I am trying to prove that $$ \mathcal{L}[J_0(\sqrt {t^2+2t})] = \frac {e^{ \sqrt {s^2+1}}}{\sqrt{s^2+1}} $$
where $ \mathcal{L}[f(t)]$ is the Laplace transform of f(t) and $J_0(f(t))$ is the p-Bessel function where p = 0
My approach
DEFINITIONS
The p-Bessel function is defined as:
$$J_p(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+p+1)}\left(\frac{x}{2}\right)^{2m+p}$$ With $p = 0$
and the Laplace Transform of a function as:
$$\mathcal{L}[f(t)](s) = \int_0^\infty f(t) e^{-st}dt$$
SOLUTION ATTEMPT
It is easy to prove that:
$$\mathcal{L}\left[J_0(t)\right](s)= \frac{1}{\sqrt{s^2 + 1}}$$
Thus we need to find a way to make $e^{\sqrt {s^2+1}}$ appear in the numerator.
For $p=0$ and $ x = \sqrt {t^2+2t} $ the Bessel function is:
$$J_0(\sqrt {t^2+2t} ) =
\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+1)}\left(\frac{\sqrt {t^2+2t} }{2}\right)^{2m} =
\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(m+1)}\left(\frac{t^2+2t}{4}\right)^{m} $$
and then I'm stuck. I am hoping that $e^{\sqrt {s^2+1}}$ will appear somehow through the Gamma function, but I have no idea how to proceed.
UPDATE
Intuitively a better approach seems to be this one:
$$ \mathcal{L}[J_0(\sqrt {t^2+2t})] = \frac {e^{ \sqrt {s^2+1}}}{\sqrt{s^2+1}} \iff \\ \mathcal{L}^{-1}[\frac {e^{ \sqrt {s^2+1}}}{\sqrt{s^2+1}}] = J_0(\sqrt {t^2+2t}) $$
The problem seems much easier now, I'll keep this post updated.
Best Answer
Your Laplace transform is incorrect, as RHS does not tends to $0$ as $s\to\infty$. The correct version should be $$\mathcal{L}[J_0(\sqrt {t^2+2t})] = \frac {e^{ s-\sqrt {s^2+1}}}{\sqrt{s^2+1}}$$
It can be proved as follows: for any $z,h$, we have $$\tag{1}(z+h)^{-\nu /2}J_\nu (\sqrt{z+h}) = \sum_{m=0}^\infty \frac{(-h/2)^m}{m!}z^{-(\nu+m)/2}J_{\nu+m}(\sqrt{z})$$ then take $z=t^2, h=2t$ gives $$J_0 (\sqrt{t^2+2t}) = \sum_{m=0}^\infty \frac{(-1)^m}{m!}J_{m}(t)$$ Taking Laplace transform both sides, using $\mathcal{L}[J_\nu(t)] = \frac{(s+\sqrt{1+s^2})^{-\nu}}{\sqrt{1+s^2}}$ completes the proof.
For a proof of $(1)$: view it as a function of $h$, expanding at $h=0$ gives $$(z+h)^{-\nu /2}J_\nu (\sqrt{z+h}) = \sum_{m=0}^\infty \frac{h^m}{m!}\frac{d^m}{dz^m}(z^{-\nu/2}J_\nu(\sqrt{z}))$$ Basic properties of Bessel functions shows that $$\frac{d}{dz}(z^{-\nu/2}J_\nu(\sqrt{z}))=-\frac{1}{2}z^{-(\nu+1)/2}J_{\nu+1}(\sqrt{z})$$ then use induction.