# Laplace Transformation

1. Sep 4, 2013

### Northbysouth

1. The problem statement, all variables and given/known data
Obtain the Laplace transformation of the fucntion defined by

f(t) = 0 t<0

= t2e-at t>=0

2. Relevant equations

3. The attempt at a solution

I'm a little unsure of what I'm doing here, so bear with me.

L {t2e-at} = ∫inf0 t2e-at dt

= ∫0inf t2e-(a+s)tdt

How do I integrate this? I tried using my TI-89 but it told me it is undefined. Any help would be greatly appreciated.

EDIT: The complete solution is required, which I think means I can't just take it out of the Laplace tables

2. Sep 4, 2013

### Ray Vickson

First of all: never, never write what you did above, which was
$$\int_0^{\infty} t^2 e^{-at} \, dt = \int_0^{\infty} t^2 e^{-(a+s)t} \, dt.$$
This is only true if s= 0.

Anyway, for your final (correct) integral, change variables from t to x = (a+s)t (assuming s > -a).

I strongly recommend that you put away the TI-89 until after you have learned how to do these problems, or at least restrict its use to numerical computation.

3. Sep 4, 2013

### Northbysouth

Sorry, I missed out a part.

There should have been an e-st

But I don't understand your point about changing the variable from t to x. Could you explain further please?

4. Sep 4, 2013

### Ray Vickson

I assume you have taken integration already. If so, you already studied change-of-variable methods (or so I hope).

If you have not done integration yet you would need a longer explanation than I am prepared to give here: we would need to go over material that takes several weeks to present in coursework! However, such information is widely available in books and on-line.

5. Sep 4, 2013

### Dick

Yes, it's just integration. But the trick you really need to handle the t^2 factor is integration by parts. By all means, review integration.

6. Sep 5, 2013

### vanhees71

It's much simpler to evaluate
$$\tilde{g}(s)=\int_0^{\infty} \mathrm{d} t \exp[-(a+s) t]$$
and then taking the 2nd derivative wrt. $s$. It's very easy to see that
$$\tilde{f}(x)=\tilde{g}''(s).$$