Laplace transform of Bessel function of order zero I'm trying to prove that the Laplace transform of the function
$$
J_0(a\sqrt{x^2+2bx})
$$
is
$$
\frac{1}{\sqrt{p^2+a^2}} \exp\left\{bp- b\sqrt{p^2+a^2} \right\}
$$
as asserted in the eqworld. This formula can also be found in the book "Tables of Integral Transforms" page 207.
My first attempt was to insert $a\sqrt{x^2+bx}$ in the series representation
of the Bessel function
$$
J_0(x) = \sum_{m=0}^\infty \frac{(-1)^m}{(m!)^2}\left(\frac{x}{2}\right)^{2m}
$$
and then integrating term by term. However, the Laplace transform of the function $(x^2+bx)^m$ is not trivial.
My second attempt was to look at a change of variable in the differential equation of $J_0$
$$
t^2J_0''(t)+tJ_0'(t)+t^2J_0(t)=0
$$
but I was not able to find it.
Thanks for any clue.

Here is the complete answer the my first question. Let $J_0$ denote the Bessel function of order zero. We want to show that
\begin{equation}
\color{red}{\mathcal{L}(J_0(a\sqrt{x^2+2bx}))} = \frac{1}{\sqrt{p^2+a^2}} \exp\left\{bp- b\sqrt{p^2+a^2} \right\}
\end{equation}
with $a$ and $b$ arbitrary real numbers.
Denoting $g(x)$ the expected result, we note that it can be expressed as
\begin{eqnarray} 
g(x) & = & \frac{1}{\sqrt{p^2+a^2}} \exp\left\{b\left(p- \sqrt{p^2+a^2}\right) \right\} \nonumber \\
     & = & \frac{1}{\sqrt{p^2+a^2}} \exp\left\{b \frac{-a^2}{p + \sqrt{p^2+a^2}}  \right\} \\
     & = &  \frac{1}{\sqrt{p^2+a^2}} \sum_{n=0}^{\infty} \frac{(-b)^n}{n^!} \frac{a^{2n}}{\left(p+\sqrt{p^2+a^2}\right)^n}. \nonumber 
\end{eqnarray}
The Bessel functions of order $n$ have the following Laplace transform
\begin{equation}
\mathcal{L}(J_n(ax)) = \frac{a^n}{\sqrt{p^2+a^2} (p+\sqrt{p^2+a^2})^n}.
\end{equation}
Thus the function $g$ can be written as $g(x) = \sum_{n=0}^{\infty} \frac{(-ab)^n}{n^!} \mathcal{L}(J_n(ax))$.
The integral representation of $J_n$ is
$$
J_n(z) = \frac{1}{2\pi i^n} \int_0^{2\pi} e^{i z \cos(t)+int}\, dt.
$$
Using this integral representation, it follows that:
\begin{eqnarray}
 \mathcal{L}^{-1}(g(p))(x) &=& \frac{1}{2\pi} \int_{0}^{2\pi} \exp\left(iax\cos(t)+ibae^{it}\right)\,dt  \nonumber  \\
 & = & \frac{1}{2\pi}\int_{0}^{2\pi}\exp\left(ia\left[\left(\frac{x}{2}+b\right)e^{it}+\frac{x}{2}e^{-it}\right]\right)\,dt \nonumber\\
 & = & \frac{1}{2\pi} \sum_{n=0}^\infty \frac{(ia)^n}{n!2^n}\int_{0}^{2\pi} \left[\left(\frac{x}{2}+b\right)e^{it}+\frac{x}{2}e^{-it}\right]^n\, dt. \label{eq:integral}
\end{eqnarray}
Now for each $n$ we can evaluate the integral as
$$
\int_{0}^{2\pi} \left[\left(\frac{x}{2}+b\right)e^{it}+\frac{x}{2}e^{-it}\right]^n\, dt
= \left\{ \begin{array}{lll}
           0   & \mbox{ if }& n = 2k+1 \\
           2\pi \dbinom{2k}{k} (x+2b)^k x^k&  \mbox{ if }& n = 2k
          \end{array}
 \right.
$$
Substituting, we arrive at the expression 
\begin{equation}
\mathcal{L}^{-1}(g(p))(x) = \sum_{k=0}^\infty \frac{(-a^2)^k (x^2+2bx)^k}{(k!)^2} = J_0\left(a\sqrt{x^2+2bx}\right).
\end{equation}
