I'm trying to calculate the Laplace transform of this function.
$$ \mathcal{L}[\frac{1-J_0(t)}{t}] $$
where $J_0(t)$ is the zeroth Bessel function.
Solution Attempt
The p-Bessel function is defined as: $$J_p(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+p+1)}\left(\frac{x}{2}\right)^{2m+p}$$
Therefore for $p=0$ the zeroth Bessel is: $$J_p(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+1)}\left(\frac{x}{2}\right)^{2m+p}$$
It is trivial to prove that: $$\mathcal{L}\left[J_0(t)\right](s)= \frac{1}{\sqrt{s^2 + 1}}$$
So, we have: $$ \mathcal{L}[\frac{1-J_0(t)}{t}] = \mathcal{L}[\frac{1-\sum_{m=0}^\infty \frac{(-1)^m}{(m!)^2}\left(\frac{x}{2}\right)^{2m+p}}{t}] $$
*I used $\Gamma(m+1) = m! $
Now, this looks like it could be transformed to some of the basic maclaurin series? Maybe $sin(t)$?
In any case I'm stuck at this point. Any ideas?
If $$\mathcal{L}_t[F(t)](s)=f(s)$$ then: $$\mathcal{L}_t\left[\frac{F(t)}{t}\right](s)=\int_s^{\infty } f(u) \, du$$ provided: $\underset{t\to 0}{\text{lim}}\frac{F(t)}{t}$ exist.
Since: $$\mathcal{L}_t[1-J_0(t)](s)=\frac{1}{s}-\frac{1}{\sqrt{1+s^2}}$$ and:$$\underset{t\to 0}{\text{lim}}\frac{1-J_0(t)}{t}=0$$ we have: $$\mathcal{L}_t\left[\frac{1-J_0(t)}{t}\right](s)=\int_s^{\infty } \left(\frac{1}{u}-\frac{1}{\sqrt{1+u^2}}\right) \, du=\ln \left(\frac{1}{2}+\frac{1}{2} \sqrt{1+\frac{1}{s^2}}\right)$$