I need to find the Laplace transform of $\frac{1}{t^{2}}$
Does the laplace transform of $\frac{1}{t^{2}}$ exist? If yes, how do we calculate it? Putting it in won't solve.
I'm not understanding in the replacement part, should I do u = -st or u = t?
$\frac{1}{s}\int_0^{s} \frac{e^{-u}}{u^2}du$?
Suggestion: $\int_0^{1} e^{-st}\frac{1}{t^{2}}$ + $ \int_1^{\infty} e^{-st}\frac{1}{t^{2}}$
Just a quick answer (be aware about my lack of rigor).
By taking $u=st$:
$$\int_0^{\infty} \frac{1}{t^2}e^{-st} dt = s \int_0^{\infty} \frac{e^{-u}}{u^2}du \geq s \int_0^{1} \frac{e^{-u}}{u^2}du\geq s\int_0^{1} \frac{e^{-1}}{u^2}du = se^{-1} \left[ \frac{-1}{u} \right]_0^1$$ and the right side evaluation doesn't exist for $u=0$. Therefore the integral does not converge.