How to find the Laplace transform of $\frac{1-\cos(t)}{t^2}$? $$ f(t)=\frac{1-\cos(t)}{t^2} $$          $$  F(S)= ? $$

Let
$$F(s) = \int_{0}^{\infty} f(t) \, e^{-st} \, dt = \int_{0}^{\infty} \frac{1-\cos t}{t^2} e^{-st} \, dt. $$
The function $f(t)$ satisfies the bound $ f(t) = O(1 \wedge t^{-2})$, thus it is absolutely integrable and we can apply Leibniz's integral to obtain
$$ F''(s) = \int_{0}^{\infty} (1-\cos t) \, e^{-st} \, dt = \frac{1}{s} - \frac{s}{s^2 + 1}. $$
Integrating and using the condition $F'(\infty) = 0$, we have
$$ F'(s) = \log s - \log \sqrt{s^2 + 1}. $$
Thus we have
$$F(s) = \int \left\{ \log s - \log \sqrt{s^2 + 1} \right\} \, ds. $$
The first term is easily integrated to yield $s \log s - s$. For the second term, note that
\begin{align*}
-\int \log \sqrt{s^2 + 1} \, ds
&= - s \log \sqrt{s^2 + 1} + \int \frac{s^2}{s^2 + 1} \, ds \\
&= - s \log \sqrt{s^2 + 1} + s - \arctan s + C.
\end{align*}
Combining, we obtain
$$ F(s) = s \log s - s \log \sqrt{s^2 + 1} - \arctan s + C. $$
But since $F(\infty) = 0$, we must have $C = \frac{\pi}{2}$ and therefore
\begin{align*}
F(s)
&= s \log s - s \log \sqrt{s^2 + 1} - \arctan s + \frac{\pi}{2} \\
&= s \log \bigg( \frac{s}{\sqrt{s^2 + 1}} \bigg) + \arctan \left(\frac{1}{s}\right).
\end{align*}
