Let f(t) be a function defined for all values of t. The Laplace Transform of f(t) is defined by: F(s)= �ç[∘,�‡](e^-st(f(t))dt). If the improper integral exists, Find the Laplace Transform for F(t)=t^2.
well, we know that L{1} = 1/s
and, we know from our handy table of transforms that
L{t^n f(t)} = (-1)^n F(n)(s)
so,
L{t^2} = 2/s^3
To find the Laplace Transform of F(t) = t^2, we need to evaluate the integral of e^(-st)(t^2)dt, which is defined as:
F(s) = ∫(e^(-st)(t^2))dt
Applying integration by parts, we have:
F(s) = -t^2/(s)e^(-st) - ∫(-2t/s)e^(-st)dt
Simplifying the expression:
F(s) = -t^2/(s)e^(-st) + 2/s * ∫(te^(-st))dt
Now, to solve the remaining integral, we can use integration by parts again:
Let u = t, dv = e^(-st)dt
Then, du = dt and v = -(1/s)e^(-st)
Substituting these values, we have:
F(s) = -t^2/(s)e^(-st) + 2/s * (-t/(s)e^(-st) - ∫(-1/s)e^(-st)dt)
Simplifying further:
F(s) = -t^2/(s)e^(-st) - 2t/(s^2)e^(-st) + 2/s^2 * ∫(e^(-st))dt
The integral of e^(-st)dt is simply (-1/s)e^(-st), so substituting this back into the expression:
F(s) = -t^2/(s)e^(-st) - 2t/(s^2)e^(-st) + 2/s^2 * (-1/s)e^(-st) + C
Simplifying and combining terms:
F(s) = (-t^2 - 2st + 2)/(s^3) * e^(-st) + C
Thus, the Laplace Transform of F(t) = t^2 is given by:
F(s) = (-t^2 - 2st + 2)/(s^3) * e^(-st) + C