Solve Laplace Transform: t{e^{ - t}}u(t - 1)

[p75213]

Homework Statement

Is there an easier way of solving this rather than doing the integral?
Find the laplace transform of:

[tex]t{e^{ - t}}u(t - 1)[/tex]

Homework Equations

The Attempt at a Solution

[tex]\int_1^\infty {t{e^{ - t}}} {e^{ - st}}dt = \int_1^\infty {t{e^{ - t(1 + s)}}} dt = \frac{{{e^{ - (s + 1)}}}}{{{{(1 + s)}^2}}} + \frac{{{e^{ - (s + 1)}}}}{{1 + s}}[/tex]

 

[LCKurtz]

Homework Statement

Is there an easier way of solving this rather than doing the integral?
Find the laplace transform of:

[tex]t{e^{ - t}}u(t - 1)[/tex]

The Attempt at a Solution

[tex]\int_1^\infty {t{e^{ - t}}} {e^{ - st}}dt = \int_1^\infty {t{e^{ - t(1 + s)}}} dt = \frac{{{e^{ - (s + 1)}}}}{{{{(1 + s)}^2}}} + \frac{{{e^{ - (s + 1)}}}}{{1 + s}}[/tex]
]

That integral isn't that tough, but if you prefer, you can use some of the theorems. For example, you have, if ##\mathcal L(f(t)) = F(s)## then ##\mathcal L(tf(t))=-F'(s)##. Also, ##\mathcal L(e^{at}f(t) = F(s-a)## and ##\mathcal L(u(t-1))=\frac{e^{-s}}{s}##.

So starting with ##\mathcal L(u(t-1))=\frac{e^{-s}}{s}= F(s)##, then ##\mathcal Le^{-t}u(t-1)=F(s+1) =\frac{e^{-(s+1)}}{s+1}## and $$L(te^{-t}u(t-1)) = -\frac d {ds}\left(\frac{e^{-(s+1)}}{s+1}\right)=\frac{e^{-s+1}(s+1)+e^{-s+1}}{(s+1)^2}$$which is the same answer.

 

[p75213]

True - it isn't that tough. But I am not so keen on integration by parts if I can avoid it.
Thanks for the reply.