Finding Laplace transform of $f(t) = t^2$. Where's my mistake?
Emily Wong 
$\mathcal{L}\{f(t)\} = \int_0^\infty{e^{-st}t^2}dt$
Integrating by parts:
$u = t^2$
$du = 2tdt$
$v = -\frac{1}{s}e^{-st}$
$dv = e^{-st}dt$
$\int_0^\infty{e^{-st}t^2}dt = -\frac{t^2}{s}e^{-st} + \frac{2}{s}\int_0^\infty{e^{-st}tdt}$
Integrating by parts on $\int_0^\infty{e^{-st}t}dt$:
$u = t$
$du = dt$
$v = -\frac{1}{s}e^{-st}$
$dv = e^{-st}dt$
$\int_0^\infty{e^{-st}t}dt = -\frac{t}{s}e^{-st}+\frac{1}{s}\int_0^\infty{e^{-st}dt}$
$\int_0^\infty{e^{-st}t}dt = -\frac{t}{s}e^{-st}+\frac{1}{s^2}$
In total:
$\int_0^\infty{e^{-st}t^2}dt = \frac{t^2}{s}e^{-st}+\frac{2}{s}[-\frac{t}{s}e^{-st}+\frac{1}{s^2}]$
I know this is incorrect, but I can't figure out where I'm messing up. Can someone help?
$\endgroup$ 21 Answer
$\begingroup$The integrals are over limits. Just substitute limits for the first term of the integration by parts term and you are through.
$\endgroup$ 2
