The Laplace transform on $3t\sin 4t$

Question

How to perform the laplace transform on $3t\sin 4t$ ? I know that for $\sin 4t$ it's $\frac{4}{s^2 +16} $, but what is it for $t\sin 4t$? And how do we get from there to $3t\sin 4t$? Apparently this should be solved using derivation but I don't understand why.

Answer 1

Use the formula $(n=1)$ $$\mathcal{L}(t^nf(t))=(-1)^n \frac {d^n}{ds^n}F(s)$$ $$\mathcal{L}(t\sin(4t))=(-1) \frac {d}{ds}(\frac 4{s^2+16})$$ Or $$\mathcal{L}(t\sin(4t))=\int_0^\infty t\sin(4t)e^{-st}dt=-\int_0^\infty \partial_s(\sin(4t)e^{-st})dt$$ $$\mathcal{L}(t\sin(4t))=-\frac {d}{ds}\int_0^\infty \sin(4t)e^{-st}dt=-\frac {d}{ds}\mathcal{L}(\sin(4t))$$