3

Mathematica gives a following identity: \begin{equation} (1) J_{1/3}(x) J_{2/3}(x) + J_{-1/3}(x)J_{-2/3}(x) = \frac{\sqrt{3}}{\pi x} \end{equation} How do I prove this identity?

I was trying to use the recurrence relations: \begin{equation} J_{n+1}(x) + J_{n-1}(x) = \frac{2 n}{x} J_n(x) \end{equation} to tackle equation (1) but I do not think this will lead to the right direction since using the recurrence will introduce Bessel functions of higher(lower) orders than the orders currently present in the identity. Can anybody give me a hint as to how to prove that identity?

Przemo
  • 11,331
  • This is a particular case ($a=-1/3$) of the general Relationship : $J_{-a}(x)J_{a+1}(x)+J_{a}(x)J_{-1-a}(x)=-2sin(pia)/(pix)$ – JJacquelin Feb 11 '14 at 17:00
  • Thank you. Can you give me a hint how to prove that? – Przemo Feb 11 '14 at 17:05
  • Supposing that the antiderivative of the product of two BesselJ functions is known. It the case of orders $a$ and $1-a$, or of orders $-a$ and $-1+a$, the results involve the same generalized hypergeometric function with oposite sign. So, in the antiderivative of the sum, the hypergeometric term disapears. The remaining term is $Coef.*ln(x)$. Finally, differentiating it leads to the result $Coef./x$. Don't ask me for calculus of the antiderivative of the product of Bessel functions: I will not go back to the good old sophomores days. I hope that someone will find a more direct way. – JJacquelin Feb 11 '14 at 19:20
  • The relation given by @JJacquelin is the well-known Wronskian relation for the Bessel $J_a$ functions: $$W{J_a(x),J_{-a}(x)} = J_{a+1}(x)J_{-a}(x)+J_a(x)J_{-(a+1)}{x} = - 2 \frac{\sin(a\pi)}{\pi x}$$ See e.g. Abramowitz and Stegun 9.1.15 or http://dlmf.nist.gov/10.5.E1. The proof can be found in textbooks on special functions, e.g. in N.N. Lebedev, Special Functions and Their Applications, Chap. 5.9 Wronskians of Pairs of Solutions of Bessel's Equation. – gammatester Feb 12 '14 at 08:14
  • Of course, "grammatester" is right : Obviously it's the Wronskian. Why I have not seen that at first sight ? – JJacquelin Feb 12 '14 at 11:09

0 Answers0