PlanetPhysics/Fourier Series in Complex Form and Fourier Integral

Fourier series in complex form

The Fourier series expansion of a Riemann integrable real [[../Bijective/|function]] $f$ on the interval \,$[-p,p]$\, is

$\begin{array}{c}f(t)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}(a_n\cos \frac{n\pi t}{p}+b_n\sin \frac{n\pi t}{p}),\end{array}$

where the coefficients are

$\begin{array}{c}{a}_n=\frac{1}{p}\underset{-p}{\overset{p}{\int }}f(x)\cos \frac{n\pi t}{p}dt,b_n=\frac{1}{p}\underset{-p}{\overset{p}{\int }}f(x)\sin \frac{n\pi t}{p}dt.\end{array}$

If one expresses the cosines and sines via Euler formulas with exponential function, the series (1) attains the form

$\begin{array}{c}f(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{c}_n{e}^{\frac{in\pi t}{p}}.\end{array}$

The coefficients $c_n$ could be obtained of $a_n$ and $b_n$, but they are comfortably derived directly by multiplying the equation (3) by $e^{-\frac{im\pi t}{p}}$ and integrating it from $-p$ to $p$.\, One obtains

$\begin{array}{c}{c}_n=\frac{1}{2p}\underset{-p}{\overset{p}{\int }}f(t)e^{\frac{-in\pi t}{p}}dt(n=0,\pm 1,\pm 2,\dots ).\end{array}$

We may say that in (3), $f(t)$ has been dissolved to sum of harmonics (elementary [[../CosmologicalConstant2/|waves]]) $c_n{e}^{\frac{in\pi t}{p}}$ with amplitudes $c_n$ corresponding the frequencies $n$.

Derivation of Fourier integral

For seeing how the expansion (3) changes when\, $p\to \mathrm{\infty }$,\, we put first the expressions (4) of $c_n$ to the series (3): $f(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{\frac{in\pi t}{p}}\frac{1}{2p}\underset{-p}{\overset{p}{\int }}f(t)e^{\frac{-in\pi t}{p}}dt$ By denoting\, ${\omega }_n:=\frac{n\pi }{p}$\, and\, ${\mathrm{\Delta }}_n\omega :={\omega }_{n+1}-{\omega }_n=\frac{\pi }{p}$,\, the last equation takes the form $f(t)=\frac{1}{2\pi }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{i{\omega }_{n}t}{\mathrm{\Delta }}_n\omega \underset{-p}{\overset{p}{\int }}f(t)e^{-i{\omega }_{n}t}dt.$ It can be shown that when\, $p\to \mathrm{\infty }$\, and thus\, ${\mathrm{\Delta }}_n\omega \to 0$,\, the limiting form of this equation is

$\begin{array}{c}f(t)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{i\omega t}d\omega \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)e^{-i\omega t}dt.\end{array}$

Here, $f(t)$ has been represented as a Fourier integral .\, It can be proved that for validity of the expansion (4) it suffices that the function $f$ is piecewise continuous on every finite interval having at most a finite amount of extremum points and that the integral $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|f(t)|dt$ converges.

For better to compare to the Fourier series (3) and the coefficients (4), we can write (5) as

$\begin{array}{c}f(t)=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}c(\omega )e^{i\omega t}d\omega ,\end{array}$

where

$\begin{array}{c}c(\omega )=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(t)e^{-i\omega t}dt.\end{array}$

Fourier transform

If we denote $2\pi c(\omega )$ as

$\begin{array}{c}F(\omega )=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-i\omega t}f(t)dt,\end{array}$

then by (5),

$\begin{array}{c}f(t)=\frac{1}{2\pi }\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{i\omega t}F(\omega )d\omega .\end{array}$

$F(\omega )$ is called the [[../FourierTransforms/|Fourier transform]] of $f(t)$.\, It is an integral transform and (9) represents its inverse transform.

N.B. that often one sees both the [[../Formula/|formula]] (8) and the formula (9) equipped with the same constant factor $\frac{1}{\sqrt{2\pi }}$ in front of the integral sign.