The traditional way to handle this is as follows:
In order to proceed, we need to have some more information about $s$; otherwise, we can't be sure that the integral will converge. So, we assume that $s$ is in the "region of convergence" (ROC) for this function. In this particular integral, we suppose that $\text{Re} \{s\}>0$, so that $\lim_{t\to\infty}e^{t(-s\pm ai)}=0$. You should be able to handle the rest.
$$
%\begin{align*}
%\left|\lim_{t\to \infty} e^{t(-s\pm ai)}\right| &=
%\lim_{t\to \infty} \left|e^{t(-s\pm ai)}\right|\\&=
%\lim_{t\to \infty} \left|e^{-st}\right|\cdot\left|e^{\pm ai}\right|\\&=
%\lim_{t\to \infty} \left|e^{-st}\right|\cdot 1=0
%\end{align*}
%$$
Not sure what you are asking with your "Is this a stroke of luck?" question...
Just in case, here's a different approach to the integral:
$$I=\int_0^\infty \frac{\sin (3x) \sin(x)}{x^2} dx=\frac{1}{2}\int_0^\infty \frac{\cos (2x)-\cos(4x)}{x^2} dx$$
$$I=\int_0^\infty \frac{\cos (x)-\cos(2x)}{x^2} dx$$
Now we also introduce a parameter, though we only need one:
$$I(a)=\int_0^\infty \frac{\cos (ax)-\cos(2ax)}{x^2} dx$$
$$I'(a)=\int_0^\infty \frac{2\sin (2ax)-\sin(ax)}{x} dx$$
The integral can now be safely separated into two terms, and each has a well known value:
$$\int_0^\infty \frac{\sin(x)}{x} dx=\frac{\pi}{2}$$
So:
$$I'(a)=\int_0^\infty \frac{2\sin (2ax)-\sin(ax)}{x} dx=\frac{\pi}{2}$$
Integrating (the constant of integration is determined by $I(0)$), we have:
$$I(a)=\frac{\pi}{2}a$$
$$I(1)=\frac{\pi}{2}$$
The proofs of the $\text{sinc}$ integral can be found elsewhere, including this site. Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?
Best Answer
My free interpretation is: how to use the Laplace transform to compute the integral $$ I=\int_{0}^{+\infty}\frac{e^{-t}\sin^2(t)}{t}\,dt $$ ? Well, since: $$\mathcal{L}\left(e^{-t}\sin^2(t)\right)=\frac{2}{(1+s)(5+2s+s^2)}\tag{1}$$ we have: $$ I = 2\int_{0}^{+\infty}\frac{ds}{(1+s)(5+2s+s^2)}=2\int_{0}^{+\infty}\frac{du}{4+e^{2u}}=\int_{0}^{+\infty}\frac{dv}{4+e^v}\tag{2} $$ or: $$ I = \int_{1}^{+\infty}\frac{dw}{w(4+w)}=\color{red}{\frac{\log 5}{4}}.\tag{3}$$