Find the Laplace transform for $$f(t) = e^{-4t}t^{\frac{3}{2}}$$
My Work So Far
The form of $f(t)$ suggests that I should use an s-axis shift, $L${$t^{3/2}$} = $F(s+4)$.
I think that the appropriate form for this would then be $f(t) = t^a$, which according to my table should lead to $F(s) = \frac{Γ(\frac{5}{2})}{(s+4)^{5/2}}$.
The denominator is left as is, and the numerator is equivalent to $\int_o^∞e^{-t}t^{3/2}dt$.
This is an integration by parts, where:
$u=e^{-t}$, $du = -e^{-t}dt$, $dv = t^{3/2}$, and $v = \frac{2}{5}t^{5/2}$
Resulting in an equation to solve:
$$\lim_{T\to ∞}\frac{2e^{-t}t^{5/2}}{5}|_0^T+ \frac{2}{5}\int t^{5/2}e^{-t}dt$$
Except...wait. The remaining integral also requires integration by parts. And that will leave behind an integral that will also require an integration by parts. And so on. This means that I've clearly messed up because I've come up with a step that I can't escape from, but I don't know where that would have been. Would anyone be able to help me figure it out?
\begin{align} \int_0^\infty e^{-t(s + 4)} t^{3/2} \, dt & = \frac{1}{(s+4)^{5/2}} \int_0^\infty e^{-u} u^{3/2} \, du \\ & = \frac{1}{(s+4)^{5/2}} \int_0^\infty e^{-u} u^{5/2-1} du \\ & = \frac{\Gamma \left(\frac{5}{2}\right)}{(s+4)^{5/2}} \end{align}