I have been tasked to show that $$\int_0^\infty J_0(ax)e^{-px} dx=\frac{1}{\sqrt{a^2+p^2}}$$ using the series expansion for $J_0(x)$ and not the Laplace transform tables.
where $$J_n(x)=\sum_{s=0}^\infty\frac{(-1)^{s}}{s!(s+n)!}\left(\frac{x}{2}\right)^{2s+n}$$
After substituting $J_0(ax)$ I'm really struggling to deal with the infinite sum inside the integral and would really grateful if someone could show me how you go about this.
$$\int_0^\infty\sum_{s=0}^\infty\frac{(-1)^{s}}{s!s!}\left(\frac{ax}{2}\right)^{2s}e^{-px}dx$$
The Laplace transform of $x^n$ is $\frac{n!}{p^{n+1}}$. So you can apply this to the series representation you have $$ J_0(a x)=\sum_{s=0}^\infty\frac{(-1)^{s}}{s!^2}\left(\frac{a x}{2}\right)^{2s} \to \sum_{s=0}^\infty\frac{(-1)^{s}a^{2s}}{s!^22^{2s}}\frac{(2s)!}{p^{2s+1}}= \mathcal{L}_{x\to p}[J_0(ax)] $$ then $$ \mathcal{L}_{x\to p}[J_0(ax)] = \frac{1}{p}\left[\sum_{s=0}^\infty\frac{(2s)!}{s!^24^{s}}X^s\right] = \frac{1}{p}\frac{1}{\sqrt{1-X}} $$ where $X=-\frac{a^2}{p^2}$, then $$ \mathcal{L}_{x\to p}[J_0(ax)] = \frac{1}{p}\frac{1}{\sqrt{1+\frac{a^2}{p^2}}} = \frac{1}{\sqrt{p^2+a^2}} $$ of course this relies on the knowledge of $$ \sum_{s=0}^\infty\frac{(2s)!}{s!^24^{s}}x^s = \frac{1}{\sqrt{1-x}} $$ which some will say is basic and others will not, but it's not from a Laplace transform table.