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Mathematics
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An identity with Bessel functions
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[QUOTE="julian, post: 6844311, member: 142346"] I am not contributing something new. I am taking a result given in Watson and arriving at the same condition that [USER=695751]@renormalize[/USER] arrived at. On page 148 of Watson is the formula: \begin{align*} J_\mu (az) J_\nu (bz) & = \dfrac{ (\frac{1}{2} az)^\mu (\frac{1}{2} b z)^\nu}{\Gamma (\nu+1)} \nonumber \\ & \times \sum_{m=0}^\infty \dfrac{(-1)^m (\frac{1}{2} az)^{2m} \;_2F_1 \left( -m , - \mu - m ; \nu+1 ; b^2/a^2 \right)}{ m! \Gamma (\mu+m+1)} \end{align*} Here [URL='https://books.google.co.uk/books?id=qy1GNv2ovHQC&lpg=PA59&dq=integral+product+bessel+functions&pg=PA59&redir_esc=y#v=onepage&q=integral%20product%20bessel%20functions&f=false']https://books.google.co.uk/books?id=qy1GNv2ovHQC&lpg=PA59&dq=integral+product+bessel+functions&pg=PA59&redir_esc=y#v=onepage&q=integral product bessel functions&f=false[/URL] they give a slightly different version of the formula: \begin{align*} J_\mu (az) J_\nu (bz) & = \dfrac{ (\frac{1}{2} az)^\mu (\frac{1}{2} b z)^\nu}{\Gamma (\mu+1)} \nonumber \\ & \times \sum_{m=0}^\infty \dfrac{(-1)^m (\frac{1}{2} bz)^{2m} \;_2F_1 \left( -m , - \nu - m ; \mu+1 ; a^2/b^2 \right)}{m! \Gamma (\nu+m+1)} \end{align*} which is obtained from the Watson formula by simultaneously performing the interchanges ##a \leftrightarrow b## and ##\mu \leftrightarrow \nu##. We'll use the second version. From which we have: \begin{align*} & \dfrac{J_{n+1} ((n+1)x)}{n+1} \dfrac{J_{n+2} ((n+2)x)}{n+2} \nonumber \\ & = \dfrac{ ((n+1)x/2)^{n+1} ((n+2)x/2)^{n+2}}{(n+1) (n+2) (n+1)!} \sum_{m=0}^\infty \dfrac{(-1)^m ((n+2) x/2)^{2m}}{ m! (n+m+2)! } \nonumber \\ & \qquad \qquad \times \;_2F_1 \left( -m , - n - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right) \end{align*} or \begin{align*} & \dfrac{J_{n+1} ((n+1)x)}{n+1} \dfrac{J_{n+2} ((n+2)x)}{n+2} \nonumber \\ & = \dfrac{(n+1)^n (n+2)^{n+1}}{(n+1)!} \sum_{m=0}^\infty \dfrac{(-1)^m (n+2)^{2m}}{ m! (n+m+2)!} \nonumber \\ & \qquad \qquad \times \;_2F_1 \left( -m , - n - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right) \left( \frac{x}{2} \right)^{2n+2m+3} \end{align*} Then, \begin{align*} x \phi (x) & = \sum_{n=0}^\infty \dfrac{J_{n+1} ((n+1)x)}{n+1} \dfrac{J_{n+2} ((n+2)x)}{n+2} \nonumber \\ & = \sum_{n=0}^\infty \sum_{m=0}^\infty \dfrac{(-1)^m (n+1)^n (n+2)^{2m+n+1}}{(n+1)! m! (n+m+2)!} \nonumber \\ & \qquad \qquad \times \;_2F_1 \left( -m , - n - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right) \left( \frac{x}{2} \right)^{2n+2m+3} \nonumber \\ & = \sum_{n=0}^\infty \sum_{m=0}^\infty s_{n,m} \left( \frac{x}{2} \right)^{2n+2m+3} \end{align*} where ##s_{n,m}## is defined by \begin{align*} s_{n,m} & = \dfrac{(-1)^m (n+1)^n (n+2)^{2m+n+1}}{(n+1)! m! (n+m+2)!} \;_2F_1 \left( -m , - n - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right) \end{align*} The double sum can be written \begin{align*} \sum_{n=0}^\infty \sum_{m=0}^\infty s_{n,m} \left( \frac{x}{2} \right)^{2n+2m+3} = \sum_{m=0}^\infty \sum_{n=0}^m s_{n,m-n} \left( \frac{x}{2} \right)^{2m+3} \end{align*} Here \begin{align*} s_{n,m-n} & = \dfrac{(-1)^{m-n} (n+1)^n (n+2)^{2m-n+1}}{(n+1)! (m-n)! (m+2)!} \;_2F_1 \left( -m + n , - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right) \end{align*} So \begin{align*} \phi (x) = \sum_{m=0}^\infty \sum_{n=0}^m \frac{1}{2} s_{n,m-n} \left( \frac{x}{2} \right)^{2m+2} . \end{align*} Writing \begin{align*} \phi (x) & = \sum_{m=0}^\infty \sum_{n=0}^m \frac{1}{2} s_{n,m-n} (-1)^m m! (m+2)! \dfrac{(-1)^m}{m! (m+2)!} \left( \frac{x}{2} \right)^{2m+2} \end{align*} we see that if ##\phi (x) = J_2 (x)## we would have \begin{align*} \sum_{n=0}^m \frac{1}{2} (-1)^m m! (m+2)! s_{n,m-n} = 1 \qquad \text{for } m \geq 0 . \end{align*} That is, if ##\phi (x) = J_2 (x)## we would have \begin{align*} \sum_{n=0}^m \dfrac{m! (-1)^{-n} (n+1)^n (n+2)^{2m-n+1} \;_2F_1 \left( n-m , - m - 2 ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right)}{2 (n+1)! (m-n)!} = 1 \end{align*} for ##m \geq 0##. Which is the same condition that [USER=695751]@renormalize[/USER] arrived at. [/QUOTE]
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An identity with Bessel functions
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