TITLE: How to find the Laplace transform of $\frac{1-\cos(t)}{t^2}$?
QUESTION [10 upvotes]: $$ f(t)=\frac{1-\cos(t)}{t^2} $$          $$  F(S)= ? $$

REPLY [2 votes]: $\newcommand{\+}{^{\dagger}}
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Laplace Transform of $\ds{\fermi\pars{t} \equiv {1 - \cos\pars{t} \over t^{2}}:\ {\large ?}}$

\begin{align}
&\color{#00f}{\large\int_{0}^{\infty}\fermi\pars{t}\expo{-st}\,\dd t}
=\int_{0}^{\infty}\bracks{1 - \cos\pars{t}}
\pars{\int_{0}^{\infty}x\expo{-tx}\,\dd x}\expo{-st}\,\dd t
\\[3mm]&=\Re\int_{0}^{\infty}\dd x\,x\int_{0}^{\infty}\bracks{%
\expo{-\pars{x + s}t} - \expo{-\pars{x + s - \ic}t}}\,\dd t
=\Re\int_{0}^{\infty}x\pars{{1 \over x + s} - {1 \over x + s - \ic}}\,\dd x
\\[3mm]&=\Re\int_{0}^{\infty}\pars{%
-\,{s \over x + s} + {s - \ic   \over x + s - \ic}}\,\dd x
=\int_{0}^{\infty}\bracks{%
-\,{s \over x + s} + {\pars{x + s}s + 1   \over \pars{x + s}^{2} + 1}}\,\dd x
\\[3mm]&=\lim_{\Lambda \to \infty}\bracks{%
\half\,s\ln\pars{\bracks{x + s}^{2} + 1 \over \bracks{x + s}^{2}}
+ \arctan\pars{x + s}}_{x = 0}^{x = \Lambda}
\\[3mm]&=\color{#00f}{\large%
-\,\half\,s\ln\pars{s^{2} + 1 \over s^{2}} + {\pi \over 2} - \arctan\pars{s}}
\end{align}