What is the Fourier transform of $f(t)=1$ or simply a constant? Wolfram alpha gives the answer to be

$$F(\omega)=\sqrt{2\pi}\delta(\omega)$$

Does that mean that the function is valued $\sqrt{2\pi}$ at all points in the frequency domain? I think this is reasonable because such function i.e. $f(t)=1$ in the time domain would be sum of all the harmonics of a sinusoid and hence would contain all the frequencies. Maybe no, the function isn't varying at all and hence the frequency is $0$. But then the Fourier transform should have been $\delta(0)$ instead of $\delta(\omega)$.

Someone please shed some light on this!
I think the clearest way to see this is by noting that we have (depending on your convention for the placement of $2 \pi$ in Fourier transforms) that
$$\mathcal{F}(\mathcal{F}(f(x))) = 2 \pi f(-x)$$
Taking the convention that
$$\tilde{f}(k) = \int_{-\infty}^\infty e^{ikx} f(x) \; dx$$
so
$$f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx} \tilde{f}(k) \; dk$$
we get
$$f(-x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-ikx} \tilde{f}(k) \; dk = \frac{1}{2\pi} \mathcal{F(\tilde{f}(k))} = \frac{1}{2\pi}\mathcal{F}(\mathcal{F}(f(x)))$$
Note we have
$$\mathcal{F}(\delta(x)) = \int_{-\infty}^\infty e^{ikx} \delta(x) \; dx = 1$$
So then
$$\mathcal{F}(1) = \mathcal{F}(\mathcal{F}(\delta(x))) = 2 \pi \delta(-x) = 2 \pi \delta(x)$$
For other constants, note by linearity we have
$$\mathcal{F}(c) = c \mathcal{F}(1) = 2 \pi c \delta(x)$$
Just use the definition. $F(w) = \int_{-\infty}^\infty e^{-iwt}dt =\delta(w)$ by how the dirac delta is defined. This is normalized to 1at zero and 0 otherwise.