For a solution $y(t)=\int_0^t \theta^2 \cos(\theta) e^{-2\theta} d\theta$, find $y(t)$ as t tends to infinity

Question

For a solution $y(t)=\int_0^t \theta^2 \cos(\theta) e^{-2\theta} d\theta$, find y(t) as t tends to infinity. I think the final value theorem could be used but I'm not entirely sure:

$\lim_{t\rightarrow\infty}f(t)=\lim_{s\rightarrow 0}s\bar f(s)$

Why would the final value theorem be applicable here in the first place?

Answer 1

All you need to do is show that

$$\tag{1}\label{eq1}\lim_{t\to\infty}\int_0^t\theta^2 \cos(\theta) e^{-2\theta}\, d\theta$$

exists and is finite. To that end, notice that

$$\int_0^t\left|\theta^2 \cos(\theta) e^{-2\theta}\right|\, d\theta\leq \int_0^t\theta^2 e^{-2\theta}\, d\theta=\frac18\int_0^{t/2}u^2e^{-u}\,du\\ =\frac18\left[-e^{-u}\left(u^2+2u+2\right)\right]_{u=0}^{t/2}.$$

Hence $\lim_{t\to\infty}\int_0^t\left|\theta^2 \cos(\theta) e^{-2\theta}\right|\, d\theta\leq \frac18\lim_{t\to\infty}\left[-e^{-u}\left(u^2+2u+2\right)\right]_{u=0}^{t/2}=\frac14$.

It follows that integral $\eqref{eq1}$ converges absolutely and hence converges.

EDIT: I suppose you want to use the theorem with $f=y$.