Bessel Function of the first kind Could you please help me understand how to prove
$$J_{(1/2)} (x) =  \sqrt{\frac2{\pi x}}\cdot \sin⁡ x$$
using,
$$J_p (x)  = \sum_{(n=0)}^\infty \frac{(-1)^n}{(n! \Gamma(n+p+1) )} \left( \frac x 2 \right)^{2n+p}$$
Thank you
 A: The Bessel Function of the first kind and order $p$ has series representation given by
$$J_p (x)  = \sum_{n=0}^\infty \frac{(-1)^n}{n! \,\Gamma(n+p+1) } \left( \frac x 2 \right)^{2n+p} $$
For $p=1/2$, we find 
$$\begin{align}
J_{1/2} (x)  &= \sum_{n=0}^\infty \frac{(-1)^n}{n! \,\Gamma(n+3/2) } \left( \frac x 2 \right)^{2n+1/2} \\\\
&=\sqrt{\frac {1}{2x}}\,\sum_{n=0}^\infty \frac{(-1)^n\,x^{2n+1}}{n! \,4^n\,\Gamma(n+3/2) } \tag 1
\end{align}$$
Applying recursively the functional relationship, $\Gamma(1+x)=x\Gamma(x)$, for the Gamma Function $\Gamma(n+3/2)$ reveals 
$$\begin{align}
\Gamma(n+3/2)&=(n+1/2)(n-1/2)(n-3/2)\cdots (3/2)(1/2)\Gamma(1/2)\\\\
&=\frac{(2n+1)!!}{2^{n+1}}\sqrt \pi\\\\
&=\frac{(2n+1)!}{2^{2n+1}\,n!}\sqrt \pi \tag2
\end{align}$$
Substituting $(2)$ into $(1)$ yields
$$\begin{align}
J_{1/2}(x)&=\sqrt{\frac{1}{2x}}\sum_{n=0}^\infty \frac{(-1)^n\,x^{2n}}{n!\,4^n\,\frac{(2n+1)!}{2^{2n+1}\,n!}\sqrt \pi }\\\\
&=\sqrt{\frac{2}{\pi x}}\sum_{n=0}^\infty \frac{(-1)^n\,x^{2n+1}}{(2n+1)!}\\\\
&=\sqrt{\frac{2}{\pi x}}\,\sin(x)
\end{align}$$
as was to be shown!
