If $u(x)$ and $v(x)$ are any two solutions of Bessel's equation of order $\alpha$, then the Wronskian $W(x;u,v)=\dfrac{c}{x}$ (See for example here).
I am trying to calculate $c$ for when $u(x)$ and $v(x)$ are $J_{\alpha}$ and $J_{-\alpha}.$ The answer is $$W(x;J_{\alpha},J_{-\alpha})=-\dfrac{2}{x} \dfrac{1}{\Gamma(\alpha)\Gamma(1-\alpha)},$$ but I can't reach to that! That is how to evaluate $$W(x;J_{\alpha},J_{-\alpha})= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \dfrac{(-1)^{m+n}}{n! m!} \dfrac{x^{2m+2n-1}}{2^{2m+2n}} \dfrac{2(n-m-\alpha)}{{\Gamma(m+\alpha+1)\Gamma(n-\alpha+1)}} $$ for $x=1$?
We have from asymptotic behavior as $x\rightarrow 0$: \begin{align} W(J_a(x),J_{-a}(x)) &= J_a(x)J_{-a-1}(x)+J_{a+1}(x)J_{-a}(x)\\ &\sim \dfrac{1}{\Gamma(a+1)}\left(\dfrac{x}{2}\right)^a \dfrac{1}{\Gamma(-a)}\left(\dfrac{x}{2}\right)^{-a-1} + \dfrac{1}{\Gamma(a+2)}\left(\dfrac{x}{2}\right)^{a+1} \dfrac{1}{\Gamma(-a+1)}\left(\dfrac{x}{2}\right)^{-a}\\ &\to \dfrac{2}{x}\dfrac{1}{a\Gamma(a)\Gamma(-a)} \end{align}