Laplace transform of $f(t)$ multiplied by $t^n$

Question

How to prove that it is $(-1)^n\frac{d^n}{ds^n}F(s)$

Answer 1

By definition of the Laplace transform, $F(s)=\int_0^{\infty}f(t)e^{-st}dt$. Differentiating wrt $s$ gives us

$$\frac{d}{ds}F(s)=\frac{d}{ds}\int_0^{\infty}f(t)e^{-st}dt\\ =\int_0^{\infty}\frac{\partial}{\partial s}\left(f(t)e^{-st}\right)dt\\ =\int_0^{\infty}\left(-t\right)f(t)e^{-st}dt.$$

Now do it again.