Fourier series in complex form and Fourier integral

0.1 Fourier series in complex form

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

$f(t)={\displaystyle \frac{a_0}{2}}+{\displaystyle \sum _{n=1}^{\mathrm{\infty }}}\left(a_n\cos {\displaystyle \frac{n\pi t}{p}}+b_n\sin {\displaystyle \frac{n\pi t}{p}}\right),$(1)

where the coefficients are

$a_n={\displaystyle \frac{1}{p}}{\displaystyle {\int }_{-p}^p}f(x)\cos {\displaystyle \frac{n\pi t}{p}}dt,b_n={\displaystyle \frac{1}{p}}{\displaystyle {\int }_{-p}^p}f(x)\sin {\displaystyle \frac{n\pi t}{p}}dt.$(2)

If one expresses the cosines and sines via Euler formulas (http://planetmath.org/ComplexSineAndCosine) with exponential function (http://planetmath.org/ComplexExponentialFunction), the series (1) attains the form

$f(t)={\displaystyle \sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}}{c}_n{e}^{\frac{in\pi t}{p}}.$(3)

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

$c_n={\displaystyle \frac{1}{2p}}{\displaystyle {\int }_{-p}^p}f(t)e^{\frac{-in\pi t}{p}}dt\mathit{\hspace{1em}\hspace{1em}}(n=0,\pm 1,\pm 2,\mathrm{\dots }).$(4)

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

0.2 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):

$$f(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{\frac{in\pi t}{p}}\frac{1}{2p}{\int }_{-p}^{p}f(t)e^{\frac{-in\pi t}{p}}𝑑t$$

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

$$f(t)=\frac{1}{2\pi }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{i{\omega }_{n}t}{\mathrm{\Delta }}_n\omega {\int }_{-p}^{p}f(t)e^{-i{\omega }_{n}t}𝑑t.$$

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

$f(t)={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{i\omega t}𝑑\omega {\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)e^{-i\omega t}𝑑t.$(5)

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

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}|f(t)|𝑑t$$

converges.

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

$f(t)={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}c(\omega )e^{i\omega t}𝑑\omega ,$(6)

where

$c(\omega )={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(t)e^{-i\omega t}𝑑t.$(7)

0.3 Fourier transform

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

$F(\omega )={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-i\omega t}f(t)𝑑t,$(8)

then by (5),

$f(t)={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{i\omega t}F(\omega )𝑑\omega .$(9)

$F(\omega )$ is called the Fourier transform of $f(t)$.  It is an integral transform and (9) its inverse transform.

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