Series related to $\,_{2p+1}F_{2p}\left(\left\{\frac{a}{2}\right\}_{2p-1},a,\frac12;a+\frac{1}{2},\left\{1+\frac{a}{2}\right\}_{2p-1};1\right)$

Question

Similar things have been discussed here. As an example, let $p=2,a=1/3$ to have $$ {}_5F_4\left ( \frac16,\frac16,\frac16,\frac13,\frac12; \frac56,\frac76,\frac76,\frac76;1 \right )=\sum_{n=0}^{\infty} \frac{\left ( \frac13 \right )_n\left ( \frac12 \right )_n }{ \left ( \frac56 \right )_n (1)_n} \frac{1}{(6n+1)^3}= =\frac{\Gamma\left ( \frac13 \right )^3\left ( 9\psi^{(1)}\left ( \frac13 \right ) -4\pi^2 \right ) }{ 144\pi\sqrt[3]{2}\sqrt{3} }. $$ The manipulation is essentially consonant with what had done in the mentioned question. We have $$ \int_{0}^{1} \left ( x^{s-1} +x^{a-s-1}\right ) {}_2F_1\left ( a,\frac12;a+\frac{1}{2};x \right )\text{d}x =\cot(\pi s)\frac{\sqrt{\pi}\,\Gamma\left ( a+\frac12 \right ) \Gamma(a-s)\Gamma\left ( \frac12-s \right ) }{ \Gamma(a)\Gamma(1-s)\Gamma\left ( a+\frac12-s \right ) } $$ whenever the integral converges(actually we can add $x^n$ terms inside to make it converge). By differentiating both sides, one can evaluate $$ \int_{0}^{1} x^{\frac{a}2-1}\ln(x)^{2p-2} {}_2F_1\left ( a,\frac12;a+\frac{1}{2};x \right )\text{d}x, $$ which turns out to be hypergeometric series. So is the computation. Besides, a challenging version seems to be essentially the same. But I couldn't come over $$ \frac{\mathrm{d} }{\mathrm{d} s}\Bigg|_{s=1/3}\int_{0}^{1} x^{s-1} \,_2F_1\left ( \frac23,\frac23;\frac43;1-x\right )\text{d}x $$ appeared in $$ \,_4F_3\left ( \frac13,\frac13,\frac23,\frac23; 1,\frac43,\frac43;1 \right ) =\frac{\Gamma\left ( \frac13 \right )^6 }{12\pi^2} + \frac{2\pi}{3\sqrt{3}\,\Gamma\left ( \frac13 \right )^3} \int_{0}^{1} \frac{\ln(x)}{x^{2/3}} \,_2F_1\left ( \frac23,\frac23;\frac43;1-x \right )\text{d}x. $$ Intuitively believe I that they have to do with($G$ denotes Catalan's constant) \begin{aligned} &\int_{0}^{1} \frac{\ln(k)^2}{\sqrt{1-k^2} }K(k)\text{d}k =\frac{\left ( 48G-\pi^2 \right )\Gamma\left ( \frac14 \right )^4 }{576\pi}\\ &\int_{0}^{1} \frac{\ln(1-k^2)^2}{\sqrt{1-k^2} }K(k)\text{d}k =\frac{\left ( 48G+5\pi^2 \right )\Gamma\left ( \frac14 \right )^4 }{144\pi}\\ &\int_{0}^{1} \frac{\ln(k)\ln(1-k^2)}{\sqrt{1-k^2} }K(k)\text{d}k =\frac{\left ( 12G-\pi^2 \right )\Gamma\left ( \frac14 \right )^4 }{144\pi}. \end{aligned} Question: Do the related ones could be shown in different techniques? Can we produce more closed-forms of these hypergeometric series?

Answer 1

We can have $$ \,_4F_3\left ( \frac14,\frac14,\frac12,\frac12;1,\frac54,\frac54;1 \right ) =\frac{\Gamma\left ( \frac14 \right )^4 }{64\pi} +\frac{\Gamma\left ( \frac14 \right )^2 }{4\sqrt{2}\pi^{5/2} } \int_{0}^{1} \frac{\ln(x)\ln\left ( \frac{1-x}{1+x} \right ) }{\sqrt{x(1-x^2)} }\text{d}x. $$