} \boxed{ \dv[n]{\tilde{f}}{k} = A \int_{-\infty}^\infty (i s x)^n f(x) \exp(i s k x) \dd{x} = \hat{\mathcal{F}}\{ (i s x)^n f(x) \} } \end{aligned}