# 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
   ${\begin{matrix}f(t)=\sum _{n=-\infty }^{\infty }c_{n}e^{\frac {in\pi t}{p}}.\end{matrix}}$
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
   ${\begin{matrix}c_{n}={\frac {1}{2p}}\int _{-p}^{\,p}f(t)e^{\frac {-in\pi t}{p}}\,dt\qquad (n=0,\,\pm 1,\,\pm 2,\,\ldots ).\end{matrix}}$
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 $n$.

### 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
   ${\begin{matrix}f(t)\,=\,{\frac {1}{2\pi }}\int _{-\infty }^{\,\infty }e^{i\omega t}d\omega \int _{-\infty }^{\,\infty }f(t)e^{-i\omega t}dt.\end{matrix}}$
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.