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?
Integrals of the Bessel function $J_0(x)$ over the intervals between its zeros
asymptoticsbessel functionsconjecturesintegrationreal-analysis
Related Solutions
Thank you for posting this question, I enjoyed trying to answer it.
Start with the expression that Mathematica gave you and replace each argument $\frac14$ of a hypergeometric function with $\frac z4$, because we will be taking limits. I will call the two hypergeometric functions $Q_1(z)$ and $Q_2(z)$. Each term can be brought to a closed form by using identity 16.6.2 from the DLMF.
Setting $a=\frac12$, $b=1-\frac\nu2$, we get $$ Q_1(z) = (1-z)^{-\frac12} F\left(\frac16, \frac36,\frac 56; 1-\frac\nu2, 1+\frac\nu2; \frac{-27 z}{4(1-z)^3} \right), $$ and setting $a=\frac{1+3\nu}{2}$, $b=1+\frac\nu2$, we get $$ Q_2(z) = (1-z)^{-\frac{1+3\nu}{2}} F\left(\frac{1+3\nu}6, \frac{3+3\nu}{6}, \frac{5+3\nu}{6}; 1+\frac\nu2, 1+\nu; \frac{-27z}{4(1-z)^3} \right). $$ (Note that there are 6 possible identities to try per function, one for each possible choice of $a$ and $b$ from the parameters, so it helps to do this on a computer.)
The reason this works is that now the point $z=1$ is a singular point of the hypergeometric functions on the right hand side, and Mathematica will succeed in finding the limits as $z\to1$. The expression for the whole integral that you have is $$ Q = \frac{2^{\frac43}\pi^{\frac12}}{\Gamma(-\frac16)\Gamma(\frac56+\frac\nu2)\Gamma(\frac56-\frac\nu2)\sin\frac{\nu\pi}2} \left( -1 + 3^{-\frac{3\nu}2}\cos\left(\frac{\nu\pi}{2}\right) \frac{\Gamma(\frac{1+3\nu}{2}) \Gamma(\frac56-\frac\nu2)}{\Gamma(\frac{1+\nu}2)\Gamma(\frac56+\frac\nu2)} \right). $$ Call the large expression in brackets $A$, and then write $$ A = -1 + B 3^{-\frac{3\nu}{2}}\cos\frac{\pi\nu}{2}, \qquad B = \frac{\Gamma(\frac{1+3\nu}{2}) \Gamma(\frac56-\frac\nu2)}{\Gamma(\frac{1+\nu}2)\Gamma(\frac56+\frac\nu2)}. $$
Now, Mathematica will not simplify $A$ or $B$ on its own, so it needs help. Set $x=\frac16+\frac\nu2$, and use the multiplication formula to get $$ \frac{\Gamma(\frac{1+3\nu}2)}{\Gamma(\frac{1+\nu}{2})\Gamma(\frac56+\frac\nu2)} = \frac{\Gamma(3x)}{\Gamma(x+\frac13)\Gamma(x+\frac23)} = \frac{\Gamma(x)}{2\pi} 3^{3x-1/2}. $$ After this, $A$ simplifies to $$ A = -1 + \frac{\Gamma(\frac16+\frac\nu2)\Gamma(\frac56-\frac\nu2)}{2\pi}\cos\frac{\pi\nu}{2} = -1 + \frac{\cos\frac{\pi\nu}{2}}{2\sin(\frac\pi6+\frac{\pi\nu}{2})}, $$ where I've also used the reflection formula for $\Gamma(z)\Gamma(1-z)$ to get rid of the gamma functions. Some further amount of manual trigonometry yields $$ A = -\frac{\sqrt{3}}{2}\frac{\sin\frac{\pi\nu}{2}}{\sin(\frac\pi6 + \frac{\nu\pi}{2})}. $$
Finally, write $$ \frac{1}{\sin(\frac\pi6+\frac{\pi\nu}{2})} = \frac{\Gamma(\frac16+\frac\nu2)\Gamma(\frac56-\frac\nu2)}{\pi}, $$ and substitute back. Lots of things cancel, and the answer is $$ Q = -\frac{3^{1/2}2^{1/3}}{\pi^{1/2}} \frac{\Gamma(\frac16+\frac\nu2)}{\Gamma(-\frac16)\Gamma(\frac56+\frac\nu2)}. $$
This closed form is equivalent to the one you gave through the use of $\Gamma(\frac16)\Gamma(-\frac16)=-12\pi$.
P.S. I would also like to note that the integral $$ I(\nu,c) = \int_0^\infty J_\nu(x)^2 J_\nu(c x)\,dx $$ and its general form $$ \int_0^\infty x^{\rho-1}J_\nu(a x) J_\mu(b x) J_\lambda(c x)\,dx $$ appear in Gradshteyn and Ryzhik, and you can find a paper "Some infinite integrals involving bessel functions, I and II" by W. N. Bailey, which evaluates this integral in terms of Appell functions, but only in the case $c>2$ ($|c|>|a|+|b|$), which is where the $F_4$ Appell function converges. DLMF 16.16.6 actually gives a way to write this integral as $$ \frac{\Gamma(\frac{1+3\nu}{2})c^{-1-2\nu}}{\Gamma(1+\nu)^2\Gamma(\frac{1-\nu}{2})} \,\,\,{}_2F_1\left( \frac{1+\nu}{2}, \frac{1+3\nu}{2}; 1+\nu; x \right)^2, \qquad x = \frac{1-\sqrt{1-4/c^2}}{2}, $$ but the issue is that this is only correct for $c>2$, and the rhs is complex for $c<2$. Appell function would only be defined by analytic continuation in this case anyway, and I didn't find anything useful about non-principal branches of Appell or hypergeometric functions.
For $c>2$, Mathematica also gives the following: $$ I(\nu,c) = \frac{\Gamma(\frac{1+3\nu}{2})c^{-1-2\nu}}{\Gamma(\frac{1-\nu}{2})\Gamma(1+\nu)^2} \,\,\,{}_3F_2\left( \frac{1+\nu}{2}, \frac{1}{2}+\nu, \frac{1+3\nu}{2}; 1+\nu, 1+2\nu; \frac{4}{c^2} \right), $$ but this is incorrect when $c<2$.
So basically, I'll evaluate a bunch of integrals, trying to avoid polylogs as much as possible.
First thing is to notice that $\displaystyle H_n-2H_{2n}+H_{4n}=\int_0^1 \frac{x^{2n}-x^{4n}}{1+x}dx$. I noticed that $H_n-2H_{2n}+H_{4n}=H_{4n}-H_{2n}-(H_{2n}-H_n)=H_{4n^{-}}-H_{{2n}^{-}}$, where $H_{n^{-}}=\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}$ is called a skew harmonic number (at least by Khristo N. Boyadzhiev. link.) Knowing they have a simple intergal representation I found the above. My answer is influenced by Boyadzhiev's work. If I make any unexplainable substitution, it's most likely $t=\frac{1-x}{1+x}$. Also, I'm not very good with Latex, so alignment should be awful. Hopefully there are no typos.
Below, easy enough to prove, is what I take for granted: $ -\ln\sin x=\ln2+\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n} ,-\ln\cos x=\ln2+\sum_{n=1}^{\infty} \frac{(-1)^n\cos(2nx)}{n} \tag{1}$
$$ \int_0^{\frac{\pi}{2}} \cos x \cos(nx)dx=\begin{cases} \frac{\pi}{4} &n=1\\0 &n \,\,\text{odd}\\ \frac{(-1)^{1+n/2}}{n^2-1} &n \,\,\text{even} \end{cases} \tag{2}$$
$$ \int_0^1 \frac{\ln(1-x)}{a+x}dx=-\operatorname{Li_2}\left(\frac1{a+1}\right)\tag{3}$$ Starting, $$\sum_{n=0}^{\infty}(H_{n}-2H_{2n}+H_{4n})^2=\sum_{n=0}^{\infty}\int_0^1\int_0^1\frac{(x^{2n}-x^{4n})(u^{2n}-u^{4n})}{(1+x)(1+u)}dxdu \\=\small\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^2)}-2\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^4)}+\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^4u^4)} \\=I_{22}-2I_{24}+I_{44}$$
Computing $I_{22}$.
Substitute $u=\frac{y}{x}$ ,change the order of integration, evaluate the inner integral, and substitue $t=\frac{1-x}{1+x}$ to get $$\begin{align} I_{22}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^2u^2)}=\int_0^1\int_0^x\frac{dydx}{(1+x)(x+y)(1-y^2)} \\=\int_0^1 \frac1{1-y^2}\int_1^y \frac{dx}{(1+x)(x+y)} dy=\int_0^1 \frac{\ln\left(\frac{(1+x)^2}{4x}\right)}{(1+x)(1-x^2)}\,dx \\=\frac{-1}{4}\int_0^1 \frac{(1+t)}{t^2}\ln(1-t^2)dt=-\frac14\int_0^1\frac{\ln(1-t^2)}{t^2}dt-\frac14\int_0^1\frac{\ln(1-t^2)}{t}dt \\=\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+1)}+\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+2)}=\frac{\ln2}{2}+\frac{\pi^2}{48}.\end{align}$$
Computing $I_{44}$.
Start the same as with $I_{22}$ to get $\displaystyle I_{44}=\int_0^1 \frac{\ln\left(\frac{(1+x)^2}{4x}\right)}{(1-x)(1-x^4)}\,dx=\frac{-1}{8}\int_0^1 \frac{\ln(1-t^2)}{t^2(1+t^2)}(1+t)^3dt$. We can calculate these integrals: $$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{1+x^2}dx=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx+\int_0^1 \frac{\ln(1-x)}{1+x^2}dx \tag{4} \\=\int_0^1 \frac{\ln(1+x)}{1+x^2}dx +\int_0^1 \frac{\ln\left(\frac{2t}{1+t}\right)}{1+t^2}dt \\=\frac{\pi}{4}\ln2+\sum_{n=0}^{\infty} (-1)^n\int_0^1\ln(t) t^{2n}dt=\frac{\pi}{4}\ln2-G. \end{align}$$ $$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{x^2(1+x^2)}dx=\int_0^1 \frac{\ln(1-x^2)}{x^2}dx-\int_0^1 \frac{\ln(1-x^2)}{1+x^2}dx \tag{5} \\=-\sum_{n=0}^{\infty} \frac1{n+1}\int_0^1 x^{2n}dx-\frac{\pi}{4}\ln2+G=G-\frac{\pi}{4}\ln2-2\ln2.\end{align}$$ $$\begin{align} \int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx=\frac12\int_0^1 \frac{\ln(1-x)}{1+x}dx \tag{6} \\=-\frac12 \operatorname{Li_2}\left(\frac12\right)=\frac{\ln^2 2}{4}-\frac{\pi^2}{24}.\end{align}$$ $$\begin{align} \int_0^1 \frac{\ln(1-x^2)}{x(1+x^2)}dx=\int_0^1 \frac{\ln(1-x^2)}{x}dx-\int_0^1 \frac{x\ln(1-x^2)}{1+x^2}dx \tag{7} \\=-\sum_{n=0}^{\infty}\frac1{n+1}\int_0^1 x^{2n+1}dx-\frac{\ln^2 2}{4}+\frac{\pi^2}{24}=-\frac{\pi^2}{24}-\frac{\ln^2 2}{4}.\end{align}$$ Altogether, $$I_{44}=\frac{-1}{8}\int_0^1 \frac{\ln(1-x^2)}{x^2(1+x^2)}(1+3x+3x^2+x^3)dx \\=-\frac{\pi}{16}\ln2+\frac{\ln2}{4}+\frac{\ln^2 2}{16}+\frac{\pi^2}{48}+\frac{G}{4}.$$
Computing $I_{24}$.
Substitute $u=\frac{y}{x^2}$, change the order of integration, let $y\to y^2$, evaluate the inner integral,and substitue $t=\frac{1-x}{1+x}$: $$\begin{align*} I_{24}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1-x^4u^2)}=\int_0^1\int_0^{x^2} \frac{dydx}{(1+x)(x^2+y)(1-y^2)} \\=\int_0^1 \frac1{1-y^2}\int_{\sqrt{y}}^1\frac{dx}{(1+x)(x^2+y)}dy=2\int_0^1\frac{y}{1-y^4}\int_{y}^1\frac{dx}{(1+x)(x^2+y^2)}dy \\=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1-x^4)}dx-2\int_0^1\frac{x\ln\left(\frac{(1+x)\sqrt{1+x^2}}{2\sqrt{2}x}\right)}{(1+x^2)(1-x^4)}dx =I_{241}-I_{242}. \end{align*}$$
Evaulation of $I_{241}$.
Substitute $t=\frac{1-x}{1+x}$ to get $\displaystyle I_{241}=\frac14\int_0^1 \frac{\tan^{-1}(t)}{t(1+t^2)^2}(1+t)^4dt$.
We can calculate these integrals.In the following, let $x=\tan\theta$: $$\begin{align} \int_0^1 \frac{\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\cos^2(\theta)d\theta=\frac{\pi^2}{64}+\frac{\pi}{16}-\frac18.\tag{8}\end{align}$$ $$\begin{align} \int_0^1 \frac{x\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\tan\theta\cos^2{\theta}d\theta=\frac12\int_0^{\frac{\pi}{4}}\theta\sin(2\theta)d\theta=\frac18.\tag{9}\end{align}$$ $$\begin{align} \int_0^1 \frac{x^2\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\sin^2(\theta)d\theta=\frac{\pi^2}{64}-\frac{\pi}{16}+\frac18.\tag{10}\end{align}$$ $$\begin{align} \int_0^1 \frac{x^3\tan^{-1}(x)}{(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\sin^3(\theta)\sec{\theta}\,d\theta \tag{11} \\=\int_0^{\frac{\pi}{4}}\theta\tan\theta \,d\theta-\int_0^{\frac{\pi}{4}}\theta\sin\theta\cos\theta \,d\theta=\frac{\pi}{8}\ln2-\frac18+\int_0^{\frac{\pi}{4}}\ln\cos\theta \,d\theta \\=\frac{\pi}{8}\ln2-\frac18-\int_0^{\frac{\pi}{4}}\ln2 \,d\theta-\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\int_0^{\frac{\pi}{4}}\cos(2n\theta)\,d\theta \\=-\frac{\pi}{8}\ln2-\frac18+\frac12\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\sin(\frac{\pi n}{2})=\frac{G}{2}-\frac{\pi}{8}\ln2-\frac18.\end{align}$$ $$\begin{align} \int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)^2}dx=\int_0^{\frac{\pi}{4}}\theta\cos^3(\theta)\csc{\theta}\,d\theta \tag{12} \\=\int_0^{\frac{\pi}{4}}\theta\cot\theta \,d\theta-\int_0^{\frac{\pi}{4}}\theta\sin\theta\cos\theta \,d\theta=-\frac18-\frac{\pi}{8}\ln2-\int_0^{\frac{\pi}{4}}\ln\sin\theta \,d\theta \\=-\frac18-\frac{\pi}{8}\ln2+\int_0^{\frac{\pi}{4}}\ln2 \,d\theta+\sum_{n=1}^{\infty}\frac1{n}\int_0^{\frac{\pi}{4}}\cos(2n\theta)\,d\theta \\=-\frac18+\frac{\pi}{8}\ln2+\frac12\sum_{n=1}^{\infty} \frac{\sin(\frac{\pi n}{2})}{n^2}=\frac{G}{2}+\frac{\pi}{8}\ln2-\frac18.\end{align}$$ Altogether, $$I_{241}=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1-x^4)}dx= \frac14\int_0^1 \frac{\tan^{-1}(x)}{x(1+x^2)^2}(1+4x+6x^2+4x^3+x^4)dx \\=\frac{\pi^2}{32}+\frac18+\frac{G}{4}$$
Evaulation of $I_{242}$.
Substitute $t=\frac{1-x}{1+x}$ to get $$ I_{242}=\frac14\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{t(1+t^2)^2}(1-t^2)(1+t)^2dt \\=\frac12\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{(1+t^2)^2}(1-t^2)dt+\frac14\int_0^1 \frac{\ln\left(\frac{\sqrt{1+t^2}}{1-t^2}\right)}{t(1+t^2)}(1-t^2)dt \\=\frac12\int_0^1 \frac{\ln\left(\frac{1+x^2}{1-x^2}\right)}{(1+x^2)^2}(1-x^2)dx-\frac14\int_0^1\frac{\ln(1+x^2)}{(1+x^2)^2}(1-x^2)dx\\+\frac18\int_0^1\frac{\ln(1+x^2)(1-x^2)}{x(1+x^2)}dx-\frac14\int_0^1\frac{\ln(1-x^2)(1-x^2)}{x(1+x^2)}dx $$ Calculating these integrals: $$\begin{align} \int_0^1 \frac{\ln\left(\frac{1+x^2}{1-x^2}\right)}{(1+x^2)^2}(1-x^2)dx=-\frac12\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{\sqrt{x}(1+x)}\frac{1-x}{1+x}dx \tag{13} \\=-\frac12\int_0^1\frac{t\ln t}{\sqrt{1-t^2}}dt=-\frac18\int_0^1\frac{\ln t}{\sqrt{1-t}}dt=-\frac18\int_0^1 t^{-1/2}\ln(1-t)dt \\=\frac14\sum_{n=0}^{\infty} \frac1{(n+1)(2n+3)}=\frac12-\frac{\ln2}{2}.\end{align}$$ $$\begin{align} \int_0^1\frac{\ln(1+x^2)(1-x^2)}{x(1+x^2)}dx=\frac12\int_0^1\frac{\ln(1+x)(1-x)}{x(1+x)}dx \tag{14} \\=\frac12\int_0^1\frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx \\=\frac12\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n+1}\int_0^1 x^n dx-\frac12\ln^2(1+x)\bigg{|}_0^1=\frac{\pi^2}{24}-\frac{\ln^2 2}{2}.\end{align}$$ $$\begin{align} \int_0^1\frac{\ln(1-x^2)(1-x^2)}{x(1+x^2)}dx=\frac12\int_0^1\frac{\ln(1-x)}{x}dx-\int_0^1\frac{\ln(1-x)}{1+x}dx \tag{15} \\=-\frac{\pi^2}{12}-\left(\frac{\ln^2 2}{2}-\frac{\pi^2}{12}\right)=-\frac{\ln^2 2}{2}.\end{align}$$ $$ \int_0^1\frac{\ln(1+x^2)}{(1+x^2)^2}(1-x^2)dx=-2\int_0^{\frac{\pi}{4}}\cos^2(\theta)(1-\tan^2\theta)\ln\cos\theta\,d\theta \tag{16} \\=-2\int_0^{\frac{\pi}{4}}\cos(2\theta)\ln\cos\theta\,d\theta=2\ln2\int_0^{\frac{\pi}{4}}\cos(2\theta)d\theta+\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\int_0^{\frac{\pi}{2}}\cos\theta \cos(n\theta)d\theta \\=\ln2-\frac{\pi}{4}+\sum_{n=1}^{\infty} \frac{(-1)^{2n}}{2n}\frac{(-1)^{n+1}}{(2n)^2-1}=\frac{\ln2}{2}-\frac{\pi}{4}+\frac12.$$ Altogether, $\displaystyle I_{242}=\frac{\pi^2}{192}+\frac{\pi}{16}+\frac{\ln^2 2}{16}-\frac{3\ln2}{8}+\frac18$,
leading to $\displaystyle I_{24}=I_{241}-I_{242}=\frac{5\pi^2}{192}-\frac{\pi}{16}-\frac{\ln^2 2}{16}+\frac{3\ln2}{8}+\frac{G}{4}$, and finally, confirming the conjecture, $$\sum_{n=0}^{\infty}(H_{n}-2H_{2n}+H_{4n})^2=I_{22}-2I_{24}+I_{44}=\frac{\pi}{8}-\frac{\pi}{16}\ln2-\frac{\pi^2}{96}+\frac{3\ln^2 2}{16}-\frac{G}{4}.$$
I don't know about higher powers. I guess the case $\mathcal A_2$ can also be done. If we start the same as with $\mathcal B_2$, writing $\mathcal A_2=J_{22}-2J_{24}+J_{44}$ we can find that $\displaystyle J_{22}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1+x^2u^2)}=2I_{44}-I_{22}=-\frac{\pi}{8}\ln2+\frac{G}{2}+\frac{\pi^2}{48}+\frac{\ln^2 2}{8}$
$J_{44}$ can be reduced to $\displaystyle =-\frac12\int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}(1+x)^3 dx$. already this i can't evaulate fully, as polylogs are inescapable. Factorizing $x^2+6x+1=(x+3+2\sqrt{2})(x+3-2\sqrt{2}).$
I can get $\displaystyle \int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}dx=-\frac{\pi^2}{12}+\frac{4-3\sqrt{2}}{16}\operatorname{Li_2}\left(\frac{2-\sqrt{2}}{4}\right)+\frac{4+3\sqrt{2}}{16}\operatorname{Li_2}\left(\frac{2+\sqrt{2}}{4}\right)$
but nothing more.
Edit 1.
After some more work and a fair amount of cancellation, we obtain $$\int_0^1 \frac{\ln(1-x^2)}{x(x^4+6x^2+1)}(1+x)^3 dx=\frac{1+2\sqrt{2}}{4}\pi\ln2-\frac{\pi^2}{24}-\frac14\ln\left(\frac{2+\sqrt{2}}{4}\right)\ln\left(\frac{2-\sqrt{2}}{4}\right) -\frac{\sqrt{2}+1}{2}\Im\operatorname{Li_2}\left(\frac{2+\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right)-\frac{\sqrt{2}-1}{2}\Im\operatorname{Li_2}\left(\frac{2-\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}\right)$$
I obtained it by calculating $\displaystyle \int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2\ln\left(\frac{a+1}{a-1}\right)+\operatorname{Li_2}\left(\frac2{1-a}\right)-\operatorname{Li_2}\left(\frac1{1-a}\right)$, which together with $(3)$ can be used to give a closed form for $\displaystyle \int_0^1\frac{\ln(1-x^2)}{x+a}dx$, which in turn, through partial fractions, can be used to give a closed form for $\displaystyle \int_0^1\frac{\ln(1-x^2)}{x^2+a^2}dx$. Fortunately, things didn't get too ugly as both $3+2\sqrt{2}$ and $3-2\sqrt{2}$ have nice square roots. I will fill in details as soon as I can.
Now we just need to evaluate $J_{24}$. Starting similarly as with $I_{24}$, we have: $$J_{24}=\int_0^1\int_0^1\frac{dxdu}{(1+x)(1+u)(1+x^4u^2)} \\=2\int_0^1\frac{\tan^{-1}\left(\frac{1-x}{1+x}\right)}{(1+x^2)(1+x^4)}dx-2\int_0^1\frac{x\ln\left(\frac{(1+x)\sqrt{1+x^2}}{2\sqrt{2}x}\right)}{(1+x^2)(1+x^4)}dx \\=J_{241}-J_{242}$$
Through $t=\frac{1-x}{1+x}$, $J_{241}$ turns to $\displaystyle \int_0^1 \frac{\tan^{-1}(x)}{(1+x^2)(x^4+6x^2+1)}(1+x)^4\,dx$. I don't have any idea about that yet. \Edit 1.
Best Answer
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 $3 \times 10^{-29}$.