Why is $\int^\infty _{0^-}\delta(t-nT)e^{-st}dt = e^{-nsT}$? I'm currently in the process of going over the $\mathcal Z$-transform and more specifically its derivation. I understand and I am able to follow it up until the final step whereby involving the Laplace transform.

The step that is confusing me is indicated as the step within the red box, derivation taken from Wikipedia page on the $\mathcal Z$-transform. I do not understand how this integral solves to $e^{-nsT}$. I have searched for possible identities relating to this but have not been able to find any help.

\begin{align}
X_q(s)&amp;=\int_{0^-}^\infty x_q(t)e^{-st} dt\\
&amp;=\int_{0^-}^\infty \sum_{n=0}^\infty x[n]\delta (t-nT)e^{-st} dt\\
&amp;=\sum_{n=0}^\infty x[n]\color{red}{\boxed{\color{black}{\int_{0^-}^\infty \delta (t-nT)e^{-st} dt}}}\\
&amp;= \sum_{n=0}^\infty x[n]\color{red}{\boxed{\color{black}{e^{-nsT}}}}
\end{align}

Thanks for taking the time for looking at my question. Any help would be greatly appreciated!

Regards
Remember the definition of the delta function: it integrates $1$ all over the t-axis but it is zero for any $t\neq 0$. This means that if we multiply any function by $\delta (t)$, then the function is going to be zero everywhere except for $t=0$. In this case, we have

$$ \delta (t-nT)e^{-st}=
\left\{
 \begin{array}{ll}
  \delta (t-nT)e^{-snT}  &amp; \mbox{if } t = nT \\
  0 &amp; \mbox{if } t\neq nT
 \end{array}
\right.$$

So if we integrate this function, we have that the integral is zero for every point except for $t=nT$. So

$$\int_0^\infty \delta (t-nT)e^{-st}dt=\int_0^\infty \delta (t-nT)e^{-snT}dt=e^{-snT}\int_0^\infty \delta (t-nT)dt$$

But by definition, $$\int_0^\infty \delta (t-nT)dt=\int_{-\infty}^\infty \delta (t-nT)dt=1$$

So $$\int_0^\infty \delta (t-nT)e^{-st}dt=e^{-snT}$$

Sidenote: For this to be true, $n$ must be positive. If it is negative, then the integral will be $0$ (as the function inside the integral will be zero for all the integration interval).