You can extend the Fourier transform to distributions like the Dirac Delta function. Taking the Fourier transform of $\delta(t)$ gives
$$\int_{-\infty}^{\infty}\delta(t)e^{-i\omega t}\;dt=1$$
because
$$\int_{-\infty}^{\infty}\delta(t)f(t)\;dt=f(0)$$
If you apply the (inverse) Fourier transform to 1 you get
$$\mathcal{F}^{-1}1=\text{p.v.}\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\;d\omega=\frac{1}{\pi}\int_{0}^{\infty}\cos\omega t\;d\omega$$
Of course this is a divergent integral, but if it is used in a convolution integral it does have a meaning and it is useful to define
$$\delta(t)=\text{p.v.}\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\;d\omega=\frac{1}{\pi}\int_{0}^{\infty}\cos\omega t\;d\omega$$
The short answer is that the $2 \pi$ comes from the inversion formula.
Here is an informal perspective which gives as hint as to how this can be made
more rigorous:
A distribution is defined by its
action on a space of nicely behaved test functions.
For a function $f$ from a restricted class of ordinary functions, we can define a distribution $T_f$ by $T_f(\phi) = \int f \phi$. For the function
$t \mapsto 1$, we get
$T_1(\phi) = \int \phi$.
The '$\delta$ function' is defined by the distribution
$T_\delta(\phi) = \phi(0)$. That is, it takes a test function $\phi$ and
returns its value at $0$.
The Fourier transform of a distribution is defined by
$\hat{T_f}(\phi) = T_f(\hat{\phi})$, where $\hat{\phi}$ is the ordinary Fourier transform of $\phi$.
We see that ${\hat{T}_1}(\phi) = T_1(\hat{\phi}) = \int \hat{\phi}$.
The standard inversion formula shows that $\int \hat{\phi} = 2 \pi \phi(0)$,
which gives ${\hat{T}_1}(\phi) = 2 \pi T_\delta(\phi)$, or more succinctly,
${\hat{T}_1} = 2 \pi T_\delta$.
The same sort of analysis shows that
${\hat{T}_{t \mapsto e^{iat}}}(\phi) = 2 \pi \phi(a) = 2 \pi T_\delta(\omega \mapsto \phi(\omega+a))$. The last expression may be written informally as
$T_{ \omega \mapsto 2 \pi \delta(\omega-a) } (\phi)$, which is the desired result.
Best Answer
The fact that the function is not in $L^1$ does not imply that it does not have a Fourier transform, as this example shows. What the classical theory tells us is that if the function is in $L^1$ then its Fourier transform is well defined.
It does not defy the scaling property: Without loss of generality forget about all the $2\pi$ factors in your derivations (both in the Fourier transform and in the function). Then $$\mathcal{F}[sin(kt)](x)=\frac{i}{2}[\delta(x+k)-\delta(x-k)],$$ so $$\frac{1}{|k|}\mathcal{F}[sin(t)](x/k)=\frac{i}{2}\Big[\frac{\delta(x/k+1)}{|k|}-\frac{\delta(x/k-1)}{|k|}\Big].$$ The function $\frac{\delta(x/k+1)}{|k|}$ is zero everywhere except at -k and $$\int_{\mathbb{R}}\frac{\delta(x/k+1)}{|k|}dx=\int_{\mathbb{R}}\delta(y+1)dy=1,$$ so it is equal to $\delta(x+k)$. The same happens with $\frac{\delta(x/k-1)}{|k|}$.
Coca