Proof $L{\rm{[}}\frac{{x(t)}}{t}{\rm{] = }}\int_s^\infty {X(u)du} $

Question

I see that we usually use the theorem to solve the Laplace transform, however i want to proof the theorem, who could give me some details!!!

$L{\rm{[}}\frac{{x(t)}}{t}{\rm{] = }}\int_s^\infty {X(u)du} $

where: $L{\rm{[x(t)] = X(s)}}$

Answer 1

Note that $\dfrac{\mathrm e^{-st}}t=\displaystyle\int_s^\infty\mathrm e^{-ut}\mathrm du$ hence, using Fubini theorem, $$\int_0^\infty\mathrm e^{-st}\frac{x(t)}t\mathrm dt=\int_0^\infty x(t)\left(\int_s^\infty\mathrm e^{-ut}\mathrm du\right)\mathrm dt=\int_s^\infty\left(\int_0^\infty\mathrm e^{-ut}x(t)\mathrm dt\right)\mathrm du, $$ that is, considering $y(t)=\dfrac{x(t)}t$ and $X=Lx$, $$ (Ly)(s)=\int_s^\infty X(u)\mathrm du. $$