2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ It is necessary to assume that $s > a$ here. $\endgroup$ – user61527 Nov 21 '13 at 1:51
  • $\begingroup$ That doesn't really help @T.Bongers; the sign we care about doesn't get switched anyway $\endgroup$ – Don Larynx Nov 21 '13 at 1:53
  • 1
    $\begingroup$ 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. $\endgroup$ – user61527 Nov 21 '13 at 1:56
  • $\begingroup$ I don't see how; the thing just changes to $$\lim_{n\to\infty}\frac{2}{(s-a)^3}e^{(s-a)t}|_0^n$$...@T.Bongers $\endgroup$ – Don Larynx Nov 21 '13 at 1:59
2
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ I DON'T SEE HOW $\lim_{n\to\infty}(s-a)n \to -\infty$????? @T.Bongers $\endgroup$ – Don Larynx Nov 21 '13 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 '13 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 '13 at 2:09
  • 2
    $\begingroup$ @DonLarynx If $(s - a) n \to \infty$, then $(a - s) n \to -\infty$. $\endgroup$ – user61527 Nov 21 '13 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 '13 at 2:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.