Verify that the transform of $y(t) = t^2e^{at}$ is $Y(s) = \frac{2}{(s-a)^3}$

Question

I made the distinction to amplify "=" 3 times for easier readability.

I tried:

$$F(s) === \int_0^\infty t^2e^{(a-s)t}dt === \frac{1}{a-s}e^{(a-s)t}t^2\Big|_0^\infty \ - \frac{2}{a-s}\int_0^\infty te^{(a-s)t}dt$$

But we have $$\int_0^\infty te^{(a-s)t}dt = \frac{1}{a-s}e^{(a-s)t}t\Big|_0^\infty - \frac{1}{a-s}\int_0^\infty e^{(a-s)t}dt$$

But the Laplace transform of $\int_0^\infty e^{(a-s)t}dt$ is just $\frac{1}{s-a}$. I tried plugging back in to get

$$\int_0^\infty te^{(a-s)t}dt = \frac{1}{a-s}e^{(a-s)t}t\Big|_0^\infty - \frac{1}{s-a}$$

and we know that integral converges when $(0 < s) \land (a < s)$, and since $t < e^t$ for large $t$ as well. Therefore the integral is just $0 - 0 + \frac{1}{a-s}$.

Then I plug back in again to get

$$\int_0^\infty t^2e^{(a-s)t}dt === \frac{1}{a-s}e^{(a-s)t}t^2\Big|_0^\infty - \frac{2}{a-s}[\frac{1}{(a-s)^2}]]$$

The integral on the RHS is just 0 as well, so the answer should be $-\frac{2}{(a-s)^3}$. But that's not the transform, it is $\frac{2}{(a-s)^3}$. Where did I go wrong?

Answer 1

$\mathbf{I}:$ $$\int t^2e^{(a-s)t}dt=\frac{t^2}{a-s}e^{(a-s)t}-\frac{2t}{(a-s)^2}e^{(a-s)t}+\frac{2}{(a-s)^3}e^{(a-s)t}$$

$\mathbf{II:}$

$$\mathcal{L}[e^{at}f(t)]=\mathcal{L}[f(t)]\Big|_{s\to s-a}$$ and you know that $\mathcal{L}(t^n)=\frac{n!}{s^{n+1}}$.

Answer 2

you may use $$\int_0^\infty x^{\alpha-1}e^{-\beta x}dx=\Gamma(\alpha)/{\beta}^{\alpha}$$

with $\alpha=3, \beta=s-a$. result is obvious. (note: $\Gamma(3)=2!=2$)