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.

We have \begin{align}f(t_{0})=:\int_{-\infty }^{\infty } \! f(t)\delta(t-t_{0}) \, \text{d}t &&(1)  \\ \hat g(\omega):=\widehat{g(\cdot)}(\omega):=\int_{-\infty }^{\infty }e^{-j \omega t}g(t)  \ \text{d}t &&(2) \end{align} 
and therefore we get:
$$\widehat{\delta(\cdot-t_0)}(\omega) \stackrel{(2)}{=} \frac{1}{\sqrt{2 \pi} }\int_{-\infty }^{\infty }  \underbrace{e^{-j \omega t}}_{=:f(t)} \delta (t-t_{0})\ \text{d}t\stackrel{(1)}{=}\frac{1}{\sqrt{2 \pi}}f(t_0)=\frac{1}{\sqrt{2 \pi}}e^{-j \omega t_{0}}$$
To be precise this is the Fourier transform of $\delta(\cdot-t_0)$ and not $\delta$. For $\delta=\delta(\cdot)$ we'd have to set $t_0=0$ in above formula and we get $\hat\delta(\omega)=(2\pi)^{-1/2}.$
