Evaluate $\int_{0}^{1} \frac{\ln(x)\ln\left ( \frac{1-x}{1+x} \right ) }{\sqrt{x(1-x^2)} }\text{d}x$

Question

In an attempt to solve the following hypergeometric series, which is originated from this question, I got the integral: $$ \,_4F_3\left ( \frac14,\frac14,\frac12,\frac12;1,\frac54,\frac54;1 \right ) =\sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )^2_n }{(1)_n^2}\frac{1}{\left ( 4n+1 \right )^2 }=\frac{\Gamma\left ( \frac14 \right )^4 }{64\pi} +\frac{\Gamma\left ( \frac14 \right )^2 }{4\sqrt{2}\pi^{5/2} } \int_{0}^{1} \frac{\ln(x)\ln\left ( \frac{1-x}{1+x} \right ) }{\sqrt{x(1-x^2)} }\text{d}x. $$ Differentiating Euler beta function gives $$ \int_{0}^{1} \frac{\ln\left ( x \right ) \ln\left ( 1-x^2 \right ) }{\sqrt{x\left ( 1-x^2 \right ) } } \text{d} x =\frac{1}{4} \left.\frac{\partial^2}{\partial s\partial t} \operatorname{B}\left ( s+\frac{1}{4},t+\frac12 \right ) \right|_{s=0,t=0} =\frac{\Gamma\left ( \frac14 \right )^2 }{8\sqrt{2\pi} } \left ( 16G-\pi^2-2\pi\ln(2) \right ). $$ But this does not work for our case. We could also expand the logarithm part to derive a twisted hypergeometric series(involving harmonic numbers) but eventually come out a mess. Explicitly we say $$ \int_{0}^{1} \frac{\ln(x)\ln(1-x)}{\sqrt{x\left ( 1-x^2 \right ) } } \text{d}x =\sum_{n=0}^{\infty} \frac{\left ( \frac{1}{2} \right )_n }{n!} \left ( \frac{4}{(4n+1)^2}H_{2n+\frac12} -\frac{2}{4n+1}\psi^{(1)}\left ( 2n+\frac32 \right ) \right ), $$ where $H_n=\gamma+\psi^{(0)}(n+1)$ are generalized harmonic numbers, $\psi^{(0)},\psi^{(1)}$ denote digamma and trigamma function respectively. It is not welcomed so I wonder whether a beautiful closed-form exists. And I am thankful for being given invaluable suggestions and insights.

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Personal Attitudes Towards the Existence of a Closed-form.
Closed-forms of the corresponding hypergeometric $L$-values $L(s)=\sum_{n=0}^{\infty} \frac{\left ( 1/2\right )^2_n }{(1)_n^2}\frac{1}{\left ( 4n+1 \right )^s }$ at odd integers are known. For instance $$ \sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )_n^2 }{(1)_n^2} \frac{1}{\left ( 4n+1 \right )^{5} } =\frac{\Gamma\left ( \frac14 \right )^4 }{32\pi^2} \left ( G^2+\beta(4) \right ) . $$ But the cases for even integers remain elusive, just $\zeta(2),\zeta(3)$ alike. Possibly we couldn't express it through conventional mathmatical constants.

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Update 1. By standard hypergeometric transformation, we have an equivalent expression. $$ \,_4F_3\left ( \frac{1}{4},\frac{1}{4} ,\frac{1}{2} , \frac12;1,\frac54,\frac54;1 \right ) =\frac{\Gamma\left ( \frac{1}{4} \right )^4 }{32\pi} -\frac{1}{\pi}\,_4F_3\left ( \left \{ 1 \right \}_4; \left \{ \frac32 \right \}_3;1 \right ). $$ Update 2. Making use of the following identity, $$ K\left ( \sqrt{1-x} \right ) =\sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )_n^2}{(1)_n^2} x^n\left ( -\frac12\ln(x)-2\left ( H_{2n}-H_n-\ln(2)\right ) \right ),\quad x\in(0,1], $$ one have $$ \frac{1}{8} \int_{0}^{1} \frac{K\left ( \sqrt{1-x} \right ) }{x^{3/4}} \text{d} x=\frac{\Gamma\left ( \frac14 \right )^4 }{32\pi} =\sum_{n=0}^{\infty} \frac{\left ( \frac12\right )_n^2}{(1)_n^2} \frac{1}{\left ( 4n+1 \right )^2 } -\sum_{n=0}^{\infty}\frac{\left ( \frac12\right )_n^2}{(1)_n^2} \frac{H_{2n}-H_{n}-\ln(2)}{4n+1}. $$ Note that two Fourier-Legendre expansions are given by $$\small \frac{\ln\left ( x(1-x) \right ) }{\sqrt{x(1-x)} } =-2\pi\sum_{n=0}^{\infty} \frac{\left ( \frac12\right )_n^2}{(1)_n^2}(4n+1)\left ( H_{2n}+2\ln(2) \right ) P_{2n}(2x-1),\qquad K\left ( \sqrt{x} \right ) =2\sum_{n=0}^{\infty} \frac{1}{2n+1}P_n(2x-1). $$ Therefore it's possible to prove $\sum_{n=0}^{\infty} \frac{\left ( 1/2\right )_n^2}{(1)_n^2}\frac{H_{2n}}{4n+1} = \frac{\Gamma\left ( \frac{1}{4} \right )^4 }{32\pi^2} (\pi-4\ln(2))$, and we leave the last formulation here. $$ \sum_{n=0}^{\infty} \frac{\left ( \frac12 \right )^2_n }{(1)_n^2\left ( 4n+1 \right ) } \left ( H_n+\frac{1}{4n+1} \right ) =\frac{\Gamma\left ( \frac14 \right )^4}{16\pi^2} \left ( \pi-3\ln(2) \right ). $$

Answer 1

This is not an answer but it is too long for comments. $$I=\int_{0}^{1} \frac{\log(x)\log\left ( \frac{1-x}{1+x} \right ) }{\sqrt{x(1-x^2)} }\,dx$$

Writing \begin{aligned} \log\left ( \frac{1-x}{1+x} \right )&=-2\sum_{n=0}^\infty \frac {x^{2n+1}}{2n+1}\\ J_n&=\int_0^1 x^{2n+1}\,\frac{\log(x) }{\sqrt{x(1-x^2)} }\,dx\\ J_n&=\frac{\sqrt{\pi }}{4}\,\frac{\Gamma \left(n+\frac{3}{4}\right)}{\Gamma \left(n+\frac{5}{4}\right)}\, \left(H_{n-\frac{1}{4}}-H_{n+\frac{1}{4}}\right)\\ I&=-\frac{\sqrt{\pi }}{2}\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{3}{4}\right)}{\Gamma \left(n+\frac{5}{4}\right)}\,\frac{\left(H_{n-\frac{1}{4}}-H_{n+\frac{1}{4}}\right) }{2n+1} \end{aligned} If $a_n$ is the summand $$\frac{a_{n+1}}{a_n}=1-\frac{5}{2 n}+\frac{45}{8 n^2}+O\left(\frac{1}{n^3}\right)$$ shows a slow convergence of the summation.

Now, could we use the fact that $$a_n=\frac{\sqrt{\pi }}{8 n^{5/2}}\,\Bigg(1-\frac{5}{4 n}+\frac{65}{64n^2}-\frac{165}{256n^3}+\frac{2829}{8192n^4}+ O\left(\frac{1}{n^5}\right)\Bigg)$$ and write $$I=\frac{(18402527372-4806289215 \pi ) \sqrt{\pi }\, \Gamma \left(\frac{3}{4}\right)}{6923211750\, \Gamma \left(\frac{5}{4}\right)}+\sum_{n=5}^\infty a_n$$ where will appear a linear combination of $\zeta \left(\frac{2m+1}{2}\right)$.

Using the above expansion, converted to decimals, this would give $I=\color{red}{1.156399}38$

Using the expansion to $O\left(\frac{1}{n^{25}}\right)$ would give $I=\color{red}{1.15639909485698}03$

Answer 2

Here's another series which is analogous to @ClaudeLeibovici 's series.

Note that the target integral $$ I = \int_{0}^{1} \frac{\ln(x)\ln\left ( \frac{1-x}{1+x} \right ) }{\sqrt{x(1-x^2)} }\text{d}x$$ can be calculated using $$I(k) = \int_0^1 \frac{ \ln (x) \ln (1-kx)}{\sqrt{x(1-x^2)}} dx$$ as the target $I = \int_{-1}^1 I'(k) dk$.

Now to find $I'(k)$ I considered another integral, $$J_k (\alpha) = \int_{0}^{1} \frac{x^{\alpha}}{(1-kx) \sqrt{x(1-x^2)}} dx$$ as $I'(k) = J'_k(1)$.

We begin by substituting $x = \sin \theta$, and then we use the geometric series expansion: $$J_k (\alpha) = \int_{0}^{1} \frac{x^{\alpha}}{(1-kx) \sqrt{x(1-x^2)}} dx \\ = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{\alpha - \frac{1}{2}} \theta}{1 - k \sin \theta} d\theta \\ = \int_{0}^{\frac{\pi}{2}} \sin^{\alpha - \frac{1}{2}} \theta \sum_{n \geq 0} k^n \sin^n \theta \\ = \sum_{n \geq 0} k^n \int_{0}^{\frac{\pi}{2}} \sin^{n+\alpha - \frac{1}{2}} \theta d \theta $$ Then use the recurrence derived here and it's solution in terms of the beta function to get the series: $$ J_k (\alpha) = \frac{1}{2} \sum_{n \geq 0} B \left( \frac{2n+2 \alpha -1}{4}, \frac{1}{2} \right) k^n$$ Differentiating with respect to $\alpha$ and then evaluating at $\alpha = 1$ one gets: $$ I'(k) = \frac{1}{2} \sum_{n \geq 0} \left[ B \left( \frac{2n+1}{4}, \frac{1}{2} \right) \left\{ \psi \left( \frac{2n+1}{4} \right) - \psi \left( \frac{2n+3}{4} \right) \right\} \right] $$

Now observe that because of the definition of the digamma function, one can write the difference in terms of the Lerch Transcendent as follows: $$ \begin{equation} \psi \left( \frac{2n+1}{4} \right) - \psi \left( \frac{2n+3}{4} \right) \\ = \int_{0}^{1} \frac{t^{\frac{2n-1}{4}} - t^{\frac{2n-3}{4}} }{1-t} dt = - \int_{0}^{1} \frac{t^{\frac{2n-3}{4}} (1 - \sqrt{t} )}{1-t} dt \\ = - \int_{0}^{1} \frac{ t^{ \frac{2n-3}{4} } }{1 + \sqrt{t}} dt = - \int_0^1 t^{ \frac{2n-3}{4} } \sum_{j \geq 0} (-1)^n t^{\frac{j}{2}} \\ = - \sum_{j \geq 0} \int_0^1 t^{ \frac{2n-3}{4} + \frac{j}{2} } dt = - \sum_{j \geq 0} \frac{1}{\frac{2n+1}{4} + \frac{j}{2}} \\ =-2 \Phi \left(-1, 1, \frac{2n+1}{2} \right) \end{equation} $$ Now using the definition $I = \int_{-1}^1 I'(k) dk$, we get: $$I = - \sum_{n \geq 0} \left[ \frac{1 - (-1)^n}{n+1} \right] \left[ B \left( \frac{2n+1}{4}, \frac{1}{2} \right) \Phi \left( -1, 1, \frac{2n+1}{2} \right) \right] $$ Lastly, observing that the first bracket is 0 whenever $n$ is even, we can replace $n$ with $2n+1$ and get: $$I = - \sum_{n \geq 0} \frac{1}{n+1} B \left( \frac{4n+3}{4}, \frac{1}{2} \right) \Phi \left( -1, 1, \frac{4n+3}{2} \right) $$ I am working on an asymptotic result as well, I will update this later :)