Assume that $f(t)$ and $tf(t)$ are bounded. $F(s)=\mathcal{L}(f(t))(s)$, show that $\mathcal{L}(tf(t))(s)=-F'(s)$.

Question

Question : Assume that $f(t)$ and $tf(t)$ are bounded. Denoting $$F(s)=\mathcal{L}(f(t))(s)$$ show that $$\mathcal{L}(tf(t))(s)=-F'(s)$$

My Try : I know that $$\mathcal{L}(f(t))(s)=\int_0^{\infty}e^{-st}f(t) \;dt$$

How to prove that Laplace Transform of $f(t)$ is equal to $-F'(s)$? I'm not sure how to solve this problem, what I know is that the Laplace Transform of $f(t)$ is $\int_0^{\infty}e^{-st}f(t) \;dt$

Answer 1

As @reuns mentioned in his comment, ‎$$\frac{d}{ds}F(s)=\frac{d}{ds}\int_0^\infty e^{-st}y(t)dt=\int_0^\infty \frac{d}{ds}e^{-st}y(t)dt=\int_0^\infty e^{-st}(-ty(t))dt={\cal L}(-tf)$$‎