Fourier Transform Units
For a given signal $g(t)$ with Fourier transform $G(\omega)$:

$$g(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} G(\omega)e^{i\omega t}\textrm{d}\omega$$

If you change variables such that $\omega=2\pi f$

$$g(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} G(2\pi f)e^{i2\pi ft}\textrm{d}(2\pi f)=\int_{-\infty}^{\infty} G(2\pi f)e^{i2\pi ft}\textrm{d}f$$

So if you present your Fourier transform as a function of $f$ instead of $\omega$, your Fourier pair of $g(t)$ will now be $\hat{G}(f)=\frac{G(2\pi f)}{2\pi}$ and:

$$g(t) =\int_{-\infty}^{\infty} \hat{G}(f)e^{i2\pi ft}\textrm{d}f$$

Note that with this definition, there is a scaling of $2\pi$ in comparison with the first equation I wrote. So now, for example, the Fourier transform of a cosine will consist of two deltas not with amplitude $\pi$, but with amplitude $\frac{\pi}{2\pi}=\frac12$, as correctly stated in your question.