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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?

Vladimir Reshetnikov
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    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. – Count Iblis May 04 '19 at 02:56
  • Check out the McMahon expansions of $j_{0,n}$ here: https://dlmf.nist.gov/10.21 , combined with the asymptotic value of the Bessel function at large $x,n$. – Alex R. May 04 '19 at 05:03
  • I think that I did improve my answer. – Claude Leibovici May 05 '19 at 03:05
  • Have a look at https://mathematica.stackexchange.com/questions/197784/how-could-this-asymptotics-expansion-be-obtained/197793#197793 – Claude Leibovici May 06 '19 at 10:21

1 Answers1

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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 $3 \times 10^{-29}$.

Claude Leibovici
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