# Inverse Laplace of t^2*u(t-a)

1. Apr 11, 2013

### mm391

1. The problem statement, all variables and given/known data

I am trying to work out the inverse Laplace transform of t^2*u(t-a).

2. Relevant equations

3. The attempt at a solution

I have been old that it starts

(t-a)^2*u(t-a)..... which I understand is substituting for (t-a). But I cannot seem to work out the rest can someone please help or explain. This isn't homework or coursework just a problem the lecturer failed to explain properly.

2. Apr 11, 2013

### HallsofIvy

Staff Emeritus
u(t-a) is the "Heaviside step function"? Then it is 0 for t< a so the Laplace transform is just
$$\int_a^\infty e^{-st}t^2 dt$$.

$t^2u(t-a)$ is NOT the same as $(t- a)^2u(t- a)$- that should be obvious- it is NOT "substituting for t- a". Substituting, say, x for t- a would mean that t= x+ a so you would have [itex](x+ a)^2u(x)[/tex]. You could as well use t for the variable, [itex](t+a)^2u(t)[/tex] but obviously if you substitute for t- a you are not going to still have t- a inside u. Making that substitution would make the integral
$$\int_0^\infty e^{-st}(t+ a)^2dt$$
which will give the same result as before.

3. Apr 11, 2013

### Ray Vickson

I don't know why you describe it as the inverse LT; what you want is the LT itself. Anyway, why not just use the definition
$$\cal{L}[u(t-a) t^2](s) = \int_{a}^{\infty} e^{-st} t^2 \, dt?$$