Proof the theorem $\int_0^\infty {{t^n}f(t)dt = {{( - 1)}^{n + 1}}\int_0^\infty {{F^{(n + 1)}}(u)du} } $

Question

Knowing that: $$ L\left[ {\int_0^\infty {\frac{{f(t)}}{t}dt} } \right] = \frac{1}{s}\int_0^\infty {F(u)du}$$

with: $L[f(t)] = F(s) $

Show that: $$\int_0^\infty {\frac{{f(t)}}{t}dt = \int_0^\infty {F(u)du} } $$

and apply it to computing: $$\int_0^\infty {\frac{{\sin u}}{u}du}$$

Show that: $$\int_0^\infty {{t^n}f(t)dt = {{( - 1)}^{n + 1}}\int_0^\infty {F^{(n + 1)}}(u)du}$$

Derivative: $${{\rm{F}}^{(n + 1)}}(u)$$