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September 3, 2014

September 3, 2014

Posted by **Mike** on Sunday, May 3, 2009 at 5:05pm.

- Math -
**Count Iblis**, Sunday, May 3, 2009 at 5:31pmf(s) = integral from t = 0 to infinity of t^2 sin(t) exp(-s t) dt

To compute this integral consider first the Laplace transform of exp(i t). Using the fact that the integral from zeo to infinity of exp(-alpha t) is 1/alpha, you find that this is:

1/(s - i)

The Laplace transform of sin(t) then follows by taking the imaginary part of this:

Im[1/s-i] =Im[(s+i)/(s^2+1)] = 1/(s^2+1)

If you differentiate this twice w.r.t. s, you bring down a factor of t^2 in the integrand of the Laplace integral. So, the Laplace thransform is given by:

8 s^2/(s^2+1)^3 - 2/(s^2+1)

- Math -
**Count Iblis**, Sunday, May 3, 2009 at 6:19pmTypo in last formula: The Laplace transform is:

8 s^2/(s^2+1)^3 - 2/(s^2+1)^2

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