Prove that L[f' ' ](s)$ = $sL[f](s)

Question

Can anyone prove this question ?

Let $f$:$\mathbb{R}$$→$$\mathbb{C}$ be continuous function such that $f$$(0)$ $=$ $0$ and that $f'$ be a piecewise continuous function and absolutely integrable on $\mathbb{R}$.

Prove that

$L[f'](s)$ $=$ $sL[f](s)$

where $L[f]$ represents the Laplace transform of f

Answer 1

$L\{f'(t)\}=\int_0^\infty e^{-st}f'(t)dt$

Integrating by parts we have,

$L\{f'(t)\}=e^{-st}f(t)|_0^\infty+s\int_0^\infty e^{-st}f(t)dt$

$L\{f'(t)\}=e^{-s(\infty)}f(\infty)-e^{-s(0)}f(0)+sL\{f(t)\}$

If $e^{-st}$ grows more rapidly than $f(t)$, we have $e^{-st}f(t)\to0$ when $t\to\infty$

$L\{f'(t)\}=sL\{f(t)\}-f(0)$

Since $f(0)=0$, this reduces to

$L\{f'(t)\}=sL\{f(t)\}$