Fourier transform of impulse function I don't understand why the Fourier transoform of $$\delta(t-t_0)$$ is equal to $$e^{-jwt_0}$$
Starting from the beginning, the FT of dirac delta is defined as:
$$ \int_{- \infty }^{ \infty  }\delta (t - t_0)e^{-j\omega t}dt$$
I have no idea how to calculate integral of dirac delta, for the second term thei ntegral is:
$$\int_{- \infty }^{ \infty  }e^{-j\omega t}dt = \frac{e^{jwt}}{jw}$$
but this far from expected result.
So could somebody make a step by step Fourier Transofrm for diract delta?
Thanks in advance :)
 A: The Dirac Delta "function" represents a distribution of functions that have an area equal to 1 with infinitely tall height and a correspondingly scaled width.  Since we don't care about the derivatives of the delta function in this problem, we can represent the shifted delta function with this sequence of rectangular pulse functions, centered at $t_0$, as the pulse width $\tau \rightarrow 0$:
\begin{equation}
  p_{\tau}(t-t_0) = \begin{cases}
    \dfrac{1}{\tau}, & t_0 - \dfrac{\tau}{2}< t < t_0 + \dfrac{\tau}{2}.\\
\\
    0, & \text{otherwise}.
  \end{cases}
\end{equation}
The rest is straightforward:
$$\begin{align*}
\mathscr{F}\left\{\delta(t-t_0)\right\} &=\int_{- \infty }^{ \infty  }\delta (t - t_0)e^{-j\omega t}dt \\
\\
&= \lim_{\tau \rightarrow 0}\space\int_{-\infty }^{\infty}p_{\tau}(t-t_0)e^{-j\omega t}dt \\
\\
&= \lim_{\tau \rightarrow 0}\space\int_{t_0-\frac{\tau}{2} }^{t_0+\frac{\tau}{2}}\dfrac{1}{\tau}e^{-j\omega t}dt \\
\\
&= \lim_{\tau \rightarrow 0}\space\dfrac{-1}{j\omega\tau}e^{-j\omega t}\biggr\rvert_{t_0-\frac{\tau}{2}}^{t_0+\frac{\tau}{2}} \\
\\
&= \lim_{\tau \rightarrow 0}\space\dfrac{-1}{j\omega\tau}\left[e^{-j\omega \left(t_0 +\frac{\tau}{2}\right)}-e^{-j\omega \left(t_0 -\frac{\tau}{2}\right)}\right] \\
\\
&= \lim_{\tau \rightarrow 0}\space\dfrac{-e^{-j\omega t_0}}{j\omega\tau}\left[e^{-j\omega\frac{\tau}{2}}-e^{j\omega\frac{\tau}{2}}\right] \\
\\
&= \lim_{\tau \rightarrow 0}\space\dfrac{e^{-j\omega t_0}}{\omega\frac{\tau}{2}}\left[\dfrac{e^{j\omega\frac{\tau}{2}}-e^{-j\omega\frac{\tau}{2}}}{2j}\right] \\
\\
&= \lim_{\tau \rightarrow 0}\space e^{-j\omega t_0}\dfrac{\sin\left(\omega\frac{\tau}{2}\right)}{\omega\frac{\tau}{2}} \\
\\
&= e^{-j\omega t_0}
\end{align*}$$
