PlanetPhysics/Fourier Series in Complex Form and Fourier Integral

Fourier series in complex form
The Fourier series expansion of a Riemann integrable real function $$f$$ on the interval \,$$[-p,\,p]$$\, is $$\begin{matrix} f(t) = \frac{a_0}{2}+\sum_{n=1}^\infty\left(a_n\cos{\frac{n\pi t}{p}}+b_n\sin{\frac{n\pi t}{p}}\right), \end{matrix}$$ where the coefficients are $$\begin{matrix} a_n = \frac{1}{p}\int_{-p}^{\,p}f(x)\cos{\frac{n\pi t}{p}}\,dt, \quad b_n = \frac{1}{p}\int_{-p}^{\,p}f(x)\sin{\frac{n\pi t}{p}}\,dt. \end{matrix}$$ If one expresses the cosines and sines via Euler formulas with exponential function, the series (1) attains the form $$\begin{matrix} f(t) = \sum_{n=-\infty}^\infty c_ne^{\frac{in\pi t}{p}}. \end{matrix}$$ 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 $$\begin{matrix} c_n = \frac{1}{2p}\int_{-p}^{\,p}f(t)e^{\frac{-in\pi t}{p}}\,dt \qquad (n = 0,\,\pm1,\,\pm2,\,\ldots). \end{matrix}$$

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

Derivation of Fourier integral
For seeing how the expansion (3) changes when\, $$p \to \infty$$,\, 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 $$f(t) = \frac{1}{2\pi}\sum_{n=-\infty}^\infty e^{i\omega_nt}\Delta_n\omega \int_{-p}^{\,p}f(t)e^{-i\omega_nt}\,dt.$$ It can be shown that when\, $$p \to \infty$$\, and thus\, $$\Delta_n\omega \to 0$$,\, the limiting form of this equation is $$\begin{matrix} f(t) \,=\, \frac{1}{2\pi}\int_{-\infty}^{\,\infty} e^{i\omega t}d\omega\int_{-\infty}^{\,\infty} f(t)e^{-i\omega t}dt. \end{matrix}$$ 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 $$\int_{-\infty}^{\,\infty}|f(t)|\,dt$$ converges.

For better to compare to the Fourier series (3) and the coefficients (4), we can write (5) as $$\begin{matrix} f(t) \,=\, \int_{-\infty}^{\,\infty}c(\omega)e^{i\omega t}d\omega, \end{matrix}$$ where $$\begin{matrix} c(\omega) \,=\, \frac{1}{2\pi}\int_{-\infty}^{\,\infty}f(t)e^{-i\omega t}dt. \end{matrix}$$

Fourier transform
If we denote $$2\pi c(\omega)$$ as $$\begin{matrix} F(\omega) \,=\, \int_{-\infty}^{\,\infty} e^{-i\omega t}f(t)\,dt, \end{matrix}$$ then by (5), $$\begin{matrix} f(t) \,=\, \frac{1}{2\pi}\int_{-\infty}^{\,\infty}e^{i\omega t}F(\omega)\,d\omega. \end{matrix}$$ $$F(\omega)$$ is called the 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 (8) and the formula (9) equipped with the same constant factor $$\frac{1}{\sqrt{2\pi}}$$ in front of the integral sign.