Integrals and Miscellaneous 12

$$I:=\int_0^1\ln^2(1+u)\ln udu=-\frac{\zeta(3)}{4}-2\ln^22+\frac{\pi^2}{6}+8\ln2-6$$PROOF.
Cauchy's product gives\begin{eqnarray*}\ln^2(1+x) &=& \left(\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}x^n\right)^2\\ &=& x^2\sum_{n=0}^\infty\sum_{k=0}^n\frac{(-1)^k}{k+1}\frac{(-1)^{n-k}}{n-k+1}x^n \\&=& \sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac{(-1)^n}{n}\left(\frac{1}{k}+\frac{1}{n-k}\right)x^n\end{eqnarray*}Therefore,\begin{equation}\ln^2(1+x)=2\sum_{n=2}^\infty\frac{(-1)^nH_{n-1}}{n}x^n\tag{1}\end{equation}Using (1) we have\begin{eqnarray}I &=& 2\sum_{n=1}^\infty\frac{(-1)^nH_n}{(n+1)(n+2)^2}\tag{A}\\&=& 2\sum_{n=1}^\infty\left[\frac{(-1)^nH_n}{n+1}-\frac{(-1)^nH_n}{n+2}-\frac{(-1)^nH_n}{(n+2)^2}\right]\tag{2}\end{eqnarray}We showed in this post that\begin{equation}\sum_{n=1}^\infty \frac{H_n}{n+1}x^n=\frac{1}{2x}\ln^2(1-x)\tag{3}\end{equation}Substitute $x=-1$ to get\begin{equation}\sum_{n=1}^\infty \frac{(-1)^nH_n}{n+1}=-\frac{1}{2}\ln^22\tag{4}\end{equation}and it follows from $H_n=H_{n+1}-\frac{1}{n+1}$ that \begin{equation}\sum_{n=0}^\infty\frac{(-1)^nH_n}{n+2}=\frac{\ln^22}{2}-2\ln2+1\tag{5}\end{equation}Recall the formula:\begin{equation}\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{(n+1)^2}=\frac{1}{8}\zeta(3)\tag{6}\end{equation}which we proved here. Together with (6) and $H_n=H_{n+1}-\frac{1}{n+1}$ leads to\begin{equation}\sum_{n=0}^\infty\frac{(-1)^nH_n}{(n+2)^2}=\frac{\zeta(3)}{8}+2-\frac{\pi^2}{12}-2\ln2\tag{7}\end{equation}Applying (4)(5)(7) to (2) we obtain$$I=-\frac{\zeta(3)}{4}-2\ln^22+\frac{\pi^2}{6}+8\ln2-6$$

\begin{eqnarray*}I:&=&\int_0^1\ln^2(1+u)\ln (1-u)du\\&=&\frac{1}{2}\zeta(3)-\frac{\pi^2}{3}(\ln2-1)+\frac{4}{3}\ln^32-4\ln^22+8\ln2-6\end{eqnarray*}PROOF.
In the same manner as "2022/11/1" ( via Cauthy's product, term-wise integration, partial fraction decomposition), we get\begin{equation}I = 2\sum_{n=1}^\infty\left[\frac{(-1)^nH_n^2}{(n+1)(n+2)}+\frac{(-1)^nH_n}{(n+1)^2(n+2)}+\frac{(-1)^nH_n}{(n+1)(n+2)^2}\right]\tag{1}\end{equation}The first term is$$\sum_{n=1}^\infty \frac{(-1)^nH_{n}^2}{(n+1)(n+2)}=\frac{1}{2}\zeta(3)-\frac{\pi^2}{12}(2\ln2-1)+\frac{2\ln^3 2}{3}-2\ln^22+2\ln2-1$$proved here. The second term of (1) can be rewritten as$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{(n+1)^2(n+2)}=-\frac{\zeta(3)}{8}+\ln^22-2\ln2+1$$proved here. We obtained the third term in "2022/11/1" above, namely$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{(n+1)(n+2)^2}=-\frac{\zeta(3)}{8}-\ln^22+\frac{\pi^2}{12}+4\ln2-3$$Hence,$$I=\frac{1}{2}\zeta(3)-\frac{\pi^2}{3}(\ln2-1)+\frac{4}{3}\ln^32-4\ln^22+8\ln2-6$$

$$I:=\int_0^1\ln^2u\ln (1+u)du =\frac{3}{2}\zeta(3)+\frac{\pi^2}{6}+4\ln2-6$$PROOF.
\begin{eqnarray*}I &=& \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\int_0^1u^n\ln^2udu \\&=& -2\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)^3} \\&=& 2\sum_{n=1}^\infty\left(\frac{(-1)^n}{n+1}+\frac{(-1)^n}{(n+1)^2}+\frac{(-1)^n}{(n+1)^3}-\frac{(-1)^n}{n}\right) \\&=& 2\sum_{n=2}^\infty\left(\frac{(-1)^{n-1}}{n}-\frac{(-1)^n}{n^2}-\frac{(-1)^n}{n^3}\right)+2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n} \\&=& 2\left(\ln2-1-\Li_2(-1)-1-\Li_3(-1)-1\right)+2\ln2\end{eqnarray*}Hence,$$\int_0^1\ln^2u\ln (1+u)du =\frac{3}{2}\zeta(3)+\frac{\pi^2}{6}+4\ln2-6$$

$h_n$ denotes alternating harmonic number $1-\frac{1}{2}+\cdots+\frac{(-1)^{n-1}}{n}$.\begin{eqnarray*}\sum_{n=1}^\infty h_n x^n &=& \frac{\ln(1+x)}{1-x}\\ \sum_{n=1}^\infty \frac{h_n}{n} x^n &=& \Li_2\left(\frac{1-x}{2}\right)-\Li_2(-x)-\ln2\ln(1-x)-\frac{\pi^2}{12}+\frac{\ln^22}{2} \\\sum_{n=1}^\infty \frac{h_{2n-1}}{n} x^{2n} &=& -\ln(1+x)\ln(1-x) \\\sum_{n=1}^\infty \frac{h_{2n}}{n} x^{2n} &=& -\ln(1+x)\ln(1-x)-\frac{1}{2}\Li_2(x^2)\\ \sum_{n=1}^\infty \frac{h_{2n-1}}{2n+1} x^{2n-1} &=& \frac{1}{x^2}\Biggl[\Li_2\left(\frac{1-x}{2}\right)+\left(\frac{1}{2}-\ln2\right)\ln(1-x)\\&&\quad\quad-\frac{\ln(1+x)}{2}+\frac{\ln^22}{2}-\frac{\pi^2}{12}+\frac{\ln(1+x)\ln(1-x)}{2}\Biggr]\\&&+\frac{\mathrm{arctanh}x-\ln(1+x)+1}{x} \\ \sum_{n=1}^\infty \frac{h_{2n-1}}{(2n+1)^2} &=& -\frac{21}{16}\zeta(3)+\frac{\pi^2}{4}\ln2-\frac{\pi^2}{8}+2-\ln2\end{eqnarray*}$$\int_0^1\ln u\ln(1+u)\ln(1-u)du=\frac{21}{8}\zeta(3)-\frac{\pi^2}{2}\ln2+\frac{5}{12}\pi^2-\ln^22+4\ln2-6$$PROOF here.

$$\int_0^\infty\frac{\arctan^2x\ln^2(1+x^2)}{x^2}dx=\frac{4}{3}\pi\ln^32+\frac{2}{3}\pi^3\ln2+\frac{\pi}{2}\zeta(3)$$PROOF here.

無限級数の収束性１(コーシー、ダランベールなど)

無限級数の収束性２(クンマーなど)

無限級数の収束性3（アーベル・ディリクレ）

From @infseriesbot . $$\int_0^\frac{\pi}{2}\sin\sin xdx=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!!^2}$$PROOF.
\begin{eqnarray*}\int_0^\frac{\pi}{2}\sin\sin xdx &=& \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\int_0^\frac{\pi}{2}\sin^{2n+1}xdx \\&=& \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}\frac{(2n)!!}{(2n+1)!!} \\&=& \sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!!^2}\end{eqnarray*}

From @infseriesbot . $$\sum_{n=1}^\infty\zeta(2n)z^{2n}=\frac{1-\pi z\cot\pi z}{2}$$PROOF.
We obtained the series representation of log-gamma here, namely\begin{equation}\log\G(1+z)=-\g z+\sum_{n=2}^\infty\frac{(-1)^n}{n}\zeta(n)z^n\tag{1}\end{equation}and we also find\begin{equation}\log\G(1-z)=\g z+\sum_{n=2}^\infty\frac{1}{n}\zeta(n)z^n\tag{2}\end{equation}(1)+(2) gives $$\log\frac{\pi z}{\sin\pi z}=2\sum_{n=1}^\infty\frac{1}{2n}\zeta(2n)z^{2n}$$After differentiating both sides, we multiply $\frac{z}{2}$ to get$$\sum_{n=1}^\infty\zeta(2n)z^{2n}=\frac{1-\pi z\cot\pi z}{2}$$

From @infseriesbot . Prove that$$\sum_{n=1}^\infty\frac{2^{2n}(n-1)!^2}{(2n)!}O_n=7\zeta(3)\quad,\; O_n:=\sum_{k=1}^n\frac{1}{2k-1}$$PROOF.
\begin{equation}\sum_{n=1}^\infty\frac{2^{2n}(n-1)!^2}{(2n)!}O_n=2\sum_{n=1}^\infty\frac{(2n-2)!!}{(2n-1)!!}\frac{O_n}{n}\tag{1}\end{equation}Recall the identity deduced here, namely$$\sum_{n=1}^\infty\frac{O_n}{n}x^{2n}=\frac{1}{4}\ln^2\frac{1-x}{1+x}$$Divide both sides by $x$ , substitute $x=\cos\t$ and integrate from $0$ to $\frac{\pi}{2}$ to find$$\sum_{n=1}^\infty\frac{O_n}{n}\int_0^\frac{\pi}{2}\cos^{2n-1}\t d\t=\int_0^\frac{\pi}{2}\frac{\ln^2\tan\frac{\t}{2}}{\cos\t}d\t$$$$\therefore\quad\sum_{n=1}^\infty\frac{(2n-2)!!}{(2n-1)!!}\frac{O_n}{n}=\int_0^\frac{\pi}{2}\frac{\ln^2\tan\frac{\t}{2}}{\cos\t}d\t$$Setting $t=\tan\frac{\t}{2}$ we have$$\sum_{n=1}^\infty\frac{(2n-2)!!}{(2n-1)!!}\frac{O_n}{n}=\int_0^1\left(\frac{\ln^2t}{1+t}+\frac{\ln^2t}{1-t}\right)dt$$Integrating the defining equation of $\Li_3(x)$ and $\Li_3(-x)$ by parts gives\begin{eqnarray*}\Li_3(x)&=&\ln x\Li_2(x)+\frac{1}{2}\ln^2 x\ln(1-x)+\frac{1}{2}\int_0^x\frac{\ln^2 t}{1-t}dt\\\Li_3(-x)&=&\ln x\Li_2(-x)+\frac{1}{2}\ln^2 x\ln(1+x)-\frac{1}{2}\int_0^x\frac{\ln^2 t}{1+t}dt\end{eqnarray*}Putting $x=1$ yields$$\int_0^1\frac{\ln^2 t}{1-t}dt=2\zeta(3)\;,\;\int_0^1\frac{\ln^2 t}{1+t}dt=\frac{3}{2}\zeta(3)$$$$\therefore\quad\sum_{n=1}^\infty\frac{(2n-2)!!}{(2n-1)!!}\frac{O_n}{n}=\frac{7}{2}\zeta(3)$$Together this with (1) we finally have$$\sum_{n=1}^\infty\frac{2^{2n}(n-1)!^2}{(2n)!}O_n=7\zeta(3)$$

From @infseriesbot .$$\sum_{n=-\infty}^\infty\frac{1}{(n+a)^2(n+b)^2} = \frac{\pi^2}{(b-a)^2}\left(\frac{1}{\sin^2\pi a}+\frac{1}{\sin^2\pi b}\right)+\frac{2\pi}{(b-a)^3}(\cot\pi b-\cot\pi a)$$Proof here:

【γ9】ディガンマ関数の相反公式・倍数公式と特殊値・ゼータ関数（ガンマ関数の基礎シリーズ9）

From @infseriesbot .$$\sum_{n=0}^\infty\left(\frac{(2n-1)!!}{(2n)!!}\right)^3=\frac{\pi}{\G^4(\frac{3}{4})}$$PROOF.
\begin{eqnarray*}\sum_{n=0}^\infty \left(\frac{(2n-1)!!}{(2n)!!}\right)^3 &=& {}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{1}{2}\\1,1\end{matrix};1\right] \\&=& {}_2F_1\left[\begin{matrix}\frac{1}{4},\frac{1}{4}\\1\end{matrix};1\right]^2\quad(\mathrm{Clausen's\: formula})\\&=&\left(\frac{\sqrt{\pi}}{\G^2(\frac{3}{4})}\right)^2\quad(\mathrm{Gauss's\: formula})\\&=&\frac{\pi}{\G^4(\frac{3}{4})}\end{eqnarray*}

From @infseriesbot .$$\sum_{n=0}^\infty(-1)^n\left(\frac{(2n-1)!!}{(2n)!!}\right)^3=\frac{\pi}{\sqrt{2}\G^2(\frac{5}{8})\G^2(\frac{7}{8})}$$PROOF.
\begin{eqnarray*}\sum_{n=0}^\infty(-1)^n\left(\frac{(2n-1)!!}{(2n)!!}\right)^3 &=& {}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{1}{2}\\1,1\end{matrix};-1\right] \\&=& {}_2F_1\left[\begin{matrix}\frac{1}{4},\frac{1}{4}\\1\end{matrix};-1\right]^2\quad(\mathrm{Clausen's\: formula})\\&=&\frac{1}{\sqrt{2}}{}_2F_1\left[\begin{matrix}\frac{1}{8},\frac{3}{8}\\1\end{matrix};1\right]^2\\&&\quad\quad\quad\quad(\mathrm{Kummer's\: quadratic\: transformations})\\&=&\frac{\pi}{\sqrt{2}\G^2(\frac{5}{8})\G^2(\frac{7}{8})}\quad(\mathrm{Gauss's\: formula})\end{eqnarray*}

*Clausen's formula was deduced here.
*Kummer's quadratic transformation was deduced here.

From @infseriesbot .$$\sum_{n=0}^\infty(8n+1)\left(\frac{(4n-3)!!!!}{(4n)!!!!}\right)^4=\frac{4}{\sqrt{2\pi}\G^2(\frac{3}{4})}$$PROOF.
$$LHS={}_5F_4\left[\begin{matrix}\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{9}{8}\\\frac{1}{8},1,1,1\end{matrix};1\right]$$Recall a deduction from Dougall's formula(proof here)\begin{eqnarray*}&&{}_5F_4\left[\begin{matrix}a,1+\frac{a}{2},b,c,d,\\\frac{a}{2},1+a-b,1+a-c,1+a-d\end{matrix};1\right]\\&&\quad\quad\quad=\frac{\G(1+a-b)\G(1+a-c)\G(1+a-d)\G(1+a-b-c-d)}{\G(1+a)\G(1+a-b-c)\G(1+a-c-d)\G(1+a-b-d)}\end{eqnarray*}Substituting $a=b=c=d=\frac{1}{4}$ yields$$\sum_{n=0}^\infty(8n+1)\left(\frac{(4n-3)!!!!}{(4n)!!!!}\right)^4=\frac{4\sqrt{\pi}}{\G(\frac{1}{4})\G^3(\frac{3}{4})}$$Using the reflection formula, we obtain$$\sum_{n=0}^\infty(8n+1)\left(\frac{(4n-3)!!!!}{(4n)!!!!}\right)^4=\frac{4}{\sqrt{2\pi}\G^2(\frac{3}{4})}$$

Is there anything wrong? It doesn't match to Wolfram's result but I cannot find what the problem is.$$\sum_{n=0}^\infty\left(\frac{(2n-1)!!}{(2n)!!}\right)^3H_{2n}\overset{?}{=}\frac{\pi}{\G(\frac{3}{4})^4}\left(\frac{\pi}{2}-\ln2\right)$$PROOF.
We define $f(c):={}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{2},c+\frac{1}{2}\\1,1-c\end{matrix};1\right]$ . From Dixon's identity:$${}_3F_2\left[\begin{matrix}a,b,c\\1+a-b,1+a-c\end{matrix};1\right]=\frac{\G(1+\frac{a}{2})\G(1+\frac{a}{2}-b-c)\G(1+a-b)\G(1+a-c)}{\G(1+a)\G(1+a-b-c)\G(1+\frac{a}{2}-b)\G(1+\frac{a}{2}-c)}$$it follows that$$\sum_{n=0}^\infty\frac{(\frac{1}{2})_n^2(c+\frac{1}{2})_n}{n!^2(1-c)_n}=\frac{\G(\frac{1}{4})}{2\sqrt{\pi}\G(\frac{3}{4})}\frac{\G(\frac{1}{4}-c)\G(1-c)}{\G(\frac{1}{2}-c)\G(\frac{3}{4}-c)}$$Differentiate both sides w.r.t. $s$ to get \begin{eqnarray*}&&\sum_{n=0}^\infty\frac{(\frac{1}{2})_n^2(c+\frac{1}{2})_n}{n!^2(1-c)_n}\left[\psi\left(c+\frac{1}{2}+n\right)-\psi\left(c+\frac{1}{2}\right)-\psi\left(1-c\right)+\psi\left(1-c+n\right)\right]\\&=&\frac{\G(\frac{1}{4})}{2\sqrt{\pi}\G(\frac{3}{4})}\frac{\G(\frac{1}{4}-c)\G(1-c)}{\G(\frac{1}{2}-c)\G(\frac{3}{4}-c)}\left[\psi\left(\frac{1}{2}-c\right)+\psi\left(\frac{3}{4}-c\right)-\psi\left(1-c\right)-\psi\left(\frac{1}{4}-c\right)\right]\end{eqnarray*}Setting $c=0$ gives$$2\sum_{n=0}^\infty\frac{(\frac{1}{2})_n^3}{n!^3}H_{2n}=\frac{\G(\frac{1}{4})^2}{2\pi\G(\frac{3}{4})^2}\left[\psi\left(\frac{1}{2}\right)+\psi\left(\frac{3}{4}\right)-\psi\left(1\right)-\psi\left(\frac{1}{4}\right)\right]$$Hence,$$\sum_{n=0}^\infty\left(\frac{(2n-1)!!}{(2n)!!}\right)^3H_{2n}=\frac{\pi}{\G(\frac{3}{4})^4}\left(\frac{\pi}{2}-\ln2\right)$$

Proof of Dixon's identity:

Dixonの定理の導出（一般化超幾何級数3F2）

$${}_3F_2\left[\begin{matrix}1,1,\frac{5}{4}\\2,\frac{7}{4}\end{matrix};1\right]=3\left(\frac{\pi}{2}-\ln2\right)$$PROOF.
\begin{eqnarray*}LHS &=& \frac{3\G(\frac{3}{4})}{\sqrt{\pi}\G(\frac{1}{4})}\int_0^1t^\frac{1}{4}(1-t)^{-\frac{1}{2}}{}_2F_1\left[\begin{matrix}1,1\\2\end{matrix};1\right]dt\\&=&\frac{3\G(\frac{3}{4})}{\sqrt{\pi}\G(\frac{1}{4})}\int_0^1t^\frac{1}{4}(1-t)^{-\frac{1}{2}}\left(-\frac{\ln(1-t)}{t}\right)dt\\&=& -\frac{3\G(\frac{3}{4})}{\sqrt{\pi}\G(\frac{1}{4})}\left.\frac{dB(\frac{1}{4},q)}{dq}\right|_{q=\frac{1}{2}} \\&=& -\frac{3\G(\frac{3}{4})}{\sqrt{\pi}\G(\frac{1}{4})}B\left(\frac{1}{4},\frac{1}{2}\right)\left[\psi(1/2)-\psi(3/4)\right]\\&=&3\left(\frac{\pi}{2}-\ln2\right)\end{eqnarray*}

From @carmath2014:$$I:=\int_0^1\frac{\ln(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx=-\frac{\G(\frac{1}{4})^2}{4\sqrt{2\pi}}(\pi-\ln2)$$PROOF.
After expanding $\ln(1-x)$, we set $y=x^2$ to get\begin{eqnarray*}I&=& -\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n}\int_0^1y^{\frac{n}{2}-\frac{3}{4}}(1-y)^{-\frac{1}{2}}dy \\&=& -\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n}B\left(\frac{n}{2}+\frac{1}{4},\frac{1}{2}\right)\\&=& -\frac{\sqrt{\pi}}{2}\sum_{n=0}^\infty\frac{1}{n+1}\frac{\G(\frac{n}{2}+\frac{3}{4})}{\G(\frac{n}{2}+\frac{5}{4})}\end{eqnarray*}Separating the series into odd and even terms, \begin{eqnarray*}&=&-\frac{\sqrt{\pi}}{2}\left(\sum_{n=0}^\infty\frac{1}{2n+1}\frac{\G(n+\frac{3}{4})}{\G(n+\frac{5}{4})}+\sum_{n=0}^\infty\frac{1}{2n+2}\frac{\G(n+\frac{5}{4})}{\G(n+\frac{7}{4})}\right)\\&=& -\frac{\sqrt{\pi}}{2}\left(\frac{4\G(\frac{3}{4})}{\G(\frac{1}{4})}{}_3F_2\left[\begin{matrix}1,\frac{1}{2},\frac{3}{4}\\\frac{3}{2},\frac{5}{4}\end{matrix};1\right]+\frac{\G(\frac{1}{4})}{6\G(\frac{3}{4})}{}_3F_2\left[\begin{matrix}1,1,\frac{5}{4}\\2,\frac{7}{4}\end{matrix};1\right]\right)\end{eqnarray*}From Dixon's identity, it follows that$${}_3F_2\left[\begin{matrix}1,\frac{1}{2},\frac{3}{4}\\\frac{3}{2},\frac{5}{4}\end{matrix};1\right]=\frac{\pi \G(\frac{1}{4})^2}{16\G(\frac{3}{4})^2}$$Euler's integral transform gives\begin{eqnarray*}{}_3F_2\left[\begin{matrix}1,1,\frac{5}{4}\\2,\frac{7}{4}\end{matrix};1\right] &=& \frac{3\G(\frac{3}{4})}{\sqrt{\pi}\G(\frac{1}{4})}\int_0^1t^\frac{1}{4}(1-t)^{-\frac{1}{2}}{}_2F_1\left[\begin{matrix}1,1\\2\end{matrix};1\right]dt\\&=&\frac{3\G(\frac{3}{4})}{\sqrt{\pi}\G(\frac{1}{4})}\int_0^1t^\frac{1}{4}(1-t)^{-\frac{1}{2}}\left(-\frac{\ln(1-t)}{t}\right)dt\\&=& -\frac{3\G(\frac{3}{4})}{\sqrt{\pi}\G(\frac{1}{4})}\left.\frac{dB(\frac{1}{4},q)}{dq}\right|_{q=\frac{1}{2}} \\&=& -\frac{3\G(\frac{3}{4})}{\sqrt{\pi}\G(\frac{1}{4})}B\left(\frac{1}{4},\frac{1}{2}\right)\left[\psi(1/2)-\psi(3/4)\right]\\&=&3\left(\frac{\pi}{2}-\ln2\right)\end{eqnarray*}Hence,$$\int_0^1\frac{\ln(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx=-\frac{\sqrt{\pi}}{4}\frac{\G(\frac{1}{4})}{\G(\frac{3}{4})}(\pi-\ln2)$$Reflection formula gives$$\int_0^1\frac{\ln(1-x)}{\sqrt{x}\sqrt{1-x^2}}dx=-\frac{\G(\frac{1}{4})^2}{4\sqrt{2\pi}}(\pi-\ln2)$$

Proof of Dixon's identity:

Dixonの定理の導出（一般化超幾何級数3F2）