Integrals and Miscellaneous 10

Nice integral from @xuefengwolaopo

$$\int_\frac{1}{4}^\frac{1}{3}\ln\frac{\G(x)}{\G(1-x)}dx=\frac{5}{144}\bigl(\g+\ln(2\pi)\bigr)-\frac{5}{24\pi^2}\zeta'(2)-\frac{\ln3}{36}$$PROOF.
Recall Kummer's formula, namely\begin{eqnarray}\log\G(x)&=&\frac{1}{2}\log2\pi+\frac{1}{2}\sum_{n=1}^\infty\frac{\cos2n\pi x}{n}\\&&+\frac{1}{\pi}\sum_{n=1}^\infty\frac{1}{n}(\g+\log2n\pi)\sin2n\pi x\tag{1}\end{eqnarray}Hence,$$\ln\frac{\G(x)}{\G(1-x)}=\frac{2}{\pi}\sum_{n=1}^\infty(\g+\ln 2n\pi)\sin2n\pi x$$Integrating from $1/4$ to $1/3$ we obtain\begin{equation}\int_\frac{1}{4}^\frac{1}{3}\ln\frac{\G(x)}{\G(1-x)}dx=\frac{1}{\pi^2}\sum_{n=1}^\infty\frac{\g+\ln2\pi+\ln n}{n^2}\left(\cos\frac{n\pi}{2}-\cos\frac{2n\pi}{3}\right)\tag{2}\end{equation}The first and second terms of (2) are$$\sum_{n=1}^\infty\frac{\g+\ln2\pi+\ln n}{n^2}\cos\frac{n\pi}{2}=-\frac{\g+\ln2\pi}{8}\zeta(2)+\frac{1}{8}\zeta'(2)$$$$\sum_{n=1}^\infty\frac{\g+\ln2\pi+\ln n}{n^2}\cos\frac{2n\pi}{3}=\left(-\frac{\g+\ln2\pi}{3}+\frac{\ln3}{6}\right)\zeta(2)+\frac{1}{3}\zeta'(2)$$Therefore,$$\int_\frac{1}{4}^\frac{1}{3}\ln\frac{\G(x)}{\G(1-x)}dx=\frac{5}{144}\bigl(\g+\ln(2\pi)\bigr)-\frac{5}{24\pi^2}\zeta'(2)-\frac{\ln3}{36}$$

$$\int_0^1\frac{x\ln(1-x)}{1+x^4}dx=\frac{\pi}{32}\ln2-\frac{G}{4}-\frac{\pi}{8}\ln(\sqrt{2}+1)$$$$\sum_{n=0}^\infty\frac{H_n}{n+2}x^n=\frac{\ln^2(1-x)}{2x^2}-\frac{1}{x}-\frac{1-x}{x^2}\ln(1-x)$$\begin{eqnarray*}\sum_{n=1}^\infty\frac{(-1)^nH_{4n}}{2n+1}=\frac{\pi}{4}\ln(\sqrt{2}+1)-\frac{\pi}{16}\ln2+\frac{\pi}{4}-\frac{\ln(\sqrt{2}+1)}{\sqrt{2}}-\frac{\pi}{2\sqrt{2}}\end{eqnarray*}

logを含む難しい積分（調和数の4倍添え字を応用）

$$\int_0^1\frac{(1-x)^{s-1}\ln(1-x)}{(1-xz)^{s+1}}dx=-\frac{1}{s(1-z)}\sum_{n=0}^\infty\frac{z^n}{s+n}$$$$\int_0^1\frac{(1-x)^{s-1}\ln(1-x)}{(1-xz)^{s+1}}dx=\frac{1}{s^2(z-1)}{}_2F_1\left[\begin{matrix} 1,s \\ s+1\end{matrix};z\right]$$$$\int_0^1\frac{(1-x)^{m-1}\ln(1-x)}{(1-xz)^{m+1}}dx=\frac{z^{-m}}{m(1-z)}\left(z+\frac{z^2}{2}+\cdots+\frac{z^{m-1}}{m-1}+\ln|1-z|\right)$$$$\int_0^1\frac{(1-x)\ln(1-x)}{(1-\frac{x}{2})^{3}}dx=2-4\ln2$$$$\int_0^1\frac{(1-x)^2\ln(1-x)}{(1+\frac{x}{2})^{4}}dx=\frac{2}{3}-\frac{16}{9}\ln\frac{3}{2}$$$$\int_0^1\frac{(1-x)^{10}\ln(1-x)}{(1-\frac{x}{10})^{12}}dx=\frac{10^{12}}{99}\left(\frac{331885624319}{3150000000000}-\ln\frac{10}{9}\right)$$$$\int_0^1\frac{\ln(1-x)}{\sqrt{1-x}(1-xz)^{\frac{3}{2}}}dx=\frac{2}{\sqrt{z}(z-1)}\ln\frac{1+\sqrt{z}}{1-\sqrt{z}}$$$$\int_0^1\frac{\ln(1-x)}{\sqrt{1-x}(1-\frac{x}{2})^{\frac{3}{2}}}dx=-8\sqrt{2}\ln(\sqrt{2}+1)$$\begin{eqnarray}\int_0^1\frac{\ln(1-x)}{(1-x)^{\frac{2}{3}}(1-xz)^{\frac{4}{3}}}dx &=& \frac{9}{z^\frac{1}{3}(z-1)}\biggl[\frac{\ln(z^\frac{2}{3}+z^\frac{1}{3}+1)}{6}-\frac{\ln(1-z^\frac{1}{3})}{3}\\&&+\frac{1}{\sqrt{3}}\left(\arctan\frac{2z^\frac{1}{3}+1}{\sqrt{3}}-\frac{\pi}{6}\right)\biggr]\end{eqnarray}$$\int_0^1\frac{\ln(1-x)}{(1-x)^{\frac{2}{3}}(1-\frac{x}{8})^{\frac{4}{3}}}dx = -\frac{144}{7}\left[\frac{\ln7}{6}+\frac{1}{\sqrt{3}}\left(\arctan\frac{2}{\sqrt{3}}-\frac{\pi}{6}\right)\right]$$

Proof:

logを含む難しい積分２（級数展開）

\begin{eqnarray*}I(s,t;a,b) &:=& \int_0^a x^{s-1}(a-x)^{t-1}\ln\bigl(1-bx(a-x)\bigr)dx \\ &=& -a^{s+t+1}bB(s+1,t+1){}_4F_3\left[\begin{matrix}1,1,s+1,t+1\\ 2,\frac{s+t+3}{2},\frac{s+t+2}{2}\end{matrix};\frac{a^2b}{4}\right]\end{eqnarray*}

\begin{eqnarray*}I(1,1;1,1) &=& \int_0^1 \ln\bigl(1-x(1-x)\bigr)dx \\ &=& \frac{\pi}{\sqrt{3}}-2\end{eqnarray*}\begin{eqnarray*}I(1,1;a,b) &=& \int_0^a \ln\bigl(1-bx(a-x)\bigr)dx \\&=& -\frac{a^{3}b}{6}{}_2F_1\left[\begin{matrix}1,1\\ \frac{5}{2}\end{matrix};\frac{a^2b}{4}\right]\\ &=& \frac{4}{\sqrt{b}}\sqrt{1-\frac{a^2b}{4}}\arcsin\frac{a\sqrt{b}}{2}-2a\end{eqnarray*}\begin{eqnarray*}I(2,1;a,b) &=& \int_0^a x\ln\bigl(1-bx(a-x)\bigr)dx \\&=& -\frac{a^{4}b}{12}{}_2F_1\left[\begin{matrix}1,1\\ \frac{5}{2}\end{matrix};\frac{a^2b}{4}\right]\\ &=& \frac{a}{2}\left(\frac{4}{\sqrt{b}}\sqrt{1-\frac{a^2b}{4}}\arcsin\frac{a\sqrt{b}}{2}-2a\right)\end{eqnarray*}\begin{eqnarray*}I(0,1;a,b) &=& \int_0^a \frac{\ln\bigl(1-bx(a-x)\bigr)}{x}dx \\&=& -\frac{a^{2}b}{2}{}_3F_2\left[\begin{matrix}1,1,1\\ 2,\frac{3}{2}\end{matrix};\frac{a^2b}{4}\right]\\ &=& -2\arcsin^2\frac{a\sqrt{b}}{2}\end{eqnarray*}\begin{eqnarray*}I(0,0;a,b) &=& \int_0^a \frac{\ln\bigl(1-bx(a-x)\bigr)}{x(a-x)}dx \\&=& -ab\:{}_3F_2\left[\begin{matrix}1,1,1\\ 2,\frac{3}{2}\end{matrix};\frac{a^2b}{4}\right]\\ &=& -\frac{4}{a}\arcsin^2\frac{a\sqrt{b}}{2}\end{eqnarray*}\begin{eqnarray*}I\left(\frac{1}{2},\frac{1}{2};a,b\right) &=& \int_0^a \frac{\ln\bigl(1-bx(a-x)\bigr)}{\sqrt{x(a-x)}}dx \\&=& -\frac{\pi a^2b}{8}{}_3F_2\left[\begin{matrix}1,1,\frac{3}{2}\\ 2,2\end{matrix};\frac{a^2b}{4}\right]\\ &=& 2\pi \ln\frac{2+\sqrt{4-a^2b}}{4}\end{eqnarray*}\begin{eqnarray*}I\left(-\frac{1}{2},-\frac{1}{2};a,b\right) &=& \int_0^a \frac{\ln\bigl(1-bx(a-x)\bigr)}{\bigl(x(a-x)\bigr)^\frac{3}{2}}dx \\&=& -b\pi\:{}_2F_1\left[\begin{matrix}1,\frac{1}{2}\\ 2\end{matrix};\frac{a^2b}{4}\right]\\ &=& -\frac{4\pi}{a^2}\left(2-\sqrt{4-a^2b}\right)\end{eqnarray*}\begin{eqnarray*}I\left(\frac{1}{2},-\frac{1}{2};a,b\right) &=& \int_0^a \frac{\ln\bigl(1-bx(a-x)\bigr)}{\sqrt{x(a-x)^3}}dx \\&=& -\frac{\pi ab}{2}{}_2F_1\left[\begin{matrix}1,\frac{1}{2}\\ 2\end{matrix};\frac{a^2b}{4}\right]\\ &=& -\frac{2\pi}{a}\left(2-\sqrt{4-a^2b}\right)\end{eqnarray*}

Proof:

logを含む難しい積分３（超幾何級数）

\begin{eqnarray*}I(a,b,c) &:=& \int_0^\infty \frac{x}{x^2+a^2}\ln\frac{x^2+2bx+c}{x^2-2bx+c}dx \\ &=& 2\pi\arctan\frac{b}{a+\sqrt{c-b^2}}\end{eqnarray*}

\begin{eqnarray*}\int_0^\infty \frac{1}{x}\ln\frac{x^2+2x+2}{x^2-2x+2}dx &=& \frac{\pi^2}{2} \\ \int_0^\infty \frac{1}{x}\ln\frac{x^2+2x+4}{x^2-2x+4}dx &=& \frac{\pi^2}{3} \\ \int_0^\infty \frac{1}{x}\ln\frac{x^2+2\sqrt{3}x+4}{x^2-2\sqrt{3}x+4}dx &=& \frac{2\pi^2}{3}\end{eqnarray*}\begin{eqnarray*}\int_0^\infty \frac{x}{x^2+1}\ln\frac{x^2+2\sqrt{3}x+7}{x^2-2\sqrt{3}x+7}dx &=& \frac{\pi^2}{3} \\\int_0^\infty \frac{x}{x^2+1}\ln\frac{x^2+2x+3}{x^2-2x+3}dx &=& \frac{\pi^2}{4} \\\int_0^\infty \frac{x}{x^2+1}\ln\frac{x^2+2(\sqrt{2}+1)x+4\sqrt{2}+7}{x^2-2(\sqrt{2}+1)x+4\sqrt{2}+7}dx &=& \frac{3\pi^2}{8} \\\int_0^\infty \frac{x}{x^2+1}\ln\frac{x^2+2(2-\sqrt{3})x+15-8\sqrt{3}}{x^2-2(2-\sqrt{3})x+15-8\sqrt{3}}dx &=& \frac{\pi^2}{12} \\\int_0^\infty \frac{x}{x^2+2}\ln\frac{x^2+4x+12}{x^2-4x+12}dx &=& \frac{\pi^2}{4}\end{eqnarray*}

Proof:

logを含む難しい積分４（複素積分演習）

$${}_2F_1\left[\begin{matrix}1,\frac{1}{2}\\2\end{matrix};x\right]=\frac{2(1-\sqrt{1-x})}{x}$$PROOF.
$$\left(x{}_2F_1\left[\begin{matrix}1,\frac{1}{2}\\2\end{matrix};x\right]\right)'=\sum_{n=0}^\infty\left(\frac{1}{2}\right)_n\frac{x^n}{n!}=\frac{1}{\sqrt{1-x}}$$Integrating both sides yields$${}_2F_1\left[\begin{matrix}1,\frac{1}{2}\\2\end{matrix};x\right]=\frac{2(1-\sqrt{1-x})}{x}$$

$${}_2F_1\left[\begin{matrix}1,1\\\frac{3}{2}\end{matrix};x\right]=\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}$$PROOF.
It follows from Euler's representation that\begin{eqnarray*}LHS &=& \frac{1}{2}\int_0^1\frac{dt}{\sqrt{1-t}(1-tx)} \\&=& \int_0^1\frac{du}{xu^2-x+1}\quad(\sqrt{1-t}=u) \\&=& \frac{1}{x}\int_0^1\frac{du}{u^2+\frac{1-x}{x}} \\&=& \frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}\end{eqnarray*}

$${}_2F_1\left[\begin{matrix}1,1\\\frac{5}{2}\end{matrix};x\right]=\frac{3}{x}\left(1-\sqrt{\frac{1-x}{x}}\arcsin\sqrt{x}\right)$$PROOF.
It follows from Euler's representation that\begin{eqnarray*}LHS &=& \frac{3}{2}\int_0^1\frac{\sqrt{1-t}}{1-tx}dt \\&=& \frac{3}{x}\int_0^1\frac{u^2du}{u^2+\frac{1-x}{x}}\quad(\sqrt{1-t}=u) \\&=& \frac{3}{x}\int_0^1\left(1-\frac{\frac{1-x}{x}}{u^2+\frac{1-x}{x}}\right)du \\&=& \frac{3}{x}\left(1-\sqrt{\frac{1-x}{x}}\arctan\sqrt{\frac{x}{1-x}}\right)\\&=&\frac{3}{x}\left(1-\sqrt{\frac{1-x}{x}}\arcsin\sqrt{x}\right)\end{eqnarray*}

$${}_2F_1\left[\begin{matrix}1,2\\\frac{5}{2}\end{matrix};x\right]=\frac{3}{2x}\left(\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}-1\right)$$PROOF.
In the same manner as above\begin{eqnarray*}LHS &=& \frac{3}{2x}\int_0^1\frac{1-u^2}{u^2+\frac{1-x}{x}}du \\&=& \frac{3}{2x}\int_0^1\left(\frac{\frac{1}{x}}{u^2+\frac{1-x}{x}}-1\right)du\\&=&\frac{3}{2x}\left[\frac{1}{\sqrt{x(1-x)}}\arctan\sqrt{\frac{x}{1-x}}-1\right]\\&=&\frac{3}{2x}\left(\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}-1\right)\end{eqnarray*}

\begin{equation}{}_3F_2\left[\begin{matrix}1,1,3\\2,\frac{7}{2}\end{matrix};x\right]=\frac{5}{4}\left[\frac{5}{3x}+\frac{1}{x^2}-\frac{\sqrt{1-x}}{x^\frac{5}{2}}(2x+1)\arcsin\sqrt{x}\right]\tag{1}\end{equation}PROOF.
Defining $LHS := g(x)$ we have\begin{eqnarray*}\left(xg(x)\right)' &=& \sum_{n=0}^\infty\frac{(3)_n}{(\frac{7}{2})_n}x^n \\&=& \frac{5}{4x}\sum_{n=1}^\infty\frac{(2)_n}{(\frac{5}{2})_n}x^n \\&=& \frac{15}{8x^2}\sum_{n=2}^\infty\frac{(1)_n}{(\frac{3}{2})_n}x^n \\&=& \frac{15}{8x^2}\left(\sum_{n=0}^\infty\frac{(1)_n}{(\frac{3}{2})_n}x^n-1-\frac{2}{3}x\right)\end{eqnarray*}Using the identity$${}_2F_1\left[\begin{matrix}1,1\\\frac{3}{2}\end{matrix};x\right]=\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}$$,which we obtained in "2022/9/27" in this page, we get$$\left(xg(x)\right)'=\frac{15}{8x^2}\left(\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}-1-\frac{2}{3}x\right)$$Let the right-hand side of (1) be f(x) and we find $(xf(x))'=(xg(x))'$. Therefore we obtain $xf(x)=xg(x)$, where the constant of integration can be calculated by $x\to +0$. Hence $f(x)=g(x)$, namely\begin{equation}{}_3F_2\left[\begin{matrix}1,1,3\\2,\frac{7}{2}\end{matrix};x\right]=\frac{5}{4}\left[\frac{5}{3x}+\frac{1}{x^2}-\frac{\sqrt{1-x}}{x^\frac{5}{2}}(2x+1)\arcsin\sqrt{x}\right]\end{equation}

$$\iiint_V zdxdydz\;,\; V=\{(x,y,z)| x^2+y^2+z^2\le 1,x^2+y^2\le x,z\ge0\}$$
Substituting $x=r\sin\t\cos\phi,y=r\sin\t\sin\phi,z=r\cos\t$ yields \begin{eqnarray*}\iiint_V zdxdydz &=& \int_0^1dr\int_0^\frac{\pi}{2}d\t\int_{-\arccos(r\sin\t)}^{\arccos(r\sin\t)}d\phi \;r^3\cos\t\sin\t \\&=& 2\int_0^1dr\cdot r^3\int_0^\frac{\pi}{2}\sin\t \arccos(r\sin\t)\cos\t d\t \\&=& 2\int_0^1dr\cdot r\int_0^r x\arccos xdx\quad(r\sin\t :=x) \\&=& \frac{1}{2}\int_0^1dr\cdot r\left[2x^2\arccos x+\arcsin x-x\sqrt{1-x^2}\right]_0^r \\&=& \int_0^1 r^3\arccos rdr+\frac{1}{2} \int_0^1 r\arcsin rdr-\frac{1}{2}\int_0^1 r^2\sqrt{1-r^2}dr \\&=& \frac{3\pi}{64}+\frac{\pi}{16}-\frac{1}{4}B\left(\frac{3}{2},\frac{3}{2}\right) \\&=& \frac{5}{64}\pi\end{eqnarray*}Hence,$$\iiint_V zdxdydz=\frac{5}{64}\pi$$

\begin{eqnarray*}\int_0^a&& x^{s+\frac{1}{2}}(a-x)^s\ln\left(1-b\sqrt{x(a-x)}\right)dx \\&&= -\frac{\sqrt{\pi}}{4^{s+1}}a^{2s+\frac{5}{2}}b\frac{\G(2s+3)}{\G(2s+\frac{7}{2})}{}_3F_2\left[\begin{matrix}1,1,2s+3\\2,2s+\frac{7}{2}\end{matrix};\frac{ab}{2}\right]\end{eqnarray*}

\begin{eqnarray*}I(-1;a,b) &=& \int_0^a \frac{\ln\left(1-b\sqrt{x(a-x)}\right)}{\sqrt{x}(a-x)}dx \\ &=& -2\sqrt{a}b\:{}_3F_2\left[\begin{matrix}1,1,1\\2,\frac{3}{2}\end{matrix};\frac{ab}{2}\right] \\&=& -\frac{4}{\sqrt{a}}\arcsin^2\sqrt{\frac{ab}{2}}\end{eqnarray*}\begin{eqnarray*}I(0;a,b) &=& \int_0^a \sqrt{x}\ln\left(1-b\sqrt{x(a-x)}\right)dx \\&=&-\frac{4}{15}a^\frac{5}{2}b{}_3F_2\left[\begin{matrix}1,1,3\\2,\frac{7}{2}\end{matrix};\frac{ab}{2}\right] \\&=& \frac{4}{3b^\frac{3}{2}}(ab+1)\sqrt{2-ab}\arcsin\sqrt{\frac{ab}{2}}-\frac{2\sqrt{a}}{9b}(5ab+6)\end{eqnarray*}\begin{eqnarray*}I\left(-\frac{5}{4};a,b\right) &=& \int_0^a x^{-\frac{3}{4}}(a-x)^{-\frac{5}{4}}\ln\left(1-b\sqrt{x(a-x)}\right)dx \\&=& -\sqrt{2}\pi b\:{}_2F_1\left[\begin{matrix}1,\frac{1}{2}\\2\end{matrix};\frac{ab}{2}\right]\\&=& \frac{4\pi}{a}\left(\sqrt{2-ab}-\sqrt{2}\right)\end{eqnarray*}\begin{eqnarray*}I\left(-\frac{3}{4};a,b\right) &=& \int_0^a x^{-\frac{1}{4}}(a-x)^{-\frac{3}{4}}\ln\left(1-b\sqrt{x(a-x)}\right)dx \\&=& -\frac{\pi ab}{2\sqrt{2}}{}_3F_2\left[\begin{matrix}1,1,\frac{3}{2}\\2,2\end{matrix};\frac{ab}{2}\right]\\&=& 2\sqrt{2}\pi\ln\frac{1+\sqrt{1-\frac{ab}{2}}}{2}\end{eqnarray*}\begin{eqnarray*}I\left(-\frac{1}{4};a,b\right) &=& \int_0^a x^{\frac{1}{4}}(a-x)^{-\frac{1}{4}}\ln\left(1-b\sqrt{x(a-x)}\right)dx \\&=& -\frac{3\pi a^2b}{16\sqrt{2}}{}_3F_2\left[\begin{matrix}1,1,\frac{5}{2}\\2,3\end{matrix};\frac{ab}{2}\right]\\&=&\frac{\pi}{\sqrt{2}b}\biggl[ab\ln\left(1+\sqrt{1-\frac{ab}{2}}\right)+2\sqrt{1-\frac{ab}{2}}+\frac{ab}{2}(1-2\ln2)-2\biggr]\end{eqnarray*}$$\int_0^a x^{-\frac{1}{2}}\ln\left(1-b\sqrt{x(a-x)}\right)dx = \frac{4}{\sqrt{b}}\left(\sqrt{2-ab}\arcsin\sqrt{\frac{ab}{2}}-\sqrt{ab}\right)$$Proof here:

logを含む難しい積分５（超幾何級数3F2）

$${}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{3}{2}\\2\end{matrix};x\right]=\frac{4}{\pi x}\left[K\left(\sqrt{x}\right)-E\left(\sqrt{x}\right)\right]$$PROOF.
\begin{eqnarray*}LHS &=& \sum_{n=0}^\infty\frac{(\frac{1}{2})_n(\frac{3}{2})_n}{(2)_n}\frac{x^n}{n!} \\&=& \frac{2}{x}\sum_{n=0}^\infty\frac{(\frac{1}{2})_{n+1}(\frac{1}{2})_{n+1}}{(n+1)!(n+1)!}\frac{2n+2}{2n+1}x^{n+1} \\&=& \frac{2}{x}\sum_{n=1}^\infty\frac{(\frac{1}{2})_{n}(\frac{1}{2})_{n}}{n!n!}\frac{2n}{2n-1}x^{n} \\&=& \frac{2}{x}\left(\sum_{n=0}^\infty\frac{(\frac{1}{2})_{n}(\frac{1}{2})_{n}}{n!n!}x^{n}-\sum_{n=0}^\infty\frac{(\frac{1}{2})_{n}(\frac{1}{2})_{n}}{n!n!}\frac{1}{1-2n}x^{n}\right) \\&=& \frac{2}{x}\left[\sum_{n=0}^\infty\left(\frac{(2n-1)!!}{(2n)!!}\right)^2(\sqrt{x})^{2n}-\sum_{n=0}^\infty\left(\frac{(2n-1)!!}{(2n)!!}\right)^2\frac{(\sqrt{x})^{2n}}{1-2n}\right] \\&=& \frac{2}{x}\left[\frac{2}{\pi}K\left(\sqrt{x}\right)-\frac{2}{\pi}E\left(\sqrt{x}\right)\right]\end{eqnarray*}Hence,$${}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{3}{2}\\2\end{matrix};x\right]=\frac{4}{\pi x}\left[K\left(\sqrt{x}\right)-E\left(\sqrt{x}\right)\right]$$

$${}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{1}{2}\\2\end{matrix};-x\right]=\frac{4\sqrt{1+x}}{\pi x}\left[K\left(\sqrt{\frac{x}{1+x}}\right)-E\left(\sqrt{\frac{x}{1+x}}\right)\right]$$PROOF.
\begin{eqnarray*}LHS &=& \frac{2}{\pi}\int_0^1t^{-\frac{1}{2}}(1-t)^\frac{1}{2}(1+tx)^{-\frac{1}{2}}dt \\&=& \frac{2}{\pi\sqrt{1+x}}\int_0^1 u^{\frac{1}{2}}(1-u)^{-\frac{1}{2}}\left(1-\frac{x}{1+x}u\right)^{-\frac{1}{2}}du\quad(t=1-u) \\&=& \frac{1}{\sqrt{1+x}}{}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{3}{2}\\2\end{matrix};\frac{x}{1+x}\right]\end{eqnarray*}(which is equivalent to Pfaff Transform.)
We showed in "2022/10/2" that $${}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{3}{2}\\2\end{matrix};x\right]=\frac{4}{\pi x}\left[K\left(\sqrt{x}\right)-E\left(\sqrt{x}\right)\right]$$Hence,$${}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{1}{2}\\2\end{matrix};-x\right]=\frac{4\sqrt{1+x}}{\pi x}\left[K\left(\sqrt{\frac{x}{1+x}}\right)-E\left(\sqrt{\frac{x}{1+x}}\right)\right]$$

$${}_3F_2\left[\begin{matrix}1,\frac{1}{2},\frac{1}{2}\\\frac{3}{2},\frac{3}{2}\end{matrix};-x^2\right]=-\frac{i}{2x}\left[\Li_2\left(ix\right)-\Li_2\left(-ix\right)\right]$$PROOF.
\begin{eqnarray*}LHS &=& \sum_{n=0}^\infty\frac{(\frac{1}{2})_n(\frac{1}{2})_n}{(\frac{3}{2})_n(\frac{3}{2})_n}(-x^2)^n \\&=& \sum_{n=0}^\infty\frac{1}{(2n+1)^2}(ix)^{2n}\\&=&-\frac{i}{x}\sum_{n=0}^\infty\frac{(ix)^{2n+1}}{(2n+1)^2} \\&=& -\frac{i}{2x}\left[\sum_{n=1}^\infty\frac{(ix)^{n}}{n^2}-\sum_{n=1}^\infty\frac{(-ix)^{n}}{n^2}\right] \\&=& -\frac{i}{2x}\left[\Li_2(ix)-\Li_2(-ix)\right]\end{eqnarray*}Or $$=-\frac{i}{x}\chi_2(ix)$$

$${}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{1}{2}\\\frac{3}{2}\end{matrix};-x\right]=\frac{\mathrm{arcsinh}\sqrt{x}}{\sqrt{x}}$$PROOF.
\begin{eqnarray*}LHS &=& \sum_{n=0}^\infty\frac{(\frac{1}{2})_n(\frac{1}{2})_n}{(\frac{3}{2})_n}\frac{(-x)^n}{n!} \\&=& \frac{1}{\sqrt{x}}\sum_{n=0}^\infty(-1)^n\frac{(2n-1)!!}{(2n)!!}\frac{\left(\sqrt{x}\right)^{2n+1}}{2n+1}\\&=&\frac{\mathrm{arcsinh}\sqrt{x}}{\sqrt{x}}\end{eqnarray*}

$${}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{5}{2}\\\frac{3}{2},3\end{matrix};-1\right]=\frac{16\sqrt{2}}{27\pi\sqrt{\pi}}\left[\frac{5\G^2(\frac{1}{4})}{8}-\frac{3\pi^2}{\G^2(\frac{1}{4})}\right]$$PROOF.
\begin{eqnarray*}LHS &=& \frac{8}{3\pi}\int_0^1 t^\frac{3}{2}(1-t)^{-\frac{1}{2}} {}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{1}{2}\\\frac{3}{2}\end{matrix};-t\right]dt\end{eqnarray*}We obtained that$${}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{1}{2}\\\frac{3}{2}\end{matrix};-x\right]=\frac{\mathrm{arcsinh}\sqrt{x}}{\sqrt{x}}$$ in "2022/10/5 A". Therefore,\begin{eqnarray}LHS &=& \frac{8}{3\pi}\int_0^1 t(1-t)^{-\frac{1}{2}} \mathrm{arcsinh}\sqrt{t}dt \\&=& \frac{16}{3\pi}\int_0^1 (1-u^2)\mathrm{arcsinh}\sqrt{1-u^2}du\quad(u=\sqrt{1-t}) \\&=& \frac{16}{3\pi}\int_0^1 \frac{(u^2-\frac{u^4}{3})du}{\sqrt{(1-u^2)(2-u^2)}}\quad(\because IBP) \\&=& \frac{8\sqrt{2}}{3\pi}\int_0^1 \frac{(u^2-\frac{u^4}{3})du}{\sqrt{(1-u^2)(1-\frac{u^2}{2})}}\tag{1}\end{eqnarray}From the definitions of $K(k)$ , $E(k)$, we have\begin{eqnarray}K\left(\frac{1}{\sqrt{2}}\right)-E\left(\frac{1}{\sqrt{2}}\right) &=& \int_0^1 \frac{1-(1-\frac{u^2}{2})}{\sqrt{(1-u^2)(1-\frac{u^2}{2})}}du \\&=& \frac{1}{2}\int_0^1 \frac{u^2}{\sqrt{(1-u^2)(1-\frac{u^2}{2})}}du\tag{2} \end{eqnarray}Now we consider$$\frac{d}{dx}\left(x\sqrt{(1-x^2)(1-\frac{x^2}{2})}\right)=\frac{3x^4-6x^2+2}{2\sqrt{(1-x^2)(1-\frac{x^2}{2})}}$$Integrating this and (2) yields\begin{equation}\int_0^1 \frac{u^4}{\sqrt{(1-u^2)(1-\frac{u^2}{2})}}du=\frac{10}{3}K\left(\frac{1}{\sqrt{2}}\right)-4E\left(\frac{1}{\sqrt{2}}\right)\tag{3}\end{equation}Hence, (1) equals $$\frac{16\sqrt{2}}{27\pi}\left[4K\left(\frac{1}{\sqrt{2}}\right)-3E\left(\frac{1}{\sqrt{2}}\right)\right]$$We calculate these elliptic integrals:\begin{eqnarray*}K\left(\frac{1}{\sqrt{2}}\right) &=& \int_0^1\frac{dx}{\sqrt{(1-x^2)(1-\frac{x^2}{2})}} \\&=& \int_0^\frac{\pi}{2}\frac{d\t}{1-\frac{\sin^2\t}{2}} \\&=& \int_0^\infty\frac{dt}{\sqrt{1+t^4}}\quad(t=\tan\frac{\t}{2}) \\&=& \frac{1}{4}\int_0^1 x^{-\frac{3}{4}}(1-x)^{-\frac{3}{4}}dx\quad(x=\frac{t^4}{1+t^4}) \\&=& \frac{\G^2(\frac{1}{4})}{4\sqrt{\pi}}\end{eqnarray*}\begin{eqnarray*}E\left(\frac{1}{\sqrt{2}}\right) &=& \int_0^1\sqrt{\frac{1-\frac{x^2}{2}}{1-x^2}}dx \\&=& \frac{1}{\sqrt{2}}\int_0^1\sqrt{\frac{1+u^2}{1-u^2}}du\quad(u=\sqrt{1-x^2}) \\&=& \frac{1}{\sqrt{2}}\int_0^1\frac{du}{\sqrt{1-u^4}}+\frac{1}{\sqrt{2}}\int_0^1\frac{u^2du}{\sqrt{1-u^4}} \\&=& \frac{1}{4\sqrt{2}}\int_0^1 s^{-\frac{3}{4}}(1-s)^{-\frac{1}{2}}ds+\frac{1}{4\sqrt{2}}\int_0^1 s^{-\frac{1}{4}}(1-s)^{-\frac{1}{2}}ds\quad(u^4=s) \\&=& \frac{\G^2(\frac{1}{4})}{8\sqrt{\pi}}+\frac{\pi\sqrt{\pi}}{\G^2(\frac{1}{4})}\end{eqnarray*}Therefore,$${}_3F_2\left[\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{5}{2}\\\frac{3}{2},3\end{matrix};-1\right]=\frac{16\sqrt{2}}{27\pi\sqrt{\pi}}\left[\frac{5\G^2(\frac{1}{4})}{8}-\frac{3\pi^2}{\G^2(\frac{1}{4})}\right]$$

$$ \int_0^a x^{s-1}(a-x)^{t-1}\ln\left(b\sqrt{x(a-x)}+\sqrt{1+b^2x(a-x)}\right)dx = a^{s+t}b\:B\left(s+\frac{1}{2},t+\frac{1}{2}\right)\:{}_4F_3\left[\begin{matrix}\frac{1}{2},\frac{1}{2},s+\frac{1}{2},t+\frac{1}{2}\\\frac{3}{2},\frac{s+t}{2}+1,\frac{s+t+1}{2}\end{matrix};-\frac{a^2b^2}{4}\right]$$

logを含む難しい積分６（4F3・楕円積分・二重対数関数）