Laplace transform of $t^2e^{at}$??

Question

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?

That doesn't really help @T.Bongers; the sign we care about doesn't get switched anyway — Don Larynx, Nov 21, 2013 at 1:53
If $s > a$, then $e^{(a - s)t} \to 0$ for $t$ positive. Then you'll get $$0 - \frac{2}{(a - s)^3} e^0$$ as desired. — user61527, Nov 21, 2013 at 1:56
I don't see how; the thing just changes to $$\lim_{n\to\infty}\frac{2}{(s-a)^3}e^{(s-a)t}|_0^[email protected] — Don Larynx, Nov 21, 2013 at 1:59

user61527user61527 · Accepted Answer · 2013-11-21 02:03:54Z

2

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

$$-\frac{2}{(a - s)^3} e^0 = -\frac{2}{(-1)^3 (s - a)^3} = \frac{2}{(s - a)^3}$$

answered Nov 21, 2013 at 2:03

user61527

$\begingroup$ I DON'T SEE HOW $\lim_{n\to\infty}(s-a)n \to -\infty$????? @T.Bongers $\endgroup$
– Don Larynx
Nov 21, 2013 at 2:06
$\begingroup$ @DonLarynx If $s - a$ is positive and you multiply it by a huge $n$, then $(s - a)n$ is still a huge number. With a slight abuse of notation, $$\lim_{n \to \infty} (s - a)n = (s - a) \lim_{n \to \infty} n = (s - a) \infty = \infty$$ $\endgroup$
– user61527
Nov 21, 2013 at 2:07
$\begingroup$ That's positive infinity, not negative infinity. Thus it does NOT converge. That is why I am so distraught. $\endgroup$
– Don Larynx
Nov 21, 2013 at 2:09
2

$\begingroup$ @DonLarynx If $(s - a) n \to \infty$, then $(a - s) n \to -\infty$. $\endgroup$
– user61527
Nov 21, 2013 at 2:10
$\begingroup$ Then my main question is, why does the latter hold? I don't know why (a - s) has to be positive. $\endgroup$
– Don Larynx
Nov 21, 2013 at 2:11

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