Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Reading the article on the Laplace Transform in Wolfram MathWorld, I found the proof that $\mathcal{L}[f'(t)] = sF(s) - f(0)$.

Laplace transform of a derivative

I understand the first and second steps, but I don't understand the third one. Why is it that $lim_{a \to \infty} [e^{-sa} f(a)] = 0$? $e^{-sa}$ does get closer to 0 when $-sa$ approaches to $\infty$, but why does $f(a)$ get closer to 0? As far as I know, $f(a)$ could be anything, so it could be possible that $lim_{x \to \infty} f(a)$ doesn't exist.

What guarantees that the limit of $f(a)$ always exists?

I hope I'm not missing some simple property of limits here.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

Recall that the Laplace transform requires $f$ to be of exponential order, e.g. $f(x)=O(e^{cx})$ as $x\to\infty$ for some $c>0$.

This means that $\frac{f(x)}{e^{ax}}$ is bounded as $x\to\infty$. Re-interpreting this in terms of the above limit, we see it is indeed $0$ when $s$ is sufficiently large (this restricts the domain of $L\{f'\}(s)$ by the way) since by taking $s>c$ (whatever $c$ is for this $f$) we then have $f=o(e^{sa})$ which means $\frac{f(a)}{e^{as}}\to0$ as $a\to\infty$ whenever $s>c$. And the limit doesn't need to exist for $f$; indeed $\lim_{a\to\infty}f(a)$ is unlikely to exist (consider $\sin a$), but that's not what is important here.

share|improve this answer
    
Oh, you're right. I forgot about the order of the function. Thanks. –  user1002327 Oct 28 '12 at 2:09
1  
I made a couple edit to help clarify the situation in more detail. –  Taylor Oct 28 '12 at 2:17
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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