Laplace transform

1. Apr 11, 2009

sara_87

1. The problem statement, all variables and given/known data

Find the laplace transform of the following:
(1-3cost)/t^2

2. Relevant equations

3. The attempt at a solution

I know many methods to find the laplace transform like the first and second shift theorems, the differentiation one, convolution, but for this particular question i dont know where to start. Any hints would be very much appreciated. thank you.

2. Apr 11, 2009

Count Iblis

Start with the Laplace transform of 1 - 3 cos(t). Then integrate both sides of the integral w.r.t. s twice.

3. Apr 11, 2009

sara_87

oh right so i integrate: (1/s) - (3s)/(s^2+1) twice wrt s.
once gives:
ln(s)-3/2(ln(s^2+1))

so i integrate this again using integration by parts to give:
s(ln(s)) - s - (3/2)s(ln(s^2+1)) + (3s-arctan(s))

is that right?

4. Apr 11, 2009

Count Iblis

It looks right, but you need toi ceck the problem again. Is it really the Laplace transform of [1 - 3 cos(t)]/t^2 that is wanted? If so, how is this Laplace transorm to beinterpreted, as the ordinary Laplace integral does not converge (due to the singularity in the integrand at t = 0).

This gives rise to the asymptotic s Log(s), behavior, which you cannot get if you compute a Laplace integral that converges, because if you let s go to infinity, the laplace transorm should go to zero.

If you had [1 - cos(t)]/t^2, then the result would be slightly different and you would find that in the large s expansion, the s Log(s) terms cancel (as well as the log(s) and s terms, only terms that approach zero for s to infinity remain).

Last edited: Apr 12, 2009
5. Apr 12, 2009

sara_87

The question is asking for [1 - 3 cos(t)]/t^2;
i was thinking exactly the same. i mean what integral limits do we put to actually evaluate the laplace transform? if the integral limits are s and infinity then log term in s(ln(s)) - s - (3/2)s(ln(s^2+1)) + (3s-arctan(s)) would diverge.

???

6. Apr 12, 2009

Count Iblis

Yes, clearly the integral of exp(-st) [1-3cos(t)]/t^2 dt from zero to infinity does not converge. However, you can imagine doing something like the following. We have a map L(f) that maps functions for which we can define the Laplace integral to their Laplace transform (as defined by that Laplace integral). Then, we can try to extend L(f) to a larger domain of functions for which the Laplace integral is not defined using some of the properties that L(f) satisfies when the domain is restricted to the Laplace integrable case.

7. Apr 12, 2009

sara_87

which properties? sorry, i dont understand what you mean. so does that mean we have to choose the limits?