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Laplace Transform of

  1. Jun 18, 2010 #1
    Laplace Transform of......

    Few months back my friend called me to help him (and his whole class!) finding the Laplace Transform of [tex]\sin \sqrt{t}[/tex].
    Since then I found myself wasting hours and days (even before non-mathematics exams!) trying to solve the problem. I also checked the provided answer by online mathematica (wolfram alpha) and found it to be correct.
    One of the most apparent way is to use Euler's formula, it does not work, however. I don't even remember the different ways I tried.
    When my friend asked his instructor (who gave him the problem), he was quick to suggest the use of series expansion, the method I personally dislike.
    Is there no other better way???
    Last edited: Jun 18, 2010
  2. jcsd
  3. Jun 18, 2010 #2

    Ben Niehoff

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    Gold Member

    Re: Laplace Transform of......

    You don't need a series expansion. It can be done using contour integration. I won't just give you the answer, but here's a description of how to do it. You should follow along with your own work so you can learn how to do it. Then check against your Mathematica answer:

    We want to integrate

    [tex]\int_0^{\infty} \sin{\sqrt{t}} \; e^{-st} \; dt[/tex]

    First do a substitution [itex]u = \sqrt{t}[/itex]. This will put an [tex]e^{-su^2}[/tex] inside your integral, but don't fret. There will also be some additional stuff in there; keep track of it.

    Next, you'll find that the integrand in terms of u is an even function. So, instead of integrating over u from 0 to [itex]\infty[/itex], you can integrate from [itex]-\infty[/itex] to [itex]\infty[/itex] and then divide by 2. This is good, because we know how to integrate [tex]e^{-su^2}[/tex] from [itex]-\infty[/itex] to [itex]\infty[/itex].

    Now, express [itex]\sin u[/itex] as [itex]\Im \{ e^{iu} \}[/itex]. You can pull the [itex]\Im[/itex] all the way outside your integral, since everything else in the integrand is a real function.

    Now you can combine your exponentials into a single exponential of a quadratic polynomial in u. Complete the square in this polynomial, so you have

    [tex]e^{-A(u - \lambda)^2 + B}[/itex]

    where A, B, and [itex]\lambda[/itex] are functions of s only.

    Now do a second substitution [itex]v = u - \lambda[/itex]. You should end up with a (finite) sum of terms of the form

    [tex]\Im \left\{ C_k(s) \int_{-\infty + i\alpha}^{\infty + i\alpha} v^k e^{-A(s) v^2} \; dv \right\}[/tex]

    where [itex]\alpha[/itex] is the imaginary part of [itex]\lambda[/itex]. This integral can be done by contour integration: we can simply shift the contour downward to the real axis, since the integrand has no poles. Effectively, we can set [itex]\alpha = 0[/itex]. Then we're left with a standard Gaussian integral, which you should look up if you don't know how to evaluate it.
    Last edited: Jun 18, 2010
  4. Jun 19, 2010 #3
    Re: Laplace Transform of......

    Thanks for solution! It worked elegantly!
    One interesting thing about this is that the transform is [tex]e^{\frac{-1}{4s}}[/tex] times the transform of [tex]\sqrt{t}[/tex] .
    Is there any link/hint to some general formula/theorem?
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