12
$\begingroup$

Let $J_0(x)$ be the Bessel function of the first kind. It has an infinite number of zeros on the positive real semi-axis. Let's denote them as $j_{0,n}$: $$j_{0,1}=2.40482...,\quad j_{0,2}=5.52007...,\quad j_{0,3}=8.65372...,\quad\small...\tag1$$ We are interested in absolute values of the integrals of $J_0(x)$ over the intervals between its consecutive zeros: $$\sigma_n=(-1)^n\int_{j_{0,n}}^{j_{0,n+1}}\!\!J_0(x)\,dx.\tag2$$ Their values are: $$\sigma_1=0.80145...,\quad\sigma_2=0.59932...,\quad\sigma_3=0.49904...,\quad\small...\tag3$$ We are interested in the asymptotic behavior of this sequence. Empirically, it seems that $$\sigma_n\,\stackrel{\color{#a0a0a0}?}\sim\,\frac{2\sqrt2}{\pi\sqrt n}\left(1-\frac1{8n}+O\!\left(\frac1{n^2}\right)\right)\!.\tag4$$ Can we prove this? Can we find next coefficients in this expansion?

$\endgroup$
  • 1
    $\begingroup$ The leading term is easily derived from the asymptotic expression $J_0(x)=\sqrt{\frac{2}{\pi x}}\cos\left(x-\frac{\pi}{4}\right)+\mathcal{O}\left(x^{-1}\right)$. So, you could try using the asymptotic expansion to higher orders to find the asymptotic expansion of the zeros and then integrate that to get a systematic expansion in powers of $n^{-1}$ for the integral. $\endgroup$ – Count Iblis May 4 at 2:56
  • $\begingroup$ Check out the McMahon expansions of $j_{0,n}$ here: dlmf.nist.gov/10.21 , combined with the asymptotic value of the Bessel function at large $x,n$. $\endgroup$ – Alex R. May 4 at 5:03
  • $\begingroup$ I think that I did improve my answer. $\endgroup$ – Claude Leibovici May 5 at 3:05
  • $\begingroup$ Have a look at mathematica.stackexchange.com/questions/197784/… $\endgroup$ – Claude Leibovici May 6 at 10:21
6
$\begingroup$

Using $$\int J_0(x)\,dx =x \, _1F_2\left(\frac{1}{2};1,\frac{3}{2};-\frac{x^2}{4}\right)$$ $$\int_0^a J_0(x)\,dx =\frac{1}{2} a (\pi \pmb{H}_0(a) J_1(a)+(2-\pi \pmb{H}_1(a)) J_0(a))$$ $$\int_0^{j_{0,n}} J_0(x)\,dx=\frac{\pi}{2}\, j_{0,n}\, \pmb{H}_0\left(j_{0,n}\right)\, J_1\left(j_{0,n}\right)$$ $$\sigma_n=(1)^n\frac{\pi}{2} \big( j_{0,n+1} \,\pmb{H}_0\left(j_{0,n+1}\right)\, J_1\left(j_{0,n+1}\right)- j_{0,n}\, \pmb{H}_0\left(j_{0,n}\right)\, J_1\left(j_{0,n}\right)\big)$$

it seems to be that $$\color{blue}{\sigma_n\,=\frac{8}{\pi\sqrt {8n}}\left(1-\frac{1}{(8 n)}+\frac{1}{(8n)^2} -\frac{1}{10}\frac{1}{(8n)^3}-\frac{2}{(8n)^4}+O\!\left(\frac1{n^5}\right)\right)}$$

This was obtained for values up to $n=100$.

For a few small values of $n$, here are the results $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 1 & 0.8012287685 & 0.8014542111 \\ 2 & 0.5992828620 & 0.5993225154 \\ 3 & 0.4990351563 & 0.4990496204 \\ 4 & 0.4365280908 & 0.4365351123 \\ 5 & 0.3928185568 & 0.3928225593 \\ 6 & 0.3600543084 & 0.3600568365 \\ 7 & 0.3343192651 & 0.3343209797 \\ 8 & 0.3134138472 & 0.3134150722 \\ 9 & 0.2959950956 & 0.2959960063 \\ 10 & 0.2811906203 & 0.2811913190 \\ 20 & 0.2000664765 & 0.2000665991 \\ 30 & 0.1636924770 & 0.1636925214 \\ 40 & 0.1419090468 & 0.1419090684 \\ 50 & 0.1270064402 & 0.1270064525 \\ 60 & 0.1159886945 & 0.1159887023 \\ 70 & 0.1074165671 & 0.1074165724 \\ 80 & 0.1005013911 & 0.1005013949 \\ 90 & 0.0947700476 & 0.0947700505 \\ 100 & 0.0899192327 & 0.0899192349 \end{array} \right)$$

Edit

Thanks to @Roman's answer to this question of mine, we now have the coefficients for $$\sigma_n=\frac{2\sqrt2}{\pi\sqrt n}\left(1+\sum_{k=1}^{18}\frac{a_k}{n^k}+O\left(\frac1{n^{19}}\right)\right)$$

Using these coefficients, the errors are smaller than $10^{-10}$ as soon as $n >3$. For $n=10$, the error is $4 \times 10^{-20}$ and for $n=100$, the error is $3 \times 10^{-29}$.

$\endgroup$

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.