Bessel function identity

Question

I was trying to find this identity of Bessel function

$$e^{-2i\gamma t} J_{\left|n\right|}(2\gamma t) = e^{\large \frac{\pi i}{2}} \sum_{k=|n|}^{\infty} \frac{(-i\gamma t)^k}{k!}\binom{2k}{k-n}$$

on some books like "Watson: theory of Bessel function" or "Abramowitz and Stegun" but I can not find it. I have also tried to derived it using the Frobenius method for the Bessel equation but there are some coefficients that don't match. Does anyone know where to find the identity or how i can prove/verify it? Any advice is welcome. Thank you in advanced

Answer 1

It will be assumed that $n$ is a positive integer.

By definition of the functions $e^z$ and $J_n(z)$ $$\begin{align} e^zJ_n(i z)&=\sum_{k=0}^\infty\frac{z^k}{k!} \sum_{l=0}^\infty\frac{(-1)^l}{l!(n+l)!}\left(\frac {iz}2\right)^{n+2l}\\ &=\left(\frac{iz}2\right)^n\sum_{m=0}^\infty z^{m} \sum_{l=0}^{\left\lfloor\frac m2\right\rfloor}\frac1{(m-2l)!\,l!\,(n+l)!\,4^l}\\ &\stackrel*=i^n\sum_{m=0}^\infty \frac1{(n+m)!}\binom{2n+2m}m\left(\frac z2\right)^{n+m}.\tag1 \end{align} $$ Substituting in $(1)$ $z=-2i\gamma t$, $m=k-n$ one obtains your identity with prefactor $i$ replaced by $i^n$ (I assume it was a typo).

Appendix:

In $(\stackrel*=)$ we used the combinatorial identity: $$ \sum_{l=0}^{\left\lfloor\frac m2\right\rfloor}\frac1{l!(n+l)!(m-2l)!4^l} =\binom{2n+2m}m\frac1{(n+m)!2^m}\\ \iff \sum_{l=0}^{\left\lfloor\frac m2\right\rfloor} \binom {n+m}l\binom{n+m-l}{m-2l}2^{m-2l} =\binom{2n+2m}m, $$ a proof of which can be found here.

Answer 2

There's a typo in the exponential factor on the right-hand side -- it should read $\exp(i\,\pi n/2).$ Mathematica can just about do the identity. Using properties of the Bessel functions for integer $n,$ I'll write (the corrected form) as

$$ (*) \quad \quad \sum_{k=n}^\infty \frac{(-1)^{n-k}}{k!} \Big(\frac{x}{2}\Big)^k \binom{2k}{k-n}=e^{-x}\,I_n{(x)} $$ where $I_n(x)$ is the modified Bessel function. Mathematica sums the left-hand side as $$ \sum_{k=n}^\infty \frac{(-1)^{n-k}}{k!} \Big(\frac{x}{2}\Big)^k \binom{2k}{k-n} = \Big(\frac{x}{2}\Big)^n \frac{1}{n!} \, {_1}F_1(n+1/2, 2n+1, -2x)$$ where ${_1}F_1(a,b,z)$ is the confluent hypergeometric function and is equivalent to the Kummer M-function $M(a,b,z).$ The numbers in the next steps refer to formulas in the Digital Libray of Mathematical Functions. Use formula 13.2.39 to determine that $M(n+1/2, 2n+1, -2x)=\exp{(-2x)}M(n+1/2, 2n+1, 2x).$ Next use 13.6.9 to determine $M(n+1/2, 2n+1, -2x)=\exp{(-x)}n!(x/2)^{-n}I_n(x).$ Formula (*) is thus true.