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Prove that

$$J_n J^{'}_{-n}-J_{-n}J{'}_n= -\frac{2 \sin(n\pi)}{\pi x}$$

I tried by substituting value of $J^{'}_n$ but it doesn't help me out. I am unable to think how to get $\sin()$ on RHS.

I also tried to substitute series form of Bessel's function but that does not lead me anywhere. Moreover I think there is some trick to solve this. ($J_n$ is Bessel function)

When $n$ is an integer, it will be true as both LHS and RHS will turn out to be zero by using $\left(J_{-n}(x)= (-1)^nJ_n(x)\right)$, so I am left with the case when $n$ is not an integer.

Any hint will be appreciated.

Thanks

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  • $\begingroup$ Is $n$ an integer? The $\sin(n\pi)=0$ and the relation follows from $J_{-n}=(-1)^n J_n$ $\endgroup$ – gammatester Mar 17 '18 at 10:14
  • $\begingroup$ @gammatester, thanks, I was typing that too. I was wondering for the case when $n$ is non integer. $\endgroup$ – User Mar 17 '18 at 10:18
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I will use the notation $J_{\nu}(z)$ for the general case and derive your equation from the Wronskian (https://dlmf.nist.gov/10.5.E1) $$J_{\nu+1}\left(z\right)J_{-\nu}\left(z\right) + J_{\nu}\left(z\right)J_{-\nu-1}\left(z\right) = -\frac{2\sin\left(\nu\pi\right)}{\pi z}$$ and the recurrence relations (https://dlmf.nist.gov/10.6.E2) $$J_{\nu+1}\left(z\right) = (\nu/z)J_{\nu}\left(z\right)-J_{\nu}'\left(z\right),$$

$$J_{\mu-1}\left(z\right) = J_{\mu}'\left(z\right) +(\mu/z) J_{\mu}\left(z\right), $$

substituting $\mu = -\nu$ in the last equation gives $$J_{-\nu-1}\left(z\right) = J_{-\nu}'\left(z\right) -(\nu/z) J_{-\nu}\left(z\right).$$ It follows, that $$J_{\nu+1}\left(z\right)J_{-\nu}\left(z\right) = \left((\nu/z)J_{\nu}\left(z\right)-J_{\nu}'\left(z\right)\right) J_{-\nu}\left(z\right)$$ and $$J_{-\nu-1}\left(z\right)J_{\nu}\left(z\right) = \left(J_{-\nu}'\left(z\right) -(\nu/z) J_{-\nu}\left(z\right)\right)J_{\nu}\left(z\right)$$ Adding both equations and canceling like terms on the RHS gives $$ J_{-\nu}'\left(z\right)J_{\nu}\left(z\right) - J_{\nu}'\left(z\right)J_{-\nu}\left(z\right) = J_{\nu+1}\left(z\right)J_{-\nu}\left(z\right) + J_{-\nu-1}\left(z\right)J_{\nu}\left(z\right)$$ Now use the Wronskian and get

$$J_{\nu}\left(z\right)J_{-\nu}'\left(z\right) - J_{-\nu}\left(z\right)J_{\nu}'\left(z\right) = -\frac{2\sin\left(\nu\pi\right)}{\pi z}$$

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