100 Laplace transform calculation

Let $\Re\lambda<\beta$ and $\beta<s$ . We substitute $z=(s-\lambda)t$ and find

\int_{0}^{\infty}t^{n}e^{\lambda t}e^{-st}\,dt=\int_{0}^{\infty}t^{n}e^{-(s-% \lambda)t}\,dt

={{1}\over{(s-\lambda)^{n+1}}}\int_{0}^{\infty}z^{n}e^{-z}dz;

this can be justified by Cauchy’s theorem from complex analysis. Integrating by parts, we obtain

={{1}\over{(s-\lambda)^{n+1}}}\Bigl{[}-z^{n}e^{-z}\Bigr{]}_{0}^{\infty}+{{n}% \over{(s-\lambda)^{n+1}}}\int_{0}^{\infty}z^{n-1}e^{-z}\,dz

=0+{{n}\over{(s-\lambda)^{n+1}}}\Bigl{[}-z^{n-1}e^{-z}\Bigr{]}_{0}^{\infty}+{{% n(n-1)}\over{(s-\lambda)^{n+1}}}\int_{0}^{\infty}z^{n-2}e^{-z}\,dz

and so until

\int_{0}^{\infty}t^{n}e^{\lambda t}e^{-st}\,dt={{n!}\over{(s-\lambda)^{n+1}}}.