Laplace transform of $\frac{\sin^2(t)}{t}$

Question

Because : $$\mathscr L\left(\frac{f(t)}{t}\right) = \int_s^∞ F(s) ds$$

and $$\mathscr L\left(\sin^2(t)\right) = \frac{2}{s\left(s^2 + 4\right)}$$

then $$∫ \frac{2}{s\left(s^2+4\right)} ds = \frac{1}{2}\ln|s|-\frac{1}{4}\ln|s^2 + 4|$$

What should I do next...?

the answer that textbook gives me is $$\frac{1}{4}\ln\left(\frac{s^2 + 4}{s^2}\right)$$

Answer 1

Using the double angle identity:

\begin{equation} \sin^2(t) = \frac{1 - \cos(2t)}{2} \end{equation}

Thus,

\begin{equation} \mathscr{L}\left[\sin^2(t)\right] = \frac{1}{2s} - \frac{s}{2\left(s^2 + 4\right)} \end{equation}

We now employ the property of Laplace Transforms:

\begin{equation} \mathscr{L}\left[ \frac{f(t)}{t} \right] = \int_{s}^{\infty}\mathscr{L}\left[ f(t) \right]_u\:ds \end{equation}

Thus,

\begin{align} \mathscr{L}\left[ \frac{\sin^2(t)}{t} \right] &= \int_{s}^{\infty}\left[ \frac{1}{2u} - \frac{u}{2\left(u^2 + 4\right)} \right]\:du = \left[ \frac{1}{2}\ln(u) - \frac{1}{4}\ln\left(u^2 + 4\right)\right]_{s}^{\infty}\\ & = \left[\frac{1}{4}\ln\left(\frac{u^2}{u^2 + 4} \right) \right]_{s}^{\infty} = \frac{1}{4}\left[0 - \ln\left(\frac{s^2}{s^2 + 4} \right) \right] = - \frac{1}{4}\ln\left(\frac{s^2}{s^2 + 4}\right)\\ & = \frac{1}{4}\ln\left(\frac{s^2 + 4}{s^2}\right) \end{align}