# Homework Help: Where did I go wrong? [Laplace Transform]

1. Sep 11, 2005

### mathwurkz

$$L$$ is Laplace Transform operator.
Question is:

Let $$f(t) = t^2$$. Derive $$L[f]$$ from $$L[1]$$

So I know $$f(1) =1$$ and $$L[1] = \frac{1}{s}$$

Carrying out the Transform...

$$L[f] = \int_{0}^{\infty} e^{-st}t^2 dt$$

Integration by parts $$u = t^2, dv = e^{-st} dt$$

I do the integration, the uv terms go to zero. after I clean up I get

$$- \int_{0}^{\infty} \frac{e^{-st}}{-s} 2t\ dt$$

Doing IBP again. $$u = t,\ dv = e^{-st} dt$$ Rinse and repeat.

$$- \frac{2}{s} \int_{0}^{\infty} \frac{e^{-st}}{-s} (1) dt \\ = - \frac{2}{s} \left( \frac{e^{-st}}{s^2}\right)_{0}^{\infty} = - \frac{2}{s^3}$$

Ok. So is that right? I don't think the negative is right at the end. But what the bigger thing that worries me is how do I use L[1] in this? Or is that a different approach?

Last edited: Sep 11, 2005
2. Sep 11, 2005

### lurflurf

It seems something like this is wanted
$$L[t^a \ f(t)]=\left(-\frac{d}{ds}\right)^aL[f]$$
thus
$$L[t^2]=L[t^2 \ 1]=\left(-\frac{d}{ds}\right)^2L[1]=\frac{d^2}{{ds}^2} \ \frac{1}{s}=\frac{2}{s^3}$$