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

 $\displaystyle f(t)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n}\cos{\frac{n% \pi t}{p}}+b_{n}\sin{\frac{n\pi t}{p}}\right),$ (1)

where the coefficients are

 $\displaystyle a_{n}=\frac{1}{p}\int_{-p}^{\,p}f(x)\cos{\frac{n\pi t}{p}}\,dt,% \quad b_{n}=\frac{1}{p}\int_{-p}^{\,p}f(x)\sin{\frac{n\pi t}{p}}\,dt.$ (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

 $\displaystyle f(t)=\sum_{n=-\infty}^{\infty}c_{n}e^{\frac{in\pi t}{p}}.$ (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

 $\displaystyle c_{n}=\frac{1}{2p}\int_{-p}^{\,p}f(t)e^{\frac{-in\pi t}{p}}\,dt% \qquad(n=0,\,\pm 1,\,\pm 2,\,\ldots).$ (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  $p\to\infty$,  we put first the expressions (4) of $c_{n}$ to the series (3):

 $f(t)=\sum_{n=-\infty}^{\infty}e^{\frac{in\pi t}{p}}\frac{1}{2p}\int_{-p}^{\,p}% f(t)e^{\frac{-in\pi t}{p}}\,dt$

By denoting  $\omega_{n}:=\frac{n\pi}{p}$  and  $\Delta_{n}\omega:=\omega_{n+1}\!-\!\omega_{n}=\frac{\pi}{p}$,  the last equation takes the form

 $f(t)=\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}e^{i\omega_{n}t}\Delta_{n}\omega% \int_{-p}^{\,p}f(t)e^{-i\omega_{n}t}\,dt.$

It can be shown that when  $p\to\infty$  and thus  $\Delta_{n}\omega\to 0$,  the limiting form of this equation is

 $\displaystyle f(t)\,=\,\frac{1}{2\pi}\int_{-\infty}^{\,\infty}e^{i\omega t}d% \omega\int_{-\infty}^{\,\infty}f(t)e^{-i\omega t}dt.$ (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

 $\int_{-\infty}^{\,\infty}|f(t)|\,dt$

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

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

where

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

0.3 Fourier transform

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

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

then by (5),

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

$F(\omega)$ is called the 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.

References

• 1 K. Väisälä: Laplace-muunnos.  Handout Nr. 163. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).
Title Fourier series in complex form and Fourier integral FourierSeriesInComplexFormAndFourierIntegral 2013-03-22 18:02:54 2013-03-22 18:02:54 pahio (2872) pahio (2872) 10 pahio (2872) Derivation msc 42A38 msc 42A16 msc 44A55 FourierTransform KalleVaisala Fourier integral