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I'm in calculus II and our teacher gave us a problem as follows: Let f(t) be a function defined for all positive values of t. The Laplace Transform of f(t) is defined by: $$F(s) = \int_0^\infty e^{-st}f(t)dt$$ Find the Laplace Transform of the function: $$f(t) = t^2$$

We haven't learned about Laplace Transforms, but I did some Google searching about it to figure out what to do. We're covering improper integrals right now. When I calculated this I got that the result is undefined for all values of s. The following is all of my steps, but I'm not sure if I made a mistake somewhere because in all of the examples I've seen , Laplace Transforms usually go to 0 depending on the sign of the constant s. Can anyone verify if I did this correctly?

I first put the integral into limit form.

$$=\lim_{b\to\infty}\int_0^b{e^{-st}t^2dt}$$

Then compute the integral using integration by parts:

$$=\lim_{b\to\infty}\left[-\dfrac{t^2e^{-st}}{s}\Bigg|_0^b + \dfrac{2}{s}\int_0^\infty te^{-st}dt\right]$$ Then using integration by parts again:

$$=\lim_{b\to\infty}\left[-\dfrac{t^2e^{-st}}{s}\Bigg|_0^b + \dfrac{2}{s}\left[ \dfrac{-te^{-st}}{s}\Bigg|_0^b + \dfrac{1}{s}\int_0^\infty e^{-st}dt\right]\space\right]$$

Then take the last integral:

$$=\lim_{b\to\infty}\left[-\dfrac{t^2e^{-st}}{s}\Bigg|_0^b + \dfrac{2}{s}\left[ \dfrac{-te^{-st}}{s}\Bigg|_0^b + \dfrac{1}{s}\times\left(\dfrac{-e^{-st}}{s}\right)\right]\space\right]$$

Simplifying that I got:

$$=\lim_{b\to\infty}-\dfrac{t^2e^{-st}}{s}\Bigg|_0^b+\dfrac{2te^{-st}}{s^2}\Bigg|_0^b - \dfrac{2e^{-st}}{s^3}\Bigg|_0^b$$

Then when you take the limit plug in infinity for b, I got: $$-\dfrac{\infty e^{-s\infty}}{s} - \dfrac{2\infty e^{-s\infty}}{s^2} - \dfrac {2e^{-s\infty}}{s^3}$$

So I said the transform was undefined for all values of s because if s is negative, you get infinity minus infinity and if it's positive, you get infinities divided infinities, both of which are undefined. Is that correct?

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  • $\begingroup$ How could I get to that answer without using a table? In other words, was the method I used above completely wrong? Since we haven't learned about this yet, I just got there by following some examples I read. $\endgroup$
    – Sabien
    Commented Mar 12, 2014 at 3:02
  • $\begingroup$ Ahh, I didn't think to pull the third part out in front of the limit. That makes sense, thanks! $\endgroup$
    – Sabien
    Commented Mar 12, 2014 at 3:33

2 Answers 2

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first of all by definition of Laplace transformation s is greater than 0. and (s does not have to be complex unless it specified to be complex). and since you did not plug zero at the final term -2/(s^3)(e^st) you miss the term 2/s^3. and as to calculating limit when b goes to infinity s>0 by definition of Laplace then realize that b/e^(bs) goes to zero since denominator grow faster than numerator.or you can get infinity over infinity and by L hopital again you get that all the terms go to zero except for 2/s^3. (b

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  • $\begingroup$ Thank you, yeah I realized the main mistake I made was not bringing the constants out in front of the limits and as you said, not plugging in 0. Thanks for the clarification. I was also wondering about what @stefano said about s being a complex number, so I'm glad to know it doesn't HAVE to be complex...that helps. $\endgroup$
    – Sabien
    Commented Mar 12, 2014 at 23:15
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First recall that $s$ is a complex number. The region of convergence is $Re(s)>0$. In this way, when you take the limit, the last expression you wrote goes to zero. Then remember also to plug in $t=0$ to get the correct answer $\frac{2}{s^3}$ (you can check it in the table provided my @Amzoti).

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  • $\begingroup$ When I originally calculated this, I did note that the third expression would go to zero with a positive s value, but here's what confuses me. The first two expresses are undefined, correct? If so, how can you isolate just the third expression? I thought any expression with an undefined part is just undefined? $\endgroup$
    – Sabien
    Commented Mar 12, 2014 at 3:17
  • $\begingroup$ The limit of the first two expressions goes to zero: for $s>0$ $lim_{t\rightarrow\infty} \frac{t^2}{e^{st}} = 0$, and similarly the second one. $\endgroup$
    – Stefano
    Commented Mar 13, 2014 at 17:53

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