Bessel Function of the first kind

Question

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

Answer 1

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!