TITLE: Integrals of the Bessel function $J_0(x)$ over the intervals between its zeros
QUESTION [12 upvotes]: 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?

REPLY [8 votes]: 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 $3 \times 10^{-29}$.