Question

Find the integral of te^-st dt from 0 to infinity.

Answer 1

the Laplace transform of \(t\) is merely \(1/s^2\)

Answer 2

I know, but I must know how to get there through work.

Answer 3

okay; first consider the indefinite integral:$$\int te^{-st}\,dt$$using integration by parts with \(u=t,dv=e^{-st}\,dt\) we get \(du=dt,v=-\dfrac1se^{-st}\) hence:$$\int te^{-st}\,dt=-\frac1ste^{-st}-\int-\frac1se^{-st}\,dt=-\frac1s te^{-st}+\frac1s\int e^{-st}\,dt=-\frac1se^{-st}\left(t+\frac1s\right)$$

Answer 4

now observe:$$\int_0^\infty te^{-st}\,dt=\lim_{a\to\infty}\int_0^a te^{-st}\,dt=-\frac1s\lim_{a\to\infty}\left[e^{-st}\left(t+\frac1s\right)\right]_0^a$$clearly evaluating that we get:$$\lim_{a\to\infty}\left(e^{-at}(a+\frac1s)-\frac1s\right)=-\frac1s$$hence:$$\int_0^\infty te^{-st}\,dt=-\frac1s\cdot-\frac1s=\frac1{s^2}$$