Finding Laplace transform of $f(t) = t^2$. Where's my mistake?

Question

$\mathcal{L}\{f(t)\} = \int_0^\infty{e^{-st}t^2}dt$

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?

Answer 1

The integrals are over limits. Just substitute limits for the first term of the integration by parts term and you are through.