# Laplace time differentiation property

1. Jul 12, 2009

### khdani

Hello,
I'm trying to prove the time differentiation property of Laplace transform.
dx(t)/dt = sX(s)

http://img10.imageshack.us/img10/290/tlaplace.jpg [Broken] http://g.imageshack.us/img10/tlaplace.jpg/1/ [Broken]

how do i continue from here ?

Last edited by a moderator: May 4, 2017
2. Jul 12, 2009

### Redbelly98

Staff Emeritus
I'm not sure what the first step here is supposed to represent. Wouldn't the first thing to do be to write

L[dx/dt] = ...

And then use the definition of the Laplace transform?

3. Jul 13, 2009

### khdani

yes, you right
the first term should be like you said
but what i have to do with the first term of the last part?
it should vanish but i don't know how? maybe by Final Value Theorem ?

4. Jul 13, 2009

### Redbelly98

Staff Emeritus
In the final integral, t is the variable of integration.

Take any term that is independent of t outside of the integral, and see what you get.

p.s. your integration limits are wrong.

5. Jul 13, 2009

### khdani

the final integral is sX(s)
but the one before should vanish somehow..., i think it's because of Final Value Theorem
why do you think the integration limits are wrong?

6. Jul 13, 2009

### Redbelly98

Staff Emeritus
Yes, good.

It looks like there are two definitions of the Laplace transform:
http://mathworld.wolfram.com/LaplaceTransform.html

The integrations limits are usually taken from 0 to +∞, but maybe your class is using the -∞ to +∞ definition instead. I didn't know about that definition before.

If the limits are 0 to +∞, you can just evaluate the x(t) e-st term at the limits. I'm not sure what to do if you're using the -∞ definition though.