Parseval's Theorem

Let $E(t)$ be a continuous function and $E(t)$ and $E_\nu$ be Fourier Transform pairs so that

$$E(t) \equiv \int_{-\infty}^\infty E_\nu e^{-2\pi i\nu t}\, d\nu$$ (1)

$$E^*(t) \equiv \int_{-\infty}^\infty {E_{\nu'}}^*e^{2\pi i\nu't}\,d\nu'.$$ (2)

Then

$\int_{-\infty}^\infty \vert E(t)\vert^2\,dt =\int_{-\infty}^\infty E(t)E^*(t)\,dt$
$\quad = \int_{-\infty}^\infty\left[{\int_{-\infty}^\infty E_\nu e^{-2\pi i\nu t}\,d\nu \int_{-\infty}^\infty {E_{\nu'}}^*e^{2\pi i\nu't}\,d\nu'}\right]\,dt$
$\quad = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty E_\nu {E_{\nu'}}^*e^{2\pi it(\nu'-\nu)}\,d\nu\,d\nu'\,dt$
$\quad = \int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty E_\nu {E_{\nu'}}^*e^{2\pi it(\nu'-\nu)}\,dt\,d\nu\,d\nu'$
$\quad = \int_{-\infty}^\infty\int_{-\infty}^\infty\delta(\nu'-\nu)E_\nu {E_{\nu'}}^*\,d\nu\,d\nu'$
$\quad = \int_{-\infty}^\infty E_\nu {E_{\nu}}^*\,d\nu = \int_{-\infty}^\infty\vert E_\nu\vert^2\,d\nu.$	(3)

where $\delta(x-x_0)$ is the Delta Function.

For finite Fourier Transform pairs $h_k$ and $H_n$,

$$\sum_{k=0}^{N-1} \vert h_k\vert^2 = {1\over N}\sum_{n=0}^{N-1} \vert H_n\vert^2.$$ (4)

If a function has a Fourier Series given by

$$f(x) = {\textstyle{1\over 2}}a_0 +\sum_{n=1}^\infty a_n\cos(nx) + \sum_{n=1}^\infty b_n\sin(nx),$$ (5)

then Bessel's Inequality becomes an equality known as Parseval's theorem. From (5),

$[f(x)]^2 = {\textstyle{1\over 4}}{a_0}^2 + a_0 \sum_{n=1}^\infty [a_n\cos(nx)+b_n\sin(nx)]$
$\quad +\sum_{n=1}^\infty \sum_{m=1}^\infty [a_na_m\cos(nx)\cos(mx)$
$\quad +a_nb_m\cos(nx)\sin(mx)+a_mb_n\sin(nx)\cos(mx)$
$\quad +b_nb_m\sin(nx)\sin(mx)].$	(6)

Integrating

$\int^{\pi}_{-\pi} [f(x)]^2\,dx = {\textstyle{1\over 4}}{a_0}^2 \int^{\pi}_{-\pi} dx$
$\quad + a_0 \int^{\pi}_{-\pi} \sum_{n=1}^\infty [a_n\cos(nx)+b_n\sin(nx)]\,dx$
$\quad +\int^{\pi}_{-\pi} \sum_{n=1}^\infty \sum_{m=1}^\infty[a_na_m\cos(nx)\cos(mx)$
$\quad +a_nb_m\cos(nx)\sin(mx)+a_mb_n\sin(nx)\cos(mx)$
$\quad +b_nb_m\sin(nx)\sin(mx)]\,dx = {\textstyle{1\over 4}}{a_0}^2 (2\pi) + 0$
$\quad + \sum_{n=1}^\infty \sum_{m=1}^\infty [a_na_m\pi \delta_{nm}+0+0+b_nb_m\pi \delta_{nm}],$	(7)

so

$${1\over\pi}\int^{\pi}_{-\pi} [f(x)]^2\,dx ={\textstyle{1\over 2}}{a_0}^2 + \sum_{n=1}^\infty ({a_n}^2+{b_n}^2).$$ (8)

For a generalized Fourier Series with a Complete Basis $\{\phi_i\}_{i=1}^\infty$, an analogous relationship holds. For a Complex Fourier Series,

$${1\over 2\pi} \int^{\pi}_{-\pi} \vert f(x)\vert^2\,dx=\sum_{n=-\infty}^\infty \vert a_n\vert^2.$$ (9)

References

Gradshteyn, I. S. and Ryzhik, I. M. Tables of Integrals, Series, and Products, 5th ed. San Diego, CA: Academic Press, p. 1101, 1979.