| Parseval's property |
$ \mathcal{X}(\omega)=\mathcal{X}(2\pi f) \ $
$ \begin{align} \int_{-\infty}^{\infty} |x(t)|^2 dt &= \frac{1}{2\pi} \int_{-\infty}^{\infty} |\mathcal{X}(2\pi f)|^2 d2\pi f \\ &= \int_{-\infty}^{\infty} |\mathcal{X}(2\pi f)|^2 df \\ &= \int_{-\infty}^{\infty} |X(f)|^2 df \end{align} $
$ Since\ X(\alpha)=\mathcal{X}(2\pi \alpha) $
