How to find the Laplace transform of $\frac{1-\cos(t)}{t^2}$? $$ f(t)=\frac{1-\cos(t)}{t^2} $$          $$  F(S)= ? $$
 A: Besides Dennis's answer, using the following fact may be helpful:

If $~\mathcal L\left\{ f(t)\right\}=F(s)~$ and Laplace of the function $~g(t)=\dfrac{f(t)}{t}~$ exists, then
$$\mathcal L \left\{\dfrac{f(t)}{t}\right\}=\int\limits_s^{\infty}F(u)\,\mathrm du$$

A: $\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}

A: 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*}
A: Probably the only way is to use the definition:
$$F(s)=\int_0^\infty e^{-st}f(t) dt$$
But, at least according to WA, it has no elementary form.
A: As we know
$$g(s)=\mathcal{L}\left\{1-\cos t\right\}=\mathcal{L}\left\{1\right\}-\mathcal{L}\left\{\cos t\right\}=\frac{1}{s}-\frac{s}{s^2+1}$$
Therefore,
$$\mathcal{L}\left\{\frac{1-\cos t}{t}\right\}=\int_{s}^{\infty}g(\sigma)\, d\sigma=\int_{s}^{\infty}
\frac{1}{\sigma}-\frac{\sigma}{\sigma^2+1}\, d\sigma$$
Since
$$\int\frac{1}{\sigma}-\frac{\sigma}{\sigma^2+1}\, d\sigma=\ln \left|\sigma\right|-\frac{1}{2}\ln\left|\sigma^2+1\right|$$
we have that
$$\mathcal{L}\left\{\frac{1-\cos t}{t}\right\}=\int_{s}^{\infty}
\frac{1}{\sigma}-\frac{\sigma}{\sigma^2+1}\, d\sigma=1-\ln \left|\frac{s}{\sqrt{s^2+1}}\right|$$
Therefore,
$$\mathcal{L}\left\{f\right\}=\int_{s}^{\infty}1-\ln \left|\frac{\sigma}{\sqrt{\sigma^2+1}}\right|\, d\sigma$$
I don't think you can go any further
