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In this MO post I have placed 4 Ramanujan-type hypergeometric series found using the LLL algorithm for fast computing of some logarithms. I could prove 3 of them by means of classical methods based on a Beta function approach that allows to work on integrals. However one of them, Eq.(2) for $\log(2)$, remains un-proven. After trying to prove this series using Au WZ seeds with no success, I have applied a similar analysis transforming this series into an integral form. But now I see that it is hard (for me) to solve it to close the proof of $I=\log(2)$. I wonder if someone here in MO can tackle it. The expression is $$I=\int_0^1P(x)\cdot\,_3F_2\left(1,a,b;c,d;z(x)\right)\cdot\frac{dx}{\sqrt{1-x}}$$ where $P(x)$ is a 5-th degree polynomial with rational coefficients and $z(x)$ is a simple polynomial argument vanishing at $x=0$ and $x=1$. Two equivalent expressions are $$\begin{equation*}I_1=\int_0^1P_1(x)\cdot\,_3F_2\left(1,\frac{1}{3},\frac{2}{3};\frac{1}{4},\frac{3}{4};\,z_1(x)\right)\cdot\frac{dx}{\sqrt{1-x}}\tag{1}\label{1} \end{equation*}$$ with $$P_1(x)=-\frac{189}{524288}x^5+\frac{189}{262144}x^4-\frac{771}{131072}x^3+\frac{5813}{196608}x^2-\frac{37819}{1572864}x+\frac{824251}{2359296}$$ and $$z_1(x)=\frac{27}{131072}\cdot\,x^4(1-x)^2$$

and this one as well,$$\begin{equation*}I_2=\int_0^1P_2(x)\cdot\,_3F_2\left(1,\frac{1}{4},\frac{3}{4};\frac{1}{6},\frac{5}{6};\,z_2(x)\right)\cdot\frac{dx}{\sqrt{1-x}}\tag{2}\label{2} \end{equation*}$$ where $$P_2(x)=\frac{1}{1944}x^5-\frac{11}{7776}x^4+\frac{569}{69120}x^3+\frac{953}{69120}x^2-\frac{95}{3456}x+\frac{17}{48}$$ and $$z_2(x)=\frac{1}{3456}\cdot\,x^3(1-x)^3=\frac 12\left(\frac{x(1-x)}{12}\right)^3$$

Any hints and suggestions are welcome.

Q: Is it possible to get a proof of either $I_1=\log(2)$ or $I_2=\log(2)$?

Since $I_1=I_2=S$, I will make extensive the bounty, to anyone proving either the integrals or the original series equal to $\log2$. The series is, $$S=\frac{1}{576}\sum_{n=0}^\infty\frac{1}{8^n}\cdot\frac{686430\,n^3+1317033\,n^2+798173\,n+153712}{(12n+1)(12n+5)(12n+7)(12n+11)}\cdot\frac{(3n)!(4n)!(6n)!}{n!(12n)!}$$

NOTE Series $S$, already un-proven, is very fast and it was used to verify the current record of decimal digits of $\log2$ up to $3\,\textrm{x}\,10^{12}$ places on Feb.12 2024, as it is reported here L1, L2 and L3

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  • $\begingroup$ Could you share what you tried to do that didn't work? Typically I would imagine one needs to use something like the Meline-Barnes integral, apply Fubini, push the Contour and compute residues. $\endgroup$
    – kindasorta
    Commented Mar 14 at 23:27
  • $\begingroup$ In case this isn't clear what I mean by Meline-Barnes, that is basically applying the inverse Meline of the Meline transform. The Meline transform of the hypergeometric function could most likely be found in Gradshteyn and Ryzhik. $\endgroup$
    – kindasorta
    Commented Mar 14 at 23:30
  • $\begingroup$ Yes, you can compute the Mellin Barnes integral or if you can put it in a Meijer-G or Fox-H form, you will be faced to an infinite sequence of poles, and the residues computed will give you a series. It is not clear if such series will be easy enough to get log(2). These integrals come from a higher hypergeometric-type series -Eq.(2) in the link beginning this post-. In fact, if any of these integrals can be evaluated to $\log2$, the source series is proven. $\endgroup$
    – Jorge Zuniga
    Commented Mar 14 at 23:47
  • $\begingroup$ I suggest you to add the explicit description of the infinite series as well. My intuition is that this sum would be expressible as the special value of some member of the family of Bessel/Hypergeometric functions of some other parameters. These sometimes 'collapse' from complicated transcendental functions to much simpler ones with the right parameters. This has to be what is happening. $\endgroup$
    – kindasorta
    Commented Mar 15 at 0:10
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    $\begingroup$ @kindasorta, done. $\endgroup$
    – Jorge Zuniga
    Commented Mar 15 at 0:40

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