14 Jun 2025
Fourier Transforms, Delta Functions and Convolution
Fourier transforms of trigonometric, Gaussian, finite-train, and elementary functions, with derivative, inverse, delta, and convolution theorems.
Use the transform pair
\[\boxed{ F(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx, \qquad f(x)=\frac1{2\pi}\int_{-\infty}^{\infty}F(k)e^{ikx}\,dk. }\]The constants and signs in every formula below follow from this convention.
Dirac delta as a Fourier integral
The inverse transform of $F(k)=1$ defines
\[\boxed{ \delta(x-x_0)=\frac1{2\pi}\int_{-\infty}^{\infty} e^{ik(x-x_0)}\,dk. }\]This equality is distributional. Multiply by a smooth test function $g(x)$, integrate over $x$, and use the inverse transform to obtain $\int\delta(x-x_0)g(x)dx=g(x_0)$.
Trigonometric functions
Since
\[\int_{-\infty}^{\infty}e^{-i(k-k_0)x}\,dx =2\pi\delta(k-k_0),\]Euler’s relations give
\[\boxed{ \mathcal F\{\cos k_0x\} =\pi[\delta(k-k_0)+\delta(k+k_0)], }\] \[\boxed{ \mathcal F\{\sin k_0x\} =\frac\pi i[\delta(k-k_0)-\delta(k+k_0)]. }\]These transforms are distributions because an infinite sinusoid is not absolutely integrable.
Gaussian transform
For $a>0$, let $f(x)=e^{-ax^2}$. Complete the square:
\[-ax^2-ikx =-a\left(x+\frac{ik}{2a}\right)^2-\frac{k^2}{4a}.\]The Gaussian integral is unchanged by the permitted contour displacement, so
\[\begin{aligned} F(k)&=e^{-k^2/(4a)} \int_{-\infty}^{\infty} e^{-a(x+ik/2a)^2}\,dx\\ &=\boxed{\sqrt{\frac\pi a}\,e^{-k^2/(4a)}}. \end{aligned}\]The dimensions are consistent: if $x$ is measured in metres, then $k$ is in $\mathrm{m}^{-1}$, $a$ is in $\mathrm{m}^{-2}$, and $F$ has the dimensions of $f$ times metres.
Finite pulse and finite wave train
For a rectangular pulse of width $L$,
\[f(x)=\begin{cases}1,&\lvert x\rvert<L/2,\\0,&\lvert x\rvert>L/2,\end{cases}\]direct integration gives
\[F(k)=\int_{-L/2}^{L/2}e^{-ikx}dx =\frac{2\sin(kL/2)}k =L\,\operatorname{sinc}\!\left(\frac{kL}{2}\right),\]where $\operatorname{sinc}u=\sin u/u$.
A finite cosine train is $f(x)=\cos(k_0x)$ for $\lvert x\rvert<L/2$ and zero otherwise. Using the two exponentials,
\[\boxed{ F(k)= \frac{\sin[(k-k_0)L/2]}{k-k_0} +\frac{\sin[(k+k_0)L/2]}{k+k_0}. }\]The apparent singularities at $k=\pm k_0$ are removable limits. Each sinc contribution by itself has first zeros $2\pi/L$ from its centre. When $k_0L\gg1$ the two contributions are well separated, so each peak has this approximate width; where they overlap, their sum need not vanish at those individual-sinc zeros.
Transform of derivatives
Assume $f$ and its required derivatives vanish at $\lvert x\rvert\to\infty$. Integration by parts gives
\[\begin{aligned} \mathcal F\{f'(x)\} &=\left[f(x)e^{-ikx}\right]_{-\infty}^{\infty} +ik\int_{-\infty}^{\infty}f(x)e^{-ikx}dx\\ &=ikF(k). \end{aligned}\]Repeated integration gives
\[\boxed{\mathcal F\{f^{(n)}(x)\}=(ik)^nF(k).}\]If the boundary terms do not vanish, they must be retained.
Inverse transform
Insert the definition of $F(k)$ into the proposed inverse:
\[\begin{aligned} \frac1{2\pi}\int F(k)e^{ikx}dk &=\int f(x')\left[ \frac1{2\pi}\int e^{ik(x-x')}dk \right]dx'\\ &=\int f(x')\delta(x-x')dx'=f(x). \end{aligned}\]Thus the inverse formula follows directly from the Fourier representation of the delta distribution.
Convolution theorem
Define
\[(f*g)(x)=\int_{-\infty}^{\infty}f(u)g(x-u)\,du.\]Its transform is
\[\begin{aligned} \mathcal F\{f*g\} &=\int dx\,e^{-ikx}\int du\,f(u)g(x-u)\\ &=\int du\,f(u)e^{-iku} \int dy\,g(y)e^{-iky}\\ &=\boxed{F(k)G(k)}, \end{aligned}\]where $y=x-u$. Conversely,
\[\boxed{\mathcal F\{f(x)g(x)\}=\frac1{2\pi}(F*G)(k).}\]The Gaussian pair, finite-train transform, derivative rule, and convolution identity are checked in the Maxima worksheet; every printed residual is zero.
Discussion