laplace transform of $t^nf(t)$

Question

I have:

$$\mathcal{L}(t^nf(t)) = \int_0^\infty t^nf(t)e^{-st}\ dt = \left(-\dfrac{d}{ds}\right)^n \int_0^\infty f(t) e^{-st}$$

I don't understand where the derivative came from

Answer 1

Calling $\mathcal{L}(f(t)) =F(s)$ and observing that $$ \frac{d^n}{ds^n}\mathrm e^{-st}= (-1)^nt^n \mathrm e^{-st} $$ we have $$F^{(n)}(s)= \int_0^\infty f(t) \left(\frac{d^n}{ds^n}e^{-st}\right)\mathrm dt =\int_0^\infty f(t) \left((-1)^nt^n \mathrm e^{-st}\right)\mathrm dt= (-1)^n \int_0^\infty t^nf(t)e^{-st}\ dt$$ that is $$(-1)^nF^{(n)}(s)= \int_0^\infty t^nf(t)e^{-st}\ dt =\mathcal{L}(t^nf(t)) $$