How to obtain the CTFT of a periodic function in terms of f in hertz (from the formula in terms of $ \omega $)
Recall:
$ x(t)=\sum^{\infty}_{k=-\infty} a_{k}e^{ikw_{0}t} $
$ \mathcal{X}(\omega)=2\pi\sum^{\infty}_{k=-\infty}a_{k}\delta(w-kw_{0}) $
To obtain X(f), use the substitution
$ \omega= 2 \pi f $.
More specifically
$ \begin{align} X(f) &=\mathcal{X}(2\pi f) \\ &=2\pi\sum^{\infty}_{k=-\infty}a_{k}\delta(2\pi f-kw_{0}) \\ &=\sum^{\infty}_{k=-\infty}a_{k}\delta(f-\frac{kw_{0}}{2\pi}) \end{align} $
$ Since\ k\delta (kt)=\delta (t),\forall k\ne 0 $
