Taylor series for $e^{a x} J_0 (b x)$ How to derive the general term for the Taylor series around $0$ for this function?
I found, using Wolfram Alpha, that:
$$e^{a x} I_0 (|a| x)= \sum_{n=0}^\infty \frac{(2n)!}{n!^3} \frac{(a x)^n}{2^n}$$
I suspect that there should be a nice series for the general function.
$$f(a,b,x)=e^{ax} J_0 (bx)$$
I see several ways to derive it:


*

*Finding an ODE for $f(a,b,x)$ and solving it with power series. Then maybe a recurrence relation will allow us to find a closed form for the coefficients.

*Using the integral representation of the Bessel function.
I'll try the second way:
$$J_0(x)= \frac{2}{\pi} \int_0^1 \frac{\cos (x u) du}{\sqrt{1-u^2}}$$
$$e^{a x} J_0(x)=\frac{2}{\pi} \int_0^1 e^{a x} \frac{\cos (x u) du}{\sqrt{1-u^2}}$$
Let's try expanding:
$$e^{a x} \cos (u x)=\frac{1}{2} \left(e^{(a+i u) x}+e^{(a-i u) x} \right)= \\ = \frac{1}{2} \sum_{n=0}^\infty \frac{(a+iu)^n+(a-iu)^n}{n!} x^n$$
So we get:
$$e^{a x} J_0 (b x)= \sum_{n=0}^\infty C_n(a,b) \frac{x^n}{n!}$$
Where:
$$C_n(a,b)=\frac{1}{\pi} \int_0^1 \frac{(a+ib u)^n+(a-ib u)^n}{\sqrt{1-u^2}} du$$

Can we simplify this expression somehow?

I suppose we could use Binomial series and then each term becomes a Beta function. On the other hand, we can directly represent this integral as a sum of two hypergeometric functions. I'll see what I can do, but I would welcome other ideas and answers.
 A: The $n$-th derivative of the product is
\begin{equation}
 \frac{d^n}{dx^n}f\left( a,b,x \right)=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^ke^{ax}\left.\frac{d^k}{dz^k}J_0(z)\right|_{z=bx}
\end{equation} 
We want to evaluate it at $x=0$, then
\begin{equation}
 C_n(a,b)=\left.\frac{d^n}{dx^n}f\left( a,b,x \right)\right|_{x=0}=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k\left.\frac{d^k}{dz^k}J_0(z)\right|_{z=0}
\end{equation} 
Using the formula for the $k$-th derivative of the Bessel function (DLMF),
\begin{equation}
\left.\frac{d^k}{dz^k}J_0(z)\right|_{z=0}=\frac{1}{2^{k}}\sum_{p=0}^{k}(-1)^{p}\binom{k}{p}%
J_{2p-k}\left(0\right)
\end{equation} 
In this sum, the only non-vanishing Bessel functions at the origin have  a zero index, i.e. $p=k/2$. Then, for integer $s$,
\begin{equation}
 \left.\frac{d^k}{dz^k}J_0(z)\right|_{z=0}=\begin{cases}
    (-1)^s2^{-2s}\binom{2s}{s} & \text{for } k=2s \\
    0 & \text{for } k=2s+1
  \end{cases}
\end{equation} 
thus
\begin{equation}
 C_n(a,b)=\sum_{s=0}^{\lfloor n/2 \rfloor}(-1)^s2^{-2s}\binom{n}{2s}\binom{2s}{s}a^{n-2s}b^{2s}
\end{equation} 
finally, using the explicit expression of the Legendre polynomials, we obtain
\begin{equation}
 C_n(a,b)=\operatorname{sign}(a)\left( a^2 +b^2\right)^{n/2}P_n\left( \frac{\left|a\right|}{\sqrt{a^2+b^2}} \right)
\end{equation} 
Taking into account the parity properties of the polynomials, this expression can be written as:
\begin{equation}
 C_n(a,b)=\left( a^2 +b^2\right)^{n/2}P_n\left( \frac{a}{\sqrt{a^2+b^2}} \right)
\end{equation} 

This result can be checked using the exponential generating function of the Legendre polynomials
\begin{equation}
\sum_{n=0}^\infty P_n(z)\frac{t^n}{n!}=e^{tz}{}_0F_1\left( ;1;\frac{t^2\left(  z^2-1\right)}{4} \right)
\end{equation} 
when $-1<x<1$. Using the hypergeometric representation of the Bessel function
\begin{equation}
J_0(u)={}_0F_1\left( ;1;\frac{-u^2}{4} \right)
\end{equation} 
By choosing $t^2\left( z^2-1 \right)=-b^2x^2$ and $tz=ax$, we obtain
\begin{equation}
z=\frac{\pm1}{\sqrt{1+\frac{b^2}{a^2}}}\quad t=\pm ax \sqrt{1+\frac{b^2}{a^2}}
\end{equation} 
then the generating function is identical to the Taylor expansion.
A: With Laplace transform
$${\cal L}\left(e^{ax}J_0(bx)\right)=\dfrac{1}{\sqrt{b^2+(s-a)^2}}=\dfrac{1}{s{\sqrt{1-2\frac{a}{\sqrt{a^2+b^2}}\frac{\sqrt{a^2+b^2}}{s}+\left(\frac{\sqrt{a^2+b^2}}{s}\right)^2}}}$$
valid for $s>\sqrt{a^2+b^2}$. With generating function of Legendre polynomials
$$e^{ax}J_0(bx)={\cal L}^{-1}\sum_{n=0}^\infty\dfrac{\sqrt{a^2+b^2}^n}{s^{n+1}}P\left(\frac{a}{\sqrt{a^2+b^2}}\right) = \sum_{n=0}^\infty\dfrac{\sqrt{a^2+b^2}^n}{n!}P\left(\frac{a}{\sqrt{a^2+b^2}}\right)x^n$$
