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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know I can use the following: $$\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a)$$ $$\mathcal{L}^{-1}\{\frac{n!}{s^{n+1}}\} = t^n$$ $$\mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t)$$

but I'm confused as how to use them. In particular, for the first inverse above, if $a$ is negative, does that mean the equation becomes $u(t+a)f(t+a)$, or does the equation stay the same if we use the problem asked? If we have $e^{-3s}\frac{1}{(s-1)^2}$, why would it be


as opposed to


If, in this problem, the $a$ is negative?


share|cite|improve this question
It doesn't matter what's the sign of $a$, that equation holds for any $a$, including negative ones. So let's say $a = -2$, then you still need to write exactly the same equations, but when substituting correct value you'll get $u(t+2)\ldots$. Also it's not clear what $u$ means. I can guess that $f(t) = \mathcal L^{-1} \{F(s)\}$, but what's $u(t)$? – Kaster Dec 14 '12 at 2:11
$u(t)$ is the unit step function. – Bailor Tow Dec 14 '12 at 2:16
Although, what you've said still doesn't make much sense to me, I'm sorry. – Bailor Tow Dec 14 '12 at 2:17
up vote 1 down vote accepted

A related problem. Note that, the Laplace transform of $ f(t)= t e^t $ is

$$ F(s) = \frac{1}{(s-1)^2} $$

Now, using the fact

$$ \mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a), $$

we have

$$ \mathcal{L}^{-1}\{e^{-3s}\frac{1}{(s-1)^2}\} = u(t-3)(t-3)e^{t-3} $$


To find the Laplace transform of $x^n g(x)$, one can use the following property

$$ \mathcal{L}(x^ng(x))=(-1)^n \frac{d^n}{ds^n} G(s), $$

where $G(s)$ is the Laplace transform of $g(x)$. For instance in your case you have the function $ t e^t $, then its Laplace transform is

$$ (-1) \frac{d}{ds}\frac{1}{s-1}=\frac{1}{(s-1)^2}. $$

share|cite|improve this answer
I think all arguments should be $t$ not $x$ – Kaster Dec 14 '12 at 2:49
@Kaster: Thanks for the comment. – Mhenni Benghorbal Dec 17 '12 at 1:38

Your Answer


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.