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I would like to find a closed form for the following series involving the Bessel function $J_k(z)$:

$$ \sum_{k=0}^{+\infty}\frac{(\mu)_{k}}{k!(\lambda)_{k}}t^k\left(\frac{z}{2}\right)^{k}J_{k+\nu}(z), $$

where $(a)_{k}$ is the Pochhammer symbol. Actually, I am trying to compute a probability density function involving such summation. Moreover, The sum

$$ \sum_{k=0}^{+\infty}\frac{(\mu)_{k}}{k!(\lambda)_{k}}\left(\frac{z}{2}\right)^{k}J_{k+\nu}(z) $$

can be expressed in closed form (see the book by Prudnikov 'Integrals and Series', for instance). Might it be useful? Should I use another approach?

Thank you in advance.

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  • $\begingroup$ How is the first sum not just the second sum with $z\to t z$? $\endgroup$ – lcv Jul 20 '17 at 18:24
  • $\begingroup$ @Icy $z$ is also argument of $J_{k+\nu}(z)$ $\endgroup$ – Johannes Trost Jul 20 '17 at 18:59
  • $\begingroup$ Could you specify what you mean by closed form ? Is an integral involving classical functions fine for you ? Last, do you have $ \mu < \lambda $ ? $\endgroup$ – Synia Jul 21 '17 at 12:41
  • $\begingroup$ @Synia I would like the infinite summation to be finite, somehow, and expressed in terms of "well-known" functions. Moreover, I have $\mu<\lambda$ and an integral involving classical functions might be fine $\endgroup$ – axl Jul 21 '17 at 13:46
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Concerning the finite sum, I am not sure you can get it (maybe for particular values of $ \nu $). Here is a possible "closed" expression for your sum. Start by writing $ \lambda := \mu + \alpha $ with $ \alpha > 0 $ and $$ \frac{ (\mu)_k }{ (\lambda)_k } = \frac{ (\mu)_k }{ (\mu + \alpha)_k } = \mathbb{E}(\beta_{\mu, \alpha}^k) $$ where $ \beta_{\mu, \alpha} $ is a random variable Beta distributed (see wikipedia for the moments and the definition ; all I do is to write your quotient of Pochammers with a Beta integral). Up to this expectation, you are left with computing $$ A := \sum_{k \geq 0} \frac{x^k }{k! } J_{k + \nu}(z) $$ Now, use the following representation (found on wikipedia) : $$ J_\alpha(z) = \frac{ (z/2)^\alpha }{ \Gamma(\alpha + \frac{1}{2}) } \int_{ [-1, 1] } e^{i sz } (1 - s^2)^{ \alpha - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } } $$ valid for $ \alpha > \frac{1}{2} $ and $ z \in \mathbb{C} $ (you must hence suppose that your $ \nu $ is greater than $ \frac{1}{2} $ ; if not, you will have to adapt).

Using dominated convergence and exchange of integral and sum, you are left with \begin{align*}%$ A & = \int_{ [-1, 1] } (z/2)^\nu \sum_{k \geq 0} \frac{ (x (1 - s^2) z/2 )^k }{k! \Gamma(\nu + k + \frac{1}{2}) } e^{i sz } (1 - s^2)^{ \nu - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } } \\ & = \int_{ [-1, 1] } (2 x (s^2 - 1) )^{-\nu} J_{\nu + 1/2} (x (s^2 - 1 ) z ) e^{i sz } (1 - s^2)^{ \nu - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } } \end{align*} where I use the definition of the Bessel function as a series (here again, see wikipedia).

You can then use the previous integral representation of the Bessel function to conclude (remember that $ x = \beta_{\mu, \alpha} t z/2 $ and you have to take the expectation in $ \beta_{\mu, \alpha} $). Your representation is then a triple integral involving classical functions.

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  • $\begingroup$ That's clever! Actually, in the book by Prudnikov 'Integrals and Series' the term $A$ is expressed in 'closed form' $\endgroup$ – axl Jul 22 '17 at 13:48
  • $\begingroup$ Even better then :) $\endgroup$ – Synia Jul 23 '17 at 16:48

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