Calculating the Laplace Transform of t^2*u(t-a)

[mm391]

Homework Statement

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

Homework Equations

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.

 

[HallsofIvy]

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 (x+ a)^2u(x)[/tex]. You could as well use t for the variable, (t+a)^2u(t)[/tex] but obviously if you substitute &lt;b&gt;for&lt;/b&gt; t- a you are not going to still have t- a inside u. Making that substitution would make the integral&lt;br /&gt; \int_0^\infty e^{-st}(t+ a)^2dt&lt;br /&gt; which will give the same result as before.

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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.

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?