Applied Mathematics/Parseval's Theorem

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

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

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

${\displaystyle X(-f)=\int _{-\infty }^{\infty }x(-t)e^{-ift}}$

${\displaystyle =\int _{-\infty }^{\infty }x(t)e^{ift}}$

${\displaystyle ={\bar {X}}(f)}$

${\displaystyle \int _{-\infty }^{\infty }|X(f)|^{2}\,df}$

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

${\displaystyle \int _{-\infty }^{\infty }{\bar {X}}(f)X(f)\,df}$

${\displaystyle =\int _{-\infty }^{\infty }\left({\frac {1}{{\sqrt {2}}\pi }}\int _{-\infty }^{\infty }x(t)e^{ift}dt\right)\left({\frac {1}{{\sqrt {2}}\pi }}\int _{-\infty }^{\infty }x(t')e^{-ift'}dt'\right)df}$

${\displaystyle =\int _{-\infty }^{\infty }x(t)x(t')\left({\frac {1}{2\pi }}e^{-if(t-t')}df\right)dtdt'}$

${\displaystyle =\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }x(t)x(t')\delta (t-t')dtdt'}$

${\displaystyle =\int _{-\infty }^{\infty }|x(t)|^{2}dt}$

Hence

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