# Applied Mathematics/Parseval's Theorem

## Parseval's theorem

$\int _{-\infty }^{\infty }|x(t)|^{2}\,dt=\int _{-\infty }^{\infty }|X(f)|^{2}\,df$

where $X(f)={\mathcal {F}}\{x(t)\}$  represents the continuous Fourier transform of x(t) and f represents the frequency component of x. The function above is called Parseval's theorem.

## Derivation

Let ${\bar {X}}(f)$  be the complex conjugation of $X(f)$ .

$X(-f)=\int _{-\infty }^{\infty }x(-t)e^{-ift}$
$=\int _{-\infty }^{\infty }x(t)e^{ift}$
$={\bar {X}}(f)$
$\int _{-\infty }^{\infty }|X(f)|^{2}\,df$

Here, we know that $X(f)$  is eqaul to the expansion coefficient of $x(t)$  in fourier transforming of $x(t)$ .
Hence, the integral of $|X(f)|^{2}$  is

$\int _{-\infty }^{\infty }{\bar {X}}(f)X(f)\,df$
$=\int _{-\infty }^{\infty }\left({\frac {1}{{\sqrt {2}}\pi }}\int _{-\infty }^{\infty }x(t)e^{ift}dt\right)\left({\frac {1}{{\sqrt {2}}\pi }}\int _{-\infty }^{\infty }x(t')e^{-ift'}dt'\right)df$
$=\int _{-\infty }^{\infty }x(t)x(t')\left({\frac {1}{2\pi }}e^{-if(t-t')}df\right)dtdt'$
$=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }x(t)x(t')\delta (t-t')dtdt'$
$=\int _{-\infty }^{\infty }|x(t)|^{2}dt$

Hence

$\int _{-\infty }^{\infty }|x(t)|^{2}\,dt=\int _{-\infty }^{\infty }|X(f)|^{2}\,df$