# Infinite summation formula of Bessel functions

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?

• How is the first sum not just the second sum with $z\to t z$?
– lcv
Jul 20 '17 at 18:24
• @Icy $z$ is also argument of $J_{k+\nu}(z)$ Jul 20 '17 at 18:59
• Could you specify what you mean by closed form ? Is an integral involving classical functions fine for you ? Last, do you have $\mu < \lambda$ ? Jul 21 '17 at 12:41
• @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
– axl
Jul 21 '17 at 13:46

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} $). Your representation is then a triple integral involving classical functions. • That's clever! Actually, in the book by Prudnikov 'Integrals and Series' the term$A\$ is expressed in 'closed form'