# Sum involving zeros of Bessel function

I came across the following sum in my work involving the infinite sum of function of zeros of Bessel functions.

$$\displaystyle I_0=\sum_{\ell=2}^{\infty}log\left(\frac{j^2_{\nu,\ell}-j^2_{\nu,1}}{j^2_{\nu,\ell}+j^2_{\nu,1}}\right),$$

where $(j_{\nu,l})$ is the sequence of zeros of the Bessel function of first kind $J_\nu$ with $\nu\geq 1/2$.\

Does anyone have any idea of how to evaluate this in term of the order and fews first zero of the Bessel function ? or at least an exploitable upper bound.

• What do you mean by "exploitable" upper bound? – Aryabhata Nov 17 '14 at 21:44
• a bound that may depend only on the order of the bessel function and the first considered zero. – Amadeus Nov 17 '14 at 21:51

To estimate $I_0$, let us set $\displaystyle M(t):=\log\left(\frac{t^2-1}{t^2+1}\right)$, thus $\displaystyle I_0:=\sum_{\ell=2}^{\infty}M\left(\frac{j_{\nu,\ell}}{j_{\nu,1}}\right)$. Now notice that $M$ is non-decreasing for $t\geq 1$ and $\displaystyle\frac{j_{\nu,\ell}}{j_{\nu,1}}\geq 1$ for all $\ell\geq 2$ so that
\begin{equation}\displaystyle I_0\leq \frac{\pi}{j_{\nu,1}}\int_{j_{\nu,2}/j_{\nu,1}}^\infty M(t)\,\textrm{d}t, \end{equation} where we have used that $j_{\nu,k+1}-j_{\nu,k}\leq \pi$ for $\nu\geq 1/2$.
On the other hand, for $a\geq 1$ $$\int_a^\infty M(t)\textrm{d}t=\log\left(\frac{a-1}{a+1}\right)+a\log\left(\frac{a^2+1}{a^2-1}\right)+2\tan^{-1}(a)-\pi$$