Conversion between DTFT in radians/sample to DTFT in cycles/sample
You can use the following :

Consider the discrete time sequence $x[n]$ and the DTFT $X(\omega)$ related by:
$$ X(\omega) = \sum_k x[n] e^{-j \omega n} \longleftrightarrow x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(\omega) e^{j\omega n} d\omega$$

Then define $X'(\mu)$ as the normalized DTFT, which is related to $X(\omega)$ as
$$X'(\mu) = X(2\pi \mu) = \sum_k x[n] e^{-j 2\pi \mu n} $$ and then
$$x[n] = \int_{-0.5}^{0.5} X'(\mu) e^{j 2\pi \mu n} d\mu $$

Applying this to your example case:
$$x[n] = 1 \longleftrightarrow X(\omega) = 2\pi \sum_r \delta(\omega - 2\pi r)$$ then since $X'(\mu) = X(2\pi \mu)$ we have

$$X'(\mu) = 2\pi \sum_r \delta(2\pi \mu - 2\pi r) = 2\pi \sum_r \delta(2\pi [\mu - r]) = 2\pi \sum_r \frac{1}{2\pi} \delta(\mu - r) $$
$$X'(\mu) = \sum_r \delta(\mu - r) $$

as expected, where we've used the property that $\delta(ax) = \frac{1}{|a|} \delta(x)$.