This question is a follow-up to MSE#3142989.

Two seemingly innocent hypergeometric series ($\phantom{}_3 F_2$) $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{(-1)^n}{2n+1}\qquad \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2\frac{1}{2^n(2n+1)}$$ can be reduced to the logarithmic integrals $$ \int_{0}^{1}\frac{\text{arctanh}(\sqrt{x})}{\sqrt{(1-x)(2-x)}}\,dx,\qquad\int_{0}^{1}\frac{\frac{1-x}{1+x}\log(x)}{\sqrt{x(1+x^2)}}\,dx\tag{A,B}$$ by FL-expansions and Fourier series, or directly by semi-integration by parts.

The issue is that neither (A) or (B) seem to be manageable through standard substitutions, the help of computer algebra systems or prayers to special deities, so I hope human beings are able to provide better insights. A promising substitution for (A) is $x=\frac{3-\cosh z}{2}$.

  • $\begingroup$ For the sake of uniformity, (A) is a multiple of $$ \int_{0}^{1}\frac{\frac{1-x}{1+x}\log(x)}{\sqrt{x(1+6x+x^2)}}\,dx.$$ $\endgroup$ – Jack D'Aurizio Mar 19 '19 at 17:12
  • $\begingroup$ (B) is a multiple of $$\int_{0}^{1}\frac{\text{arctanh}(\sqrt{x})}{\sqrt{(1-x)(1+x)}}\,dx.$$ $\endgroup$ – Jack D'Aurizio Mar 19 '19 at 17:15
  • $\begingroup$ In particular, in order to find (B) it is sufficient to find $$ \int_{0}^{1}\log\left(1+\sqrt{\sin\theta}\right)\,d\theta. $$ $\endgroup$ – Jack D'Aurizio Mar 19 '19 at 17:52
  • 1
    $\begingroup$ Are you sure that here $\int_{0}^{1}\log\left(1+\sqrt{\sin\theta}\right)\,d\theta$ the upper bound is $1$? $\endgroup$ – Zacky Mar 19 '19 at 19:01
  • $\begingroup$ @Zacky: of course the upper bound should have been $\frac{\pi}{2}$, thanks for pointing it out. $\endgroup$ – Jack D'Aurizio Mar 19 '19 at 19:07

Long Comment: Notes on evaluating $I_A$:

I first found the equivalent integral $$I_A=2 \int_0^1 \frac{\sinh ^{-1}(x)}{x \sqrt{1-x^2}} \, dx\tag{1}$$ Integrating (1) by parts I then found $$I_A=2 \int_0^1 \frac{\log \left(\frac{\sqrt{1-x^2}+1}{x}\right)}{\sqrt{x^2+1}} \, dx\tag{2}$$

Then since $$\frac{1}{\sqrt{x^2+1}}=\sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\, x^{2 n}$$

we obtain using (2) $$I_A=2 \sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\, \int_0^1 x^{2 n}\log \left(\frac{\sqrt{1-x^2}+1}{x}\right) \, dx $$

Then using Mathematica

$$\int_0^1 x^{2 n} \log \left(\frac{\sqrt{1-x^2}+1}{x}\right) \, dx=\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{3}{2}\right)}{(2 n+1)^2\, \Gamma (n+1)}$$

$$I_A=2 \sum _{n=0}^{\infty } \frac{(-1)^n \binom{2 n}{n}}{4^n}\,\frac{\sqrt{\pi }\, \Gamma \left(n+\frac{3}{2}\right)}{(2 n+1)^2\, \Gamma (n+1)}=\pi \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};-1\right)$$

UPDATE 23/03/2019

In thinking about Jack D'Aurizio's comment below: I think the relationship between the integrals and Euler sums can perhaps be found by proving and then making use of (3) and (4) $$\int_0^1 x^n \tanh ^{-1}\left(\sqrt{x}\right) \, dx=\frac{1}{n+1}\sum _{k=1}^{n+1} \frac{1}{2 k-1}\tag{3}$$

$$\int_0^1 x^{2n} \tanh ^{-1}\left(\sqrt{x}\right) \, dx=\frac{1}{2n+1}\sum _{k=1}^{2n+1} \frac{1}{2 k-1}\tag{4}$$

for Integrals A and B respectively.

In regard to integral A this looks difficult because it involves simplifying something like:

$$I_A=\frac{1}{\sqrt{2}} \int_0^1 \left(\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^n}{4^n}\right) \left(\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^n}{8^n} \right) \tanh ^{-1}\left(\sqrt{x}\right) \,dx$$

I get stuck trying to simplify the Cauchy Product.

The proof may be easier in regard to integral B since

$$\frac{1}{\sqrt{1-x^2}}=\sum _{n=0}^{\infty } \binom{2 n}{n} \frac{x^{2 n}} {4^n} $$


$$I_B=\sqrt{2} \int_0^1\frac{ \tanh ^{-1}\left(\sqrt{x}\right)}{\sqrt{(1-x) (x+1)}}\; dx =\sqrt{2}\; \sum _{n=0}^{\infty } \frac{\binom{2 n}{n} \sum _{k=1}^{2 n+1} \frac{1}{2 k-1}}{4^n (2 n+1)}$$

  • $\begingroup$ Perhaps there is a $_3F_2(...;\pm1)$ identity that could help us... $\endgroup$ – clathratus Mar 21 '19 at 4:06
  • $\begingroup$ @clathratus: Since $$2 \int_0^1 \frac{\sin ^{-1}(x)}{x \sqrt{1-x^2}} \, dx=2 \int_0^1 \frac{\log \left(\frac{\sqrt{1-x^2}+1}{x}\right)}{\sqrt{1-x^2}} \, dx=4G$$ where $G$ is Catalan's Constant; we then have $$2 \int_0^1 \frac{\log \left(\frac{\sqrt{1-x^2}+1}{x}\right)}{\sqrt{1\pm x^2}} \, dx=\pi \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};1,\frac{3}{2};\mp 1\right)$$ $\endgroup$ – James Arathoon Mar 21 '19 at 7:48
  • $\begingroup$ Thanks for your work. Actually I and Marco know a lot of hypergeometric series turning to be equivalent to $I_A$ or $I_B$, our main concern is that none of them seem to have a closed form in terms of standard or alternating Euler sums. My suspect is that one has to consider Euler sums over the Gaussian integers to write $I_A$ or $I_B$ in a explicit way. $\endgroup$ – Jack D'Aurizio Mar 21 '19 at 14:37
  • $\begingroup$ Numerically calculated in Mathematica it appears that $$I_B=\sqrt{2} \int_0^1 \log \left(\frac{\sqrt{1-x^2}+1}{x}\right) \left(\frac{\sqrt{x}}{2 (1-x)}+\tanh ^{-1}\left(\sqrt{x}\right)\right) \, dx$$. However swapping the order of integral and summation as above fails when applied to $I_B$. $\endgroup$ – James Arathoon Mar 21 '19 at 18:44
  • $\begingroup$ The result I get is $$\frac{\sqrt{\frac{\pi }{2}} \left(\Gamma \left(\frac{1}{4}\right)^2 \, _3F_2\left(\frac{3}{4},1,\frac{5}{4};\frac{7}{4},\frac{7}{4};1\right)+36 \Gamma \left(\frac{3}{4}\right)^2 \, _3F_2\left(\frac{1}{4},\frac{3}{4},1;\frac{5}{4},\frac{5}{4};1\right)\right)}{9 \Gamma \left(\frac{1}{4}\right) \Gamma \left(\frac{3}{4}\right)}$$ Possibly correct, but I am not sure how to simplify this further. $\endgroup$ – James Arathoon Mar 21 '19 at 18:57

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