The following question is the second part to this previous question:
Prove that $$J_p(x)J_{-p}^{\prime}(x)-J_{-p}(x)J_{p}^{\prime}(x)=-\frac{2}{\pi x}\sin(p\pi)\tag{1}$$ from $$\frac{\mathrm{d}}{\mathrm{d}x}\left[x\left(J_pJ_{-p}^{\prime}-J_{-p}J_{p}^{\prime}\right)\right]=0\qquad\longleftarrow\text{(Proved in Part 1)}$$ it follows that $$J_pJ_{-p}^{\prime}-J_{-p}J_{p}^{\prime}=\frac{c}{x}\tag{2}$$ To find $c$ use $$\fbox{$J_p(x)=\sum_{n=0}^\infty\frac{(-1)^n}{\Gamma(n+1)\Gamma(n+1+p)}\left(\frac{x}{2}\right)^{2n+p}$}\tag{3}$$ for each of the $4$ functions and pick out the $\dfrac{1}{x}$ terms in the products. Then use $$\fbox{$\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin(p\pi)}$}\tag{4}$$
Writing out the $4$ functions:
$$J_p(x)=\sum_{n=0}^\infty\frac{(-1)^n\cdot x^{2n+p}}{\Gamma(n+1)\Gamma(n+1+p)\cdot 2^{2n+p}}\tag{a}$$
$$J_p^{\prime}(x)=\sum_{n=0}^\infty\frac{(-1)^n(2n+p)\cdot x^{2n+p}}{\Gamma(n+1)\Gamma(n+1+p)\cdot 2^{2n+p}}\cdot\frac{1}{x}\tag{b}$$
$$J_{-p}(x)=\sum_{n=0}^\infty\frac{(-1)^n\cdot x^{2n-p}}{\Gamma(n+1)\Gamma(n+1-p)\cdot 2^{2n-p}}\tag{c}$$
$$J_{-p}^{\prime}(x)=\sum_{n=0}^\infty\frac{(-1)^n(2n-p)\cdot x^{2n-p}}{\Gamma(n+1)\Gamma(n+1-p)\cdot 2^{2n-p}}\cdot\frac{1}{x}\tag{d}$$
Insertion of $(\mathrm{a})$, $(\mathrm{b})$, $(\mathrm{c})$, $(\mathrm{d})$ into $(2)$ gives
$$\bbox[#AFF]{ \sum_{n=0}^\infty\frac{(-1)^n\cdot x^{2n+p}}{\Gamma(n+1)\Gamma(n+1+p)\cdot 2^{2n+p}}\cdot\sum_{n=0}^\infty\frac{(-1)^n(2n-p)\cdot x^{2n-p}}{\Gamma(n+1)\Gamma(n+1-p)\cdot 2^{2n-p}}\cdot\frac{1}{x}\quad-\\\sum_{n=0}^\infty\frac{(-1)^n\cdot x^{2n-p}}{\Gamma(n+1)\Gamma(n+1-p)\cdot 2^{2n-p}}\cdot\sum_{n=0}^\infty\frac{(-1)^n(2n+p)\cdot x^{2n+p}}{\Gamma(n+1)\Gamma(n+1+p)\cdot 2^{2n+p}}\cdot\frac{1}{x}=\frac{c}{x} }$$
So unless I'm mistaken this means that the highlighted equation above requires that $$J_pJ_{-p}^{\prime}-J_{-p}J_{p}^{\prime}=c\tag{5}$$
If this is the case then insertion of $(5)$ into $(2)$ implies that $$J_pJ_{-p}^{\prime}-J_{-p}J_{p}^{\prime}=\frac{J_pJ_{-p}^{\prime}-J_{-p}J_{p}^{\prime}}{x}\implies x=1$$
I'm not sure if what I have done so far is correct, also I have no idea how to make use of $$\fbox{$\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin(p\pi)}$}$$
Could anyone please provide some feedback, hints or tips on how I can prove that $$\fbox{$\color{red}{J_p(x)J_{-p}^{\prime}(x)-J_{-p}(x)J_{p}^{\prime}(x)=-\frac{2}{\pi x}\sin(p\pi)}$}$$
Much appreciated.
We want to find $c$ such that $$x\left[J_{p}\left(x\right)J'_{-p}\left(x\right)-J'_{p}\left(x\right)J_{-p}\left(x\right)\right]=c.$$ We use $$J_{p}\left(x\right)=\sum_{m\geq0}\frac{\left(-1\right)^{m}}{\Gamma\left(m+1\right)\Gamma\left(p+m+1\right)}\left(\frac{x}{2}\right)^{2m+p}$$ $$ J_{-p}\left(x\right)=\sum_{k\geq0}\frac{\left(-1\right)^{k}}{\Gamma\left(k+1\right)\Gamma\left(-p+k+1\right)}\left(\frac{x}{2}\right)^{2k-p}$$ $$ J'_{p}\left(x\right)=\sum_{l\geq0}\frac{\left(-1\right)^{l}\left(2l+p\right)}{\Gamma\left(l+1\right)\Gamma\left(p+l+1\right)}\left(\frac{x}{2}\right)^{2l+p-1}$$ $$ J'_{-p}\left(x\right)=\sum_{n\geq0}\frac{\left(-1\right)^{n}\left(2n-p\right)}{\Gamma\left(n+1\right)\Gamma\left(-p+n+1\right)}\left(\frac{x}{2}\right)^{2n-p-1}$$ We observe that $$x\left[J_{p}\left(x\right)J'_{-p}\left(x\right)-J'_{p}\left(x\right)J_{-p}\left(x\right)\right]$$ $$=\sum_{n\geq0}\sum_{m\geq0}\frac{\left(-1\right)^{n+m}\left(2n-p\right)}{\Gamma\left(n+1\right)\Gamma\left(m+1\right)\Gamma\left(-p+n+1\right)\Gamma\left(p+m+1\right)}\left(\frac{x}{2}\right)^{2n+2m} $$ $$-\sum_{k\geq0}\sum_{l\geq0}\frac{\left(-1\right)^{k+l}\left(2l+p\right)}{\Gamma\left(k+1\right)\Gamma\left(l+1\right)\Gamma\left(-p+k+1\right)\Gamma\left(p+l+1\right)}\left(\frac{x}{2}\right)^{2p+2l} $$ $$=-{\sum_{u\geq0}\sum_{v\geq0}\frac{\left(-1\right)^{u+v}{\left(2p-2u-2v\right)}}{\Gamma\left(u+1\right)\Gamma\left(v+1\right)\Gamma\left(p+u+1\right)\Gamma\left(-p+v+1\right)}\left(\frac{x}{2}\right)^{2u+2v}}\tag{1}$$ where $(1)$ follows from the fact that the we have two double series with the same addends and opposing sign. So $$\begin{align} {\lim_{x\rightarrow0}x\left[J_{p}\left(x\right)J'_{-p}\left(x\right)-J'_{p}\left(x\right)J_{-p}\left(x\right)\right]}= & -\frac{2p}{\Gamma\left(p+1\right)\Gamma\left(1-p\right)} \\ = & -\frac{2}{\Gamma\left(p\right)\Gamma\left(1-p\right)} \\ = & -\frac{2\sin\left(p\pi\right)}{\pi}. \end{align}$$ Note that the calculation of the limit follows from the fact that if we take $x\rightarrow 0$ the only non vanish term is when $m=n=0$ since it doesn't depend on $x$.