I need to find a Laplace Transform but have problems

Question

Using Laplace transform find: $e^{-10t}u(t)$.

According to the Laplace table, that is equal to $1/s+10$, but how can I prove it?

Answer 1

The First Translation Theorem:

If $\mathcal{L}\{f(t)\}=F(s)$ and $a\in\mathbb{R}$, then $\mathcal{L}\{e^{at}f(t)\}=F(s-a)$.

The Second Translation Theorem:

If $F(s)=\mathcal{L}\{f(t)\}$ and $a>0$, then $\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$.

Using these two in conjunction, we easily deduce that $\mathcal{L}\{e^{-10t}u(t)\}=\frac{1}{s+10}$.

OR

$\mathcal{L}\{e^{-10t}u(t)\}=\int_0^\infty e^{-st}\left(e^{-10t}u(t)\right)dt$

$=\int_0^\infty e^{-(10+s)t}u(t)dt$

$=\int_0^\infty e^{-(10+s)t}u(t)dt$

$=\int_0^\infty e^{-(10+s)t}dt$
$\because$ $u(t)=1$ when $t\geq0$

$=\frac{e^{-(10+s(\infty))}}{-(10+s)}-\frac{e^{-(10+s(0))}}{-(10+s)}$

$=\frac{1}{s+10}$