Asked by Mike
What is the laplace transform of (t^2)sin(t)?
Answers
Answered by
Count Iblis
f(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)
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)
Answered by
Count Iblis
Typo in last formula: The Laplace transform is:
8 s^2/(s^2+1)^3 - 2/(s^2+1)^2
8 s^2/(s^2+1)^3 - 2/(s^2+1)^2