$\renewcommand{\Re}{\operatorname{Re}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$

Fourier transform, Fourier integral

  1. Heuristics
  2. Definitions and Remarks
  3. $\cos $- and $\sin$-Fourier transform and integral
  4. Discussion: pointwise convergence of Fourier integrals and series


In the previous Lecture 16 we wrote Fourier series in the complex form

\begin{equation} f(x)= \sum_{n=-\infty}^\infty c_n e^{\frac{i\pi nx}{l}} \label{eq-1} \end{equation} with \begin{equation} c_n= \frac{1}{2l}\int_{-l}^l f(x)e^{-\frac{i\pi n x}{l}}\,dx \qquad n=\ldots,-2, -1, 0,1,2,\ldots\label{eq-2} \end{equation} and \begin{equation} 2l\sum_{n=-\infty}^\infty |c_n|^2=\int_{-l}^l|f(x)|^2\,dx. \label{eq-3} \end{equation}

From this form we formally without any justification deduct Fourier integral.

First we introduce \begin{equation} \omega_n := \frac{\pi n}{l}\qquad \text{and}\qquad \Delta \omega_n = \omega_{n}-\omega_{n-1}= \frac{\pi}{l} \label{eq-4} \end{equation} and rewrite (\ref{eq-1}) as \begin{equation} f(x)= \sum_{n=-\infty}^\infty C(\omega_n) e^{i\omega_n x}\Delta \omega_n \label{eq-5} \end{equation} with \begin{equation} C(\omega)= \frac{1}{2\pi}\int_{-l}^l f(x)e^{-i\omega x}\,dx \label{eq-6} \end{equation} where we used $C(\omega_n) := c_n /(\Delta\omega_n)$; (\ref{eq-3}) should be rewritten as \begin{equation} \int_{-l}^l|f(x)|^2\,dx=2\pi\sum_{n=-\infty}^\infty |C(\omega_n)|^2\Delta \omega_n. \label{eq-7} \end{equation} Now we formally set $l\to +\infty$; then integrals from $-l$ to $l$ in the right-hand expression of (\ref{eq-6}) and the left-hand expression of (\ref{eq-7}) become integrals from $-\infty$ to $+\infty$.

Meanwhile, $\Delta\omega_n \to +0$ and Riemannian sums in the right-hand expressions of (\ref{eq-5}) and (\ref{eq-7}) become integrals: \begin{equation} f(x)= \int_{-\infty}^\infty C(\omega ) e^{i\omega x}\,d \omega \label{eq-8} \end{equation} with \begin{equation} C(\omega)= \frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx; \label{eq-9} \end{equation} (\ref{eq-3}) becomes \begin{equation} \int_{-\infty}^\infty |f(x)|^2\,dx=2\pi\int_{-\infty}^\infty |C(\omega )|^2\,d\omega. \label{eq-10} \end{equation}

Definitions and Remarks

  1. (\ref{eq-9}) gives us a Fourier transform of $f(x)$, it usually is denoted by "hat":\begin{equation} \hat{f}(\omega)= \frac{1}{2\pi}\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx; \tag{FT} \end{equation} sometimes it is denoted by "tilde" ($\tilde{f}$), and seldom just by a corresponding capital letter $F(\omega)$.

  2. (\ref{eq-8}) is a Fourier integral aka inverse Fourier transform : \begin{equation*} f(x)= \int_{-\infty}^\infty \hat{f}(\omega) e^{i\omega x}\,d \omega \label{FI} \end{equation*} aka \begin{equation*} \check{F}(x)= \int_{-\infty}^\infty F(\omega) e^{i\omega x}\,d \omega \label{IFT} \end{equation*}

  3. (\ref{eq-10}) is Plancherel theorem \begin{equation} \int_{-\infty}^\infty |f(x)|^2\,dx=2\pi\int_{-\infty}^\infty |\hat{f}(\omega )|^2\,d\omega. \tag{PT} \end{equation}

  4. Sometimes expoments of $\pm i\omega x$ is replaced by $\pm 2\pi i\omega x$ and factor $1/(2\pi)$ dropped. Sometimes factor $\frac{1}{\sqrt{2\pi}}$ is placed in both Fourier transform and Fourier integral: \begin{align} &\hat{f}(\omega)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-i\omega x}\,dx; \tag{FT*}\\ &f(x)= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \hat{f}(\omega) e^{i\omega x}\,d \omega \tag{FI*} \end{align} Then FT and IFT differ only by $i$ replaced by $-i$ and Plancherel theorem becomes \begin{equation} \int_{-\infty}^\infty |f(x)|^2\,dx= \int_{-\infty}^\infty |\hat{f}(\omega )|^2\,d\omega. \tag{PT*} \end{equation}

$\cos $- and $\sin$-Fourier transform and integral

Applying the same arguments but as in the previous Lecture 16 we can rewrite (\ref{eq-8})--(\ref{eq-10}) as \begin{equation} f(x)= \int_0^\infty \bigl( A(\omega ) \cos (\omega x) +B(\omega ) \sin (\omega x)\bigr) \,d \omega \label{eq-11} \end{equation} with \begin{align} & A(\omega)= \frac{1}{\pi}\int_{-\infty}^\infty f(x)\cos (\omega x) \,dx, \label{eq-12}\\ & B(\omega)= \frac{1}{\pi}\int_{-\infty}^\infty f(x)\sin (\omega x) \,dx, \label{eq-13} \end{align} and \begin{equation} \int_{-\infty}^\infty |f(x)|^2\,dx=\pi\int_0^\infty \bigl( |A(\omega )|^2+|B(\omega )|^2\bigr)\,d\omega. \label{eq-14} \end{equation}

$A(\omega)$ and $B(\omega)$ are $\cos$- and $\sin$- Fourier transforms and

  1. $f(x)$ is even function iff $B(\omega)=0$;
  2. $f(x)$ is odd function iff $A(\omega)=0$.


  1. Each function on $[0,\infty)$ ] could be decomposed into $\cos$-Fourier integral \begin{equation} f(x)= \int_0^\infty A(\omega ) \cos (\omega x) \,d \omega \label{eq-15} \end{equation} with \begin{equation} A(\omega)= \frac{2}{\pi}\int_0^\infty f(x)\cos (\omega x) \,dx. \label{eq-16} \end{equation}
  2. Each function on $[0,\infty)$ ] could be decomposed into $\sin$-Fourier integral \begin{equation} f(x)= \int_0^\infty B(\omega ) \sin (\omega x) \,d \omega \label{eq-17} \end{equation} with \begin{equation} B(\omega)= \frac{2}{\pi}\int_0^\infty f(x)\sin (\omega x) \,dx. \label{eq-18} \end{equation}

Discussion: pointwise convergence of Fourier integrals and series

Recall Theorem 15.2

Let $f$ be a piecewise continuously differentiable function. Then the Fourier series \begin{equation} \frac{a_0}{2}+\sum_{n=1}^\infty \Bigl(a_n\cos \bigl(\frac{\pi n x}{l}\bigr) + a_n\cos \bigl(\frac{\pi n x}{l}\bigr)\Bigr) \label{eq-19} \end{equation} converges to

(b) $\frac{1}{2}\bigl(f(x+0)+f(x-0)\bigr)$ if $x$ is internal point and $f$ is discontinuous at $x$.

Exactly the same statement holds for Fourier Integral in the real form \begin{equation} \int_0^\infty \Bigl(A(\omega) \cos (\omega x) + B(\omega)\sin (\omega x)\Bigr)\,d\omega \label{eq-20} \end{equation} where $A(\omega)$ and $B(\omega)$ are $\cos$- and $\sin$-Fourier transforms.

None of them however holds for Fourier series or Fourier Integral in the complex form: \begin{gather} \sum_{n=-\infty}^\infty c_n e^{i\frac{\pi n x}{l}},\label{eq-21}\\ \int_{-\infty}^\infty C(\omega)e^{i\omega x}\,d\omega.\label{eq-22} \end{gather}

Why and what remedy do we have? If we consider definition of the partial sum of (\ref{eq-19}) and then rewrite in the complex form and similar deal with (\ref{eq-22}) we see that in fact we should look at \begin{gather} \lim_{N\to \infty} \sum_{n=-N}^N c_n e^{i\frac{\pi n x}{l}},\label{eq-23}\\ \lim_{N\to \infty} \int_{-N}^N C(\omega)e^{i\omega x}\,d\omega\label{eq-24}. \end{gather} Meanwhile convergence in (\ref{eq-21}) and (\ref{eq-22}) means more than this: \begin{gather} \lim_{M,N\to \infty} \sum_{n=-M}^N c_n e^{i\frac{\pi n x}{l}},\label{eq-25}\\ \lim_{M,N\to \infty} \int_{-M}^N C(\omega)e^{i\omega x}\,d\omega\label{eq-26} \end{gather} where $M,N$ tend to $\infty$ independently. So the remedy is simple: understand convergence as in (\ref{eq-23}), (\ref{eq-24}) rather than as in (\ref{eq-25}), (\ref{eq-26}).

For integrals such limit is called essential value of integral and is denoted by \begin{equation*} \operatorname{ess}-\int_{-\infty}^\infty G(\omega)\,d\omega \end{equation*} or \begin{equation*} \operatorname{vrai}-\int_{-\infty}^\infty G(\omega)\,d\omega \end{equation*} BTW similarly is defined \begin{equation*} \operatorname{ess}-\int_{a}^b G(\omega)\,d\omega:= \lim_{\varepsilon\to +0} \Bigl(\int_a^{c-\varepsilon}G(\omega)\,d\omega+ \int_{c+\varepsilon}^bG(\omega)\,d\omega\Bigr) \end{equation*} if there is a singularity at $c\in (a,b)$.

This is more general than the improper integrals studied in the end of Calculus I (which in turn generalize Riemann integrals). Those who took Complex Variables encountered such notion.