# How to calculate the following Laplace transform: L[(1-J_0(t))/(t)]?

How to calculate the following Laplace transform: $\mathcal{L}\left[\frac{1-{J}_{0}\left(t\right)}{t}\right]$? where ${J}_{0}\left(t\right)$ is the zeroth Bessel function.
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Mateo Tate
If
${\mathcal{L}}_{t}\left[F\left(t\right)\right]\left(s\right)=f\left(s\right)$
then:
${\mathcal{L}}_{t}\left[\frac{F\left(t\right)}{t}\right]\left(s\right)={\int }_{s}^{\mathrm{\infty }}f\left(u\right)\phantom{\rule{thinmathspace}{0ex}}du$
provided: $\underset{t\to 0}{\text{lim}}\frac{F\left(t\right)}{t}$ exist.
Since:
${\mathcal{L}}_{t}\left[1-{J}_{0}\left(t\right)\right]\left(s\right)=\frac{1}{s}-\frac{1}{\sqrt{1+{s}^{2}}}$
and:
$\underset{t\to 0}{\text{lim}}\frac{1-{J}_{0}\left(t\right)}{t}=0$
we have:
${\mathcal{L}}_{t}\left[\frac{1-{J}_{0}\left(t\right)}{t}\right]\left(s\right)={\int }_{s}^{\mathrm{\infty }}\left(\frac{1}{u}-\frac{1}{\sqrt{1+{u}^{2}}}\right)\phantom{\rule{thinmathspace}{0ex}}du=\mathrm{ln}\left(\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{{s}^{2}}}\right)$