I want to find the laplace transform for the function:

$$f(t) = \left\{\begin{array}tt,\quad t\lt 2 \\ t^2 , \quad t\geq 2 \end{array} \right.$$

So I thought that the proper setup was:

$$\mathcal{L}(f(t)) = t - (t)u(t-2) + (t^2)u(t-2)$$

$$\mathcal{F}(s) = \frac{1}{s^2}- \frac{e^{-2s}}{s^2} +\frac{2e^{-2s}}{s^3}$$

$$= \frac{s-se^{-2s} + 2e^{-2s}}{s^3}$$

But it would seem I don't know how to setup the heaviside unit step function properly.

Why is the above wrong?


You don't need to use the language of the Heaviside function. Just compute the integral directly:

$$\int_0^{\infty} dt \, f(t) e^{-s t} = \int_0^2 dt\, t\, e^{-s t} + \int_2^{\infty} dt \, t^2\, e^{-s t} $$

The first integral may be evaluated easily by differentiating a simpler integral with respect to $s$:

$$\int_0^2 dt\, t\, e^{-s t} = -\frac{d}{ds} \int_0^2 dt\, e^{-s t} = -\frac{d}{ds} \frac{1-e^{-2 s}}{s} = \frac{1-e^{-2 s}}{s^2} - 2 \frac{e^{-2 s}}{s}$$

We can use a similar trick for the other integral:

$$\begin{align}\int_2^{\infty} dt \, t^2\, e^{-s t} &= \frac{d^2}{ds^2} \int_2^{\infty} dt \, e^{-s t}\\ &= \frac{d^2}{ds^2}\frac{e^{-2 s}}{s}\\ &= \frac{d}{ds} \left (-2 \frac{e^{-2 s}}{s} - \frac{e^{-2 s}}{s^2} \right ) \\ &= -2 \left (-2 \frac{e^{-2 s}}{s} - \frac{e^{-2 s}}{s^2} \right ) - 4\frac{e^{-2 s}}{s^2} - 4 \frac{e^{-2 s}}{s^3} \\ &= \left (\frac{4}{s}-\frac{2}{s^2}-\frac{2}{s^3} \right ) e^{-2 s} \end{align}$$

The LT is then the sum of the two integrals:

$$F(s) = \frac{1-e^{-2 s}}{s^2} + 2 \left (\frac1{s}-\frac1{s^2}-\frac1{s^3} \right ) e^{-2 s} $$

  • 1
    $\begingroup$ Oh I see, that is very easy. Thank you! Can this be done with Heaviside also? $\endgroup$ – Calculus Nov 19 '14 at 1:32
  • $\begingroup$ @Calculus: Well, yes, but all the Heaviside does is provide notation; the result will be the same in terms of the two integrals above. $\endgroup$ – Ron Gordon Nov 19 '14 at 1:34
  • $\begingroup$ Oh I didn't know that, shame on me for falling into the computation(human algorithm) trap with that one! Thank you once again $\endgroup$ – Calculus Nov 19 '14 at 1:36

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