Dirac delta distribution and fourier transform Dirac delta distribution is defined as 
$f(t_{0})=\int_{-\infty }^{\infty } \! f(t)\delta(t-t_{0}) \, dt $ where $f(t)$ is smooth function. Then my question is:
:Calculate Fourier transform $\hat \delta(\omega)$ from $\delta (t-t_{0})$
Solution:
$$\hat \delta(\omega)=\frac{1}{\sqrt{2 \pi} }\int_{-\infty }^{\infty } \! \delta (t-t_{0}) e^{-j \omega t}\, dt   $$

$$\hat \delta(\omega)=\frac {1}{\sqrt{2 \pi}}e^{-j \omega t_{0}}$$

Can someone explain me how they got this solution and write what are the steps between? On internet I always find some general formulas and I don't know how to use them.

You indeed have to start with the definition. The action of the Fourier transform of the distribution $f$ on a test function $\phi$ is defined as
$$(\mathcal F[f], \phi) = (f, \mathcal F[\phi]),$$
which is valid because if $\phi$ is a test function of rapid decay, then so is $\mathcal F[\phi]$, and the functional defined in this way is linear and continuous. From this we obtain
$$(\mathcal F[\delta], \phi) =
(\delta, \mathcal F[\phi]) =
\mathcal F[\phi](0) =
\frac 1 {\sqrt {2 \pi}} \int_{-\infty}^\infty \phi(t) dt,$$
which means that $\mathcal F[\delta]$ is the same functional as $1/\sqrt {2 \pi}$.
The transform of $\delta(t - t_0)$ can be found either in the same way directly or by applying the translation property, which is valid for singular distributions because a linear transformation is defined so as to be consistent with a linear transformation of a regular distribution.
