How to prove that $ \mathcal{L}[J_0(\sqrt {t^2+2t})] = \frac {e^{ \sqrt {s^2+1}}}{\sqrt{s^2+1}} $

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.

Solution 1:

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.

Solution 2:

If we complete the square under the square root and then make the substitution $u = t+1$, we get $$\begin{align} \int_{0}^{\infty} e^{-st} J_{0} (\sqrt{t^{2}+2t}) \, \mathrm dt &= \int_{0}^{\infty} e^{-st} J_{0} \left( \sqrt{(t+1)^{2}-1}\right) \, \mathrm dt \\ &= e^{s} \int_{1}^{\infty} e^{-su} J_{0} (\sqrt{u^{2}-1}) \, \mathrm du. \tag{1}\end{align}$$

Integral $(1)$ is the case $\beta=1$ of the integral formula $$\int_{1}^{\infty} e^{-st}J_{0} (\beta\sqrt{t^{2}-1}) \, \mathrm dt = \frac{e^{-\sqrt{s^{2}+\beta^{2}}}}{\sqrt{s^{2}+\beta^{2}}}, \quad \left(s\ge 0, \ \beta \in \mathbb{R} \right). $$

See this answer.

Therefore, $$\mathcal{L} \left[J_0(\sqrt {t^2+2t})\right](s) = \frac{e^{s-\sqrt{s^{2}+1}}}{\sqrt{s^{2}+1}}, \quad s \ge 0. $$

More generally, we have $$\mathcal{L} \left[J_0(\sqrt {t^2+2 rt})\right](s) = \frac{e^{r(s-\sqrt{s^{2}+1})}}{\sqrt{s^{2}+1}}, \quad s,r \ge 0, $$ since $$\int_{r}^{\infty} e^{-st}J_{0} (\beta \sqrt{t^{2}-r^{2}}) \, \mathrm dt = \frac{e^{-r\sqrt{s^{2}+\beta^{2}}}}{\sqrt{s^{2}+\beta^{2}}}.$$