How to obtain the Parseval's property in terms of f in hertz (from the formula in terms of $ \omega $)
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \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) $
