# Summation of a series of bessel functions

1. Jul 16, 2009

### JertLankster

The problem is to prove the following:

$$\sum_{m>0}J_{j+m}(x)J_{j+m+n}(x) = \frac{x}{2n}\left(J_{j+1}(x)J_{j+n}(x) - J_{j}(x)J_{j+n+1}(x)\right).$$

Now for the rambling...

I've been reading for a while, but this is my first post. Did a quick search, but I didn't find anything relevant. I could have missed it, though.

I was reading a paper on quantum spin chains (Antal et al, Phys. Rev E, 59, 4912 (1999)) and a series of Bessel functions suddenly turned into a compact expression (shown above) without any mention of what was done. If you have access to the journal, this appears in the appendix on the last page... eq. (B2).

As a good chunk of the paper dealt with the details of evaluating the asymptotics of another series, it makes me think I really missed something simple since they just inserted this result without even a reference to Abramowitz/Stegun or Gradshteyn. After a month or so of being involved with this stuff, I came up with an extremely cumbersome "proof." I am by no means an expert on Bessel functions, so I just thought I'd ask if anyone who is more familiar with these manipulations knows a slick way to sum that series that doesn't take 2 pages. As an interesting corollary, if you differentiate the series, use some recurrence relations, and then equate it to the given result, you wind up with

$$\frac{1}{2}\int_{0}^{x}\left(J_{j+1}(x')J_{j+n}(x')+J_{j}(x')J_{j+n+1}(x')\right)dx' = \frac{x}{2n}\left(J_{j+1}(x)J_{j+n}(x)-J_{j}(x)J_{j+n+1}(x)\right).$$

I haven't been able to find a direct proof for this either... don't know which part is easier to attack, but I thought it was worth a mention.

Thanks for reading.

Last edited: Jul 16, 2009
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