Calculate Inverse Laplace transform Can anybody help me with the answer of this question?

Find the inverse Laplace transform of  $$\frac {1}{(s-3)^4}$$

 A: This will surely help: 

Let $F(t)=\mathscr L^{-1}\{f(s)\}$. Then $$\mathscr L^{-1}\{f(s-a)\}=e^{at} F(t)$$

A: Take advantage of formal properties of the Laplace transform:


*

*If $\mathcal{L}_s(f(x)) = g(s)$, then $\mathcal{L}_s(\mathrm{e}^{a x} f(x)) = g(s-a)$


Also use the table entry:


*

*$\mathcal{L}_s(x^n) = (n-1)! s^{-n}$ 

A: A related problem. The inverse Laplace transform is defined by,
$$ f(x) = \frac{1}{2\pi i}\int_C F(s) {\rm e}^{sx} ds \,$$
where $F(s)$ is the Laplace transform of $f(x)$, and $C$ is the Bromwich contour (see below). 
$$ f(x) = \frac{1}{2\pi i} \int_C  \frac{{\rm e}^{sx}}{(s-2)^4} ds = \frac{1}{3!} \frac{d^3 }{ds^3} {\rm e}^{sx} |_{s=2} = \frac{1}{6} x^3 {\rm e}^{2x}\,,$$
since the integral has a pole of order $4$ at $s=2$, and using the residue theorem.

A: $$ \mathcal{L}[t^n]= \frac{\Gamma(n+1)}{s^{n+1}}   $$
 if n is natural number  we can write : 
$$  \mathcal{L}[t^n]= \frac{n!}{s^{n+1}}   $$
and we know: 
$$ \mathcal{L}[e^{at}f(t)]=F(s-a) $$
therefor: 
$$ \frac {1}{(s-3)^4} = \frac { \frac{3!}{3!} }{(s-3)^{3+1}} =\frac{1}{3!}\frac{3!}{(s-3)^{3+1}}$$
$$ \mathcal{L^{-1}} \left[ \frac{1}{3!}\frac{3!}{(s-3)^{3+1}} \right] = \frac{1}{3!} t^3 e^{3t} $$
