Prove that
$$\int_0^\infty J_0(ax) e^{-px} dx = \frac{1}{\sqrt{a^2+p^2}}$$
using power series expansion of Bessel function $J_n(x)$, without using Laplace transform.
My attempt:
$$ J_0(ax) = 1-\frac{x^2a^2}{2^2}+\frac{x^4a^4}{2^2 \cdot 4^2}-\cdots$$
Hence
$$\int_0^\infty J_0(ax) e^{-px} dx =\int^\infty_0 \left\{ 1-\frac{x^2a^2}{2^2}+\frac{x^4a^4}{2^2\cdot 4^2}-\cdots\right\}e^{-px} dx.$$
How to continue from here? I have no idea.
Original image:
For starters, the $a$ parameter is just synctatic sugar, it can be removed through a suitable substitution, so the problem boils down to finding the Laplace transform of $J_0(x)$.
$J_0(x)$ is an entire function with the following power series:
$$ J_0(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n}}{4^n n!^2} $$ and since for any $p>0$ $$\int_{0}^{+\infty}x^{2n}e^{-px}\,dx = \frac{(2n)!}{p^{2n+1}}$$ we formally have: $$ \int_{0}^{+\infty}J_0(x)e^{-px}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{4^n p^{2n+1}}\binom{2n}{n}=\frac{1}{p}\sum_{n\geq 0}\binom{-1/2}{n}\frac{1}{p^{2n}} $$ or $$ \int_{0}^{+\infty}J_0(x)e^{-px}\,dx = \frac{1}{p\sqrt{1+\frac{1}{p^2}}}=\frac{1}{\sqrt{p^2+1}}$$ due to the extended binomial theorem.