Evaluating the integral with help of Laplace transforms

Question

I came across an integral that is to be solved by the Laplace transform method.
$$ \int_0^{\infty} \frac{\sin^2{x}}{x^2}$$
My Approch
let $ f(t) = \sin^{2}t$ and $$\sin^{2}t = \frac{1-\cos2t}{2}$$
also we know from the properties of Laplace transforms
$$\frac{\mathcal{L}{f(t)}}{t} = \int_s^\infty F(s) ds $$
so $$ \mathcal{L}\left[\frac{1-\cos2t}{2}\right] = \frac{1}{2} \left[ \frac{1}{s} – \frac{s}{s^2+4} \right]$$
now, $$\mathcal{L}\left[\frac{1-\cos2t}{2t}\right] = \frac{1}{2}\int_s^\infty \left(\frac{1}{s} -\frac{s}{s^2+4} \right) ds$$
after integrating
$$ \mathcal{L} \left[\frac{1-cos2t}{2t}\right]= \frac{1}{2}\left[\ln(s) – \frac{\ln(s^2+4)}{2}\right]\big{|}_s^\infty$$
$$ \mathcal{L}\left[\frac{1-\cos2t}{2t}\right] =\frac{1}{2} \left[\ln(\frac{s}{(s^2+4)^{1/2}})\right] \big{|}_s^\infty $$
but at this point the integral becomes not-defined,
How can I approach this question as it has to be solved by the method of Laplace transforms
Hints are appreciated

Answer 1

First, there's a missing $2$ in the denominator of $\frac{1-\cos 2t}{t}$. I suppose it's a typo? Next, $$\mathcal{L}\left[\frac{1-\cos2t}{2t}\right] = \frac{1}{2}\int_0^\infty \left(\frac{1}{s} -\frac{s}{s^2+4} \right) ds $$ can't be true because the RHS is possibly a number (the integral actually diverges) while the LHS is a function. What we have instead is $$\mathcal{L}\left[\frac{\sin^2 t}{t}\right]=\mathcal{L}\left[\frac{1-\cos2t}{2t}\right]=\frac 12\int_s^{\infty}\left(\frac 1y-\frac{y}{y^2+4}\right)dy=\frac 14\log\frac{s^2+4}{s^2} $$ Now, if we let $g(t)=\sin^2 t/t$, we've found $\mathcal{L}[g(t)]$. So $$\int_0^{\infty}\frac{\sin^2 x}{x^2}dx=\int_0^{\infty}\frac{g(x)}{x}dx=\frac 14\int_0^{\infty}\log\frac{p^2+4}{p^2}dp $$ Can you continue from here? Use integration by parts.