I'm trying to prove that $$\mathcal{L}\{t^2e^{at}\} = \frac{2}{(s-a)^3}.$$
I've gotten to the last integration by parts where
$$ \lim_{n\to\infty}\int_0^n\frac{1}{(a-s)^22e^{(a-s)t}}dt = \left. \lim_{n\to\infty}\frac{2}{(a-s)^3}e^{(a-s)t} \right|_0^n. $$
Now what do I do? I can't find a way to make that last term converge?
$\endgroup$ 41 Answer
$\begingroup$Evaluating the integral for a fixed $n$ gives
$$\frac{2}{(a - s)^3} e^{(a - s)n} - \frac{2}{(a - s)^3} e^0$$
Assume that $s > a$ and let $n$ go to infinity. Then since $(a - s) n \to -\infty$, the first term disappears and the limit is
$$-\frac{2}{(a - s)^3} e^0 = -\frac{2}{(-1)^3 (s - a)^3} = \frac{2}{(s - a)^3}$$
$\endgroup$ 7
