ℒ{f(t)}(s)=0∫∞f(t)e-stdt; f(t)=t3e3t.
ℒ{f(t)}(s)=0∫∞t3e3te-stdt=0∫∞t3e(3-s)tdt.
Integrate by parts:
Let u=t3, du=3t2dt; dv=e(3-s)tdt, v=(1/(3-s))e(3-s)t. We'll consider the integration limits later.
F(s)=ℒ{f(t)}(s)=∫udv=∫t3e(3-s)tdt=(t3/(3-s))e(3-s)t-(3/(3-s))∫t2e(3-s)tdt. (This is simply a rearrangement of d(uv)=vdu+udv or uv=∫vdu+∫udv, ∫udv=uv-∫vdu.)
Let J=∫t2e(3-s)tdt, u=t2, du=2tdt; v=(1/(3-s))e(3-s)t as before.
J=t2e(3-s)t/(3-s)-(2/(3-s))∫te(3-s)tdt.
Let K=∫te(3-s)tdt, u=t, du=dt; J=t2e(3-s)t/(3-s)-2K/(3-s).
K=te(3-s)t/(3-s)-(1/(3-s))∫e(3-s)tdt=te(3-s)t/(3-s)-e(3-s)t/(3-s)2.
Assume 3-s<0, i.e., s>3. All terms containing e(3-s)t are zero when t→∞ and are 1 when t=0, which zeroises (cancels) all t terms.
Now apply the integration limits (indicated by |0∞):
K|0∞=1/(3-s)2; J|0∞=2K|0∞/(3-s)=2/(3-s)3;
F(s)=[t3e(3-s)t/(3-s)-3J|0∞/(3-s)]t=0t=∞=6/(3-s)4, which can also be written 3!/(s-3)4.
So ℒ{t3e3t}=3!/(s-3)4.
ℒ{tneat}=n!/(s-a)n+1 in general.