\documentclass[12pt]{article}
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\begin{document}

{\bf Question}

Prove that if $\ds J_0(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^2}
\left(\frac{x}{2}\right)^{2n}$

then $\ds \int_0^\infty e^{-ax}J_0(x)dx=\frac{1}{\sqrt{1+a^2}}$


\vspace{0.25in}

{\bf Answer}

$\ds J_0(x)=\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^2}
\left(\frac{x}{2}\right)^{2n}$

${}$

$\ds \int_0^\infty e^{-ax}J_0(x)=\int_0^\infty
e^{-ax}\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^2}\left(\frac{x}{2}\right)^{2n}
=\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^2}\frac{1}{4^n}\int_0^\infty
e^{-ax}x^{2n}dx$

${}$

$\ds
=\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^2}\frac{1}{4^n}\frac{1}{a^{2n+1}}\int_0^\infty
e^{-y}x^{2n}dy$

${}$

$\ds
=\sum_{n=0}^\infty\frac{(-1)^n}{(n!)^2}\frac{1}{4^n}\frac{1}{a^{2n}}\Gamma(2n+1)$

${}$

$\ds
=\frac{1}{a}\sum_{n=0}^\infty\frac{(-1)^n(2n)!}{(n!)^2}\frac{1}{4^n}\frac{1}{a^{2n}}
=\frac{1}{\sqrt{1+a^2}}$


\end{document}