6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What do you mean by "exploitable" upper bound? $\endgroup$ – Aryabhata Nov 17 '14 at 21:44
  • $\begingroup$ a bound that may depend only on the order of the bessel function and the first considered zero. $\endgroup$ – Amadeus Nov 17 '14 at 21:51
3
$\begingroup$

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 $a\geq 1$ $$\int_a^\infty M(t)\textrm{d}t=\log\left(\frac{a-1}{a+1}\right)+a\log\left(\frac{a^2+1}{a^2-1}\right)+2\tan^{-1}(a)-\pi$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.