I am given $f(t) = t^2e^{-2t}$ for the Laplace transform. I used IBP twice resulting in
$$[t^2e^{-t(s+2)}-\frac{1}{2}te^{-t(s+2)}-\frac{1}{(s+2)}e^{-t(s+2)}]_0^{\infty}$$
First, is my IBP wrong? Second, how do I take the limit (I am reviewing limits now).
I was taught that its $b-a$ for limits. Since $b=\infty$ wouldn they all go to infinity?
But if I plug in $0$ for $t$ then we have $\frac{1}{s+2}$ which is wrong! So what do I do?
In terms of classical methods: \begin{align} \mathcal{L}\{t^{2} \, e^{-a t}\} &= \int_{0}^{\infty} t^{2} \, e^{-(a+s) t} \, dt \\ &= \left[- t^{2} \, \frac{e^{-(a+s) t}}{s+a} \right]_{0}^{\infty} + \frac{2}{s+a} \, \int_{0}^{\infty} t \, e^{-(a+s) t} \, dt \\ &= (-1)\left[t^{2} \, \frac{e^{-(a+s) t}}{s+a} + \frac{2 \, t \, e^{-(a+s) t}}{(s+a)^2} \right]_{0}^{\infty} + \frac{2}{(s+a)^2} \, \int_{0}^{\infty} e^{-(a+s) t} \, dt \\ &= (-1)\left[t^{2} \, \frac{e^{-(a+s) t}}{s+a} + \frac{2 \, t \, e^{-(a+s) t}}{(s+a)^2} + \frac{2 \, e^{-(a+s) t}}{(s+a)^3} \right]_{0}^{\infty} \\ &= \frac{2}{(s+a)^3} \end{align}
A faster method: \begin{align} \mathcal{L}\{t^{2} \, e^{-a t}\} &= \partial_{s}^{2} \, \int_{0}^{\infty} e^{-(a+s) t} \, dt = \partial_{s}^{2} \, \left(\frac{1}{s+a} \right) \\ &= \frac{2}{(s+a)^{3}} \end{align}
Laplace transform property: $$\mathcal{L}\{t^{n} \, e^{-a t} \, f(t), s \} = (-1)^{n} \, \partial_{s}^{n} \, \mathcal{L}\{f(t), s+a\}.$$