Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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 that they probably treated $\displaystyle f(s,t) = e^{-st} f(t)$ so the integration/differentiation thing doesn't matter, but what confuses me is when they got rid of the derivative and how the "$t$" pop out? It's taking the derivative with respect to $s$ not $t$.

Proof: Consider the identity $$\frac{dF(s)}{ds} = \frac{d}{ds} \int_0^{\infty} e^{-st} f(t) dt.$$ Because of the assumptions on $f(t)$, we can apply a theorem from advanced calculus (sometimes called Leibniz's rule) to interchange the order of integration and differentiation: $$ \begin{align} \frac{dF(s)}{ds} & = \frac{d}{ds} \int_0^{\infty} e^{-st} f(t) dt\\ & = \int_0^{\infty} \frac{d \left(e^{-st} \right)}{ds} f(t) dt\\ & = - \frac{d}{ds} \int_0^{\infty} t e^{-st} f(t) dt\\ & = - \mathcal{L} \{ tf(t) \}(s). \end{align} $$ Thus, $$\mathcal{L} \{ tf(t) \}(s) = (-1) \frac{dF(s)}{ds}$$

share|cite|improve this question
And the $t$ then is considered a constant. The chain rule... e.g. $(e^{2s})'=e^{2s}\cdot2$. – David Mitra May 8 '12 at 3:42
What happens if you differentiate $e^{-s t}$ with respect to $s$? – J. M. May 8 '12 at 3:42
Thank you for catching my minor brain bug my rep is going down lol – Hawk May 8 '12 at 3:44
P.S. you'll sometimes see "differentiation under the integral sign" instead of "Leibniz's rule" in some contexts. – J. M. May 8 '12 at 3:46
up vote 2 down vote accepted

In general, $\frac{d}{dx} e^{ax} = a e^{ax}$

Thus, since $t$ is independent of $s$, $\frac{d}{ds} e^{-st} = -te^{-st}$

$f(t)$ doesn't depend on $s$, so it's a constant with respect to differentiation by $s$, and they pulled the negative outside the integral.

Does that make sense?

share|cite|improve this answer
Put another way: differentiating with respect to $s$ did not change the fact that $t$ was still a dummy variable within the Laplace transform integral... – J. M. May 8 '12 at 3:44

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.