show that $\displaystyle \mathcal{L} \left \{ \frac d{dt} f(t) \right \} = sF(s) - f(0)$

how???

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- Jan 29th 2009, 04:07 PMrazorfeverThe Laplace Transform of a Derivative
show that $\displaystyle \mathcal{L} \left \{ \frac d{dt} f(t) \right \} = sF(s) - f(0)$

how??? - Jan 29th 2009, 09:46 PMJhevon
By definition: $\displaystyle F(s) = \mathcal{L} \{ f(t) \} = \int_0^\infty e^{-st} f(t)~dt$

Thus, $\displaystyle \mathcal {L} \{ f'(t) \} = \int_0^\infty e^{-st} f'(t)~dt$

Now proceed using integration by parts with $\displaystyle u = e^{-st}$ and $\displaystyle dv = f'(t)~dt$