2
$\begingroup$

I would like to know if there is some relation between Laplace Transform and similar definite integrals. For instance, if I know that

$$\mathcal{L}\{f(t)\}(s)=F(s),$$

have I some information about $\int_a^be^{-st}f(t)dt$?

It would be useful on functions defined by parts etc.

Many thanks!

$\endgroup$
  • 1
    $\begingroup$ by definition $\mathcal{L}\left\{f(t)\right\}=\int_{0}^{\infty}e^{-st}f(t)dt=F(s)$ . If I understand, you want to know if you have $F(s)$, do I have information about $f(t)$. The answer is yes, it´s called the inverse transform, $\mathcal{L}^{-1}\left\{F(s)\right\}$. $\endgroup$ – Ricardo770 Aug 10 at 18:41
  • $\begingroup$ @Ricardo770, many thanks! In fact, I have $f(t)$ and $F(s)$, but I have troubles to calculate $\int_a^be^{-st}f(t)dt$. So, I'd like to know if having $F(s)$ this integral has some property. Thank you so much. $\endgroup$ – Still_waters Aug 11 at 0:48
  • 1
    $\begingroup$ You should look for the region of convergence of the laplace transform to see when its valid. $\endgroup$ – Ricardo770 Aug 11 at 1:28
  • $\begingroup$ Say $F(s)$ exists for any $s\geq 0$. I have some information for instance to $\int_{5}^{100}e^{-st}f(t)dt$...? Many thanks! $\endgroup$ – Still_waters Aug 11 at 1:39
  • 1
    $\begingroup$ It will depend on how well behaved $f(t)$ is. $\endgroup$ – Ricardo770 Aug 11 at 1:43
1
+50
$\begingroup$

Do you know that $\mathcal{L}(f(t)g(t)) = lim_{T \rightarrow \infty} \frac{1}{2\pi i} \int_{c - iT}^{c+iT}{F(\sigma)G(\tau - \sigma) d\sigma}$ (where the integration is done along a line in the complex plane which is contained in the domain of convergence of $F,G$)? If you know $F(s)$ you could plug in $g$ as the indicator function $1_{[a,b]}$ which has a simple transform, assuming $[a,b]\subseteq \mathbb{R}_{+}$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Many thanks. Your answer involving indicator function make me search a little about Laplace transform of step functions and this help me a lot. $\endgroup$ – Still_waters Aug 15 at 0:00
  • $\begingroup$ Thats great, sounds like a $step$ in the right direction badum tss $\endgroup$ – Tom Ariel Aug 15 at 12:57

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.