Integrals and Miscellaneous 11

$${}_3F_2\left[\begin{matrix}1,1,\frac{5}{2}\\\frac{3}{2},3\end{matrix};x\right]=-\frac{2}{3x}-\frac{2}{3}\left(\frac{1}{x^2}+\frac{1}{x}\right)\ln(1-x)$$PROOF.
\begin{eqnarray*}LHS &=& \frac{2}{3}\sum_{n=0}^\infty\frac{2n+3}{(n+1)(n+2)}x^n \\&=& \frac{2}{3}\sum_{n=0}^\infty\left(\frac{1}{n+1}+\frac{1}{n+2}\right)x^n \\&=& \frac{2}{3x}\sum_{n=1}^\infty\frac{x^n}{n}+\frac{2}{3x^2}\sum_{n=2}^\infty\frac{x^n}{n} \\&=& -\frac{2}{3x}\ln(1-x)+\frac{2}{3x^2}\left[-\ln(1-x)-x\right]\end{eqnarray*}Hence,$${}_3F_2\left[\begin{matrix}1,1,\frac{5}{2}\\\frac{3}{2},3\end{matrix};x\right]=-\frac{2}{3x}-\frac{2}{3}\left(\frac{1}{x^2}+\frac{1}{x}\right)\ln(1-x)$$

$${}_3F_2\left[\begin{matrix}1,1,\frac{7}{2}\\\frac{3}{2},4\end{matrix};x\right]=-\frac{1}{5x^3}\left[\frac{7}{2}x^2+3x+(3x^2+2x+3)\ln(1-x)\right]$$PROOF.
\begin{eqnarray*}LHS &=& \frac{2}{5}\sum_{n=0}^\infty\frac{(2n+3)(2n+5)}{(n+1)(n+2)(n+3)}x^n \\&=& \frac{2}{5}\sum_{n=0}^\infty\left(\frac{\frac{3}{2}}{n+1}+\frac{1}{n+2}+\frac{\frac{3}{2}}{n+3}\right)x^n \\&=& \frac{3}{5x}\sum_{n=1}^\infty\frac{x^n}{n}+\frac{2}{5x^2}\sum_{n=2}^\infty\frac{x^n}{n}+\frac{3}{5x^3}\sum_{n=3}^\infty\frac{x^n}{n} \\&=& -\frac{3}{5x}\ln(1-x)+\frac{2}{5x^2}(-\ln(1-x)-x)+\frac{3}{5x^3}\left(-\ln(1-x)-x-\frac{x^2}{2}\right)\end{eqnarray*}Hence,$${}_3F_2\left[\begin{matrix}1,1,\frac{7}{2}\\\frac{3}{2},4\end{matrix};x\right]=-\frac{1}{5x^3}\left[\frac{7}{2}x^2+3x+(3x^2+2x+3)\ln(1-x)\right]$$

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

$${}_2F_1\left[\begin{matrix}1,\frac{1}{2}\\\frac{3}{2}\end{matrix};x\right]=\frac{\mathrm{artanh}\sqrt{x}}{\sqrt{x}}$$The proof is similar to "2022/10/7 C".

$${}_2F_1\left[\begin{matrix}1,\frac{3}{2}\\\frac{5}{2}\end{matrix};x\right]=3\frac{\mathrm{artanh}\sqrt{x}-\sqrt{x}}{x\sqrt{x}}$$PROOF. We define $LHS=f(x)$ to see $$f(x)=\sum_{n=0}^\infty\frac{3}{2n+3}x^n$$$$\therefore\quad\left(x^3 f(x^2)\right)'=\frac{3x^2}{1-x^2}$$Hence, we find the formula.

\begin{eqnarray*}\int_0^a && \frac{x^{-\frac{1}{2}}}{\sqrt{1+b^2\sqrt{x(a-x)}}}\ln\left(b\sqrt[4]{x(a-x)}+\sqrt{1+b^2\sqrt{x(a-x)}}\right)dx \\&&= \frac{\pi}{b\sqrt{2}}\ln\left(1+\frac{ab^2}{2}\right)\end{eqnarray*}\begin{eqnarray*}\int_0^a && \frac{x^{\frac{1}{2}}}{\sqrt{1+b^2\sqrt{x(a-x)}}}\ln\left(b\sqrt[4]{x(a-x)}+\sqrt{1+b^2\sqrt{x(a-x)}}\right)dx \\&&=\frac{3\pi a^2b}{16\sqrt{2}}\:{}_3F_2\left[\begin{matrix}1,1,\frac{5}{2}\\\frac{3}{2},3\end{matrix};-\frac{ab^2}{2}\right] \\&&=\frac{\pi}{4\sqrt{2}b^3}\left[ab^2+(ab^2-2)\ln\left(1+\frac{ab^2}{2}\right)\right]\end{eqnarray*}\begin{eqnarray*}\int_0^a && \frac{x(a-x)^{\frac{1}{2}}}{\sqrt{1+b^2\sqrt{x(a-x)}}}\ln\left(b\sqrt[4]{x(a-x)}+\sqrt{1+b^2\sqrt{x(a-x)}}\right)dx \\&&=\frac{5\pi a^3b}{64\sqrt{2}}\:{}_3F_2\left[\begin{matrix}1,1,\frac{7}{2}\\\frac{3}{2},4\end{matrix};-\frac{ab^2}{2}\right] \\&&=\frac{\pi}{64\sqrt{2}b^5}\left[ab^2(7ab^2-12)+2(3a^2b^4-4ab^2+12)\ln\left(1+\frac{ab^2}{2}\right)\right]\end{eqnarray*}\begin{eqnarray*}\int_0^a && \frac{x^{-\frac{1}{2}}(a-x)^{-1}}{\sqrt{1+b^2\sqrt{x(a-x)}}}\ln\left(b\sqrt[4]{x(a-x)}+\sqrt{1+b^2\sqrt{x(a-x)}}\right)dx \\&&=\sqrt{2}b\pi \:{}_2F_1\left[\begin{matrix}1,\frac{1}{2}\\\frac{3}{2}\end{matrix};-\frac{ab^2}{2}\right] \\&&=\frac{2\pi}{\sqrt{a}}\arctan\left(b\sqrt{\frac{a}{2}}\right)\end{eqnarray*}

PROOF:

logを含む難しい積分７（arcsinhの利用）

$${}_2F_1\left[\begin{matrix}\frac{1}{4},\frac{3}{4}\\1\end{matrix};x^2\right]=\frac{2K\left(\sqrt{\frac{2x}{1+x}}\right)}{\pi\sqrt{1+x}}$$PROOF.
Kummer's quadratic transformation is$${}_2F_1\left[\begin{matrix}a,b\\2b\end{matrix};2z\right]=(1-z)^{-a}{}_2F_1\left[\begin{matrix}\frac{a}{2},\frac{1+a}{2}\\b+\frac{1}{2}\end{matrix};\frac{z^2}{(1-z)^2}\right]$$Setting $a=b=\frac{1}{2}$ , $\frac{z}{1-z}=x$ yields$${}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{1}{2}\\1\end{matrix};\frac{2x}{1+x}\right]=\sqrt{1+x}{}_2F_1\left[\begin{matrix}\frac{1}{4},\frac{3}{4}\\1\end{matrix};x^2\right]$$The complete elliptic integral of the first kind is$$K(k)={}_2F_1\left[\begin{matrix}\frac{1}{2},\frac{1}{2}\\1\end{matrix};k^2\right]$$Hence,$${}_2F_1\left[\begin{matrix}\frac{1}{4},\frac{3}{4}\\1\end{matrix};x^2\right]=\frac{2K\left(\sqrt{\frac{2x}{1+x}}\right)}{\pi\sqrt{1+x}}$$Especially, substituting $x=\frac{1}{3}$ yields$${}_2F_1\left[\begin{matrix}\frac{1}{4},\frac{3}{4}\\1\end{matrix};\frac{1}{9}\right]=\frac{\sqrt{3}}{\pi}K\left(\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{3\pi}}{2\G^2(\frac{3}{4})}$$The special value of $K(k)$ is derived here

Integrals and Miscellaneous 10

\begin{eqnarray*}\sum_{n=1}^\infty\frac{(2)_n(n+1)}{(\frac{5}{2})_n}\frac{H_n}{2^n} &=& 3 \\ \sum_{n=1}^\infty\frac{(4)_n(n+1)}{(\frac{7}{2})_n}\frac{H_n}{2^n} &=& \frac{20}{3} \\ \sum_{n=1}^\infty\frac{H_n}{2^n} &=& 2\ln 2 \\ \sum_{n=1}^\infty (n+1)\frac{H_n}{2^n} &=& 2+4\ln 2 \\ \sum_{n=1}^\infty\frac{H_n}{2^n} &=& 2\ln 2 \\ \sum_{n=1}^\infty (n+1)(n+4)\frac{H_n}{2^n} &=& 18+24\ln 2 \\ \sum_{n=1}^\infty\frac{(\frac{1}{2})_n(n+1)}{(\frac{7}{4})_n}\frac{H_n}{2^n} &=&\frac{3}{8}\pi-\frac{3}{4}\ln2 \\ \sum_{n=1}^\infty \binom{2n}{n}^2\frac{H_n}{32^n} &=& \frac{\G^2(\frac{1}{4})}{4\sqrt{\pi}}\left(1-\frac{4\ln2}{\pi}\right) \\ \sum_{n=1}^\infty \frac{(3n)!}{(n!)^3}\frac{H_n}{54^n} &=& \frac{\G^3(\frac{1}{3})}{2^\frac{7}{3}\pi}\left(\sqrt{3}-\frac{9\ln3}{2\pi}\right)\end{eqnarray*}PROOF here:

調和数と超幾何級数1

From @infseriesbot\begin{eqnarray}\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{1}{2n+1}&=&\frac{\pi}{2}\tag{1}\\\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^2}&=& \frac{\pi}{2}\ln2\tag{2}\\\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^3}&=&\frac{\pi}{4}\ln^22+\frac{\pi^3}{48}\tag{3} \\\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^4}&=&\frac{\pi}{8}\zeta(3)+\frac{\pi^3}{48}\ln2+\frac{\pi}{12}\ln^32\tag{4}\end{eqnarray}PROOF.
We start from \begin{equation}\arcsin x=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}\tag{5}\end{equation}Immediately we can show (1). Divide (5) by $x$ and integrate from $0$ to $1$ to get\begin{eqnarray}\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^2}&=&\int_0^1\frac{\arcsin x}{x}dx\\&=&-\int_0^\frac{\pi}{2}\ln\sin xdx \\&=&\frac{\pi}{2}\ln2\tag{6}\end{eqnarray}Hence we have (2).
Next, Divide (5) by $x$ , integrate from $0$ to $x$ , divide by $x$ again and integrate from $0$ to $1$ we find\begin{eqnarray}\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^3}&=&\int_0^1\frac{dx}{x}\int_0^x\frac{\arcsin y}{y}dy \\&=& \int_0^1\frac{\arcsin y}{y}dy\int_y^1\frac{dx}{x}\\&=& -\int_0^1\frac{\arcsin y}{y}\ln ydy \\&=& \frac{1}{2}\int_0^1\frac{\ln^2y}{\sqrt{1-y^2}}dy\\&=& \frac{1}{16}\int_0^1\frac{\ln^2x}{\sqrt{x}\sqrt{1-x}}dx\\&=&\frac{1}{16}\left.\frac{\partial^2B}{\partial x^2}(x,y)\right|_{x=y=1/2} \\&=& \frac{\pi}{16}\left[\left(\psi(1/2)-\psi(1)\right)^2+\psi'(1/2)-\psi'(1)\right]\\&=&\frac{\pi}{16}(4\ln^22+\frac{\pi^2}{3})\tag{7}\end{eqnarray}to show (3).
In the end, we prove (4) in the same manner.\begin{eqnarray}\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{1}{(2n+1)^4}&=&\int_0^1\frac{dx}{x}\int_0^x\frac{dy}{y}\int_0^y\frac{\arcsin z}{z}dz\\&=&\int_0^1\frac{\arcsin z}{z}dz\int_z^1\frac{dy}{y}\int_y^1\frac{dx}{x}\\&=&\frac{1}{2}\int_0^1\frac{\arcsin z}{z}\ln^2zdz\\&=&-\frac{1}{96}\int_0^1\frac{\ln^3x}{\sqrt{x}\sqrt{1-x}}\\&=&-\frac{1}{96}\left.\frac{\partial^3B}{\partial x^3}(x,y)\right|_{x=y=1/2} \\&=&-\frac{\pi}{96}\left(-8\ln^32-6\ln2\cdot\frac{\pi^2}{3}-12\zeta(3)\right)\end{eqnarray}Then we get (4).

\begin{eqnarray*}&&\sum_{n=1}^\infty \binom{2n}{n}^2\frac{H_n}{16^n}x^n = K\left(\sqrt{1-x}\right)+\frac{K(\sqrt{x})}{\pi}\ln\frac{x}{16(1-x)} \\&& \sum_{n=1}^\infty \frac{(3n)!}{(n!)^3}\frac{H_n}{27^n}\left(\frac{3\sqrt{3}-5}{4}\right)^n = \frac{(2+\sqrt{3})^\frac{1}{4}}{3^\frac{1}{8}}\frac{\Gamma^2(\frac{1}{4})}{(2\pi)^\frac{3}{2}}\left[\pi+\ln(\sqrt{3}-1)-\frac{\ln2}{2}-\frac{9}{4}\ln3\right]\\&& \sum_{n=1}^\infty \binom{4n}{2n}\binom{2n}{n}\frac{H_n}{64^n}x^n = \sqrt{\frac{2}{1+\sqrt{1-x}}}K\left(\sqrt{\frac{2\sqrt{1-x}}{1+\sqrt{1-x}}}\right)+\frac{\ln\frac{x}{64(1-x)}}{\pi\sqrt{1+\sqrt{x}}}K\left(\sqrt{\frac{2\sqrt{x}}{1+\sqrt{x}}}\right)\end{eqnarray*}

PROOF here:

調和数と超幾何級数2

\begin{equation}\int_0^\infty\frac{\ln x}{(x^2+4)^2}dx=\frac{\pi}{32}(\ln2-1)\end{equation}\begin{equation}\int_0^\infty\frac{x^{-b}}{x+a}dx=\frac{\pi}{a^b\sin\pi b}\end{equation}, where $a>0$ , $0<b<1$ .
PROOF:

複素積分演習（logの分岐点と切断）

From @nasya_tw .

$$I:=\int_0^\frac{\pi}{2}\frac{\ln\cos x \ln(1-\sin x)}{\sin x}dx=?$$

Substitute $t=\tan\frac{x}{2}$ to get \begin{eqnarray}I &=& \int_0^1\ln\frac{1-t^2}{1+t^2}\ln\frac{(1-t)^2}{1+t^2}\frac{dt}{t}\\ &=& 2\int_0^1\frac{\ln(1-t^2)\ln(1-t)}{t}dt-\int_0^1\frac{\ln(1-t^2)\ln(1+t^2)}{t}dt\\&&-2\int_0^1\frac{\ln(1+t^2)\ln(1-t)}{t}dt+\int_0^1\frac{\ln^2(1+t^2)}{t}dt\tag{1}\end{eqnarray}Now we evaluate above four integrals. The first is\begin{eqnarray}2\int_0^1\frac{\ln(1-t^2)\ln(1-t)}{t}dt &=& -2\sum_{n=1}^\infty\frac{1}{n}\int_0^1 t^{2n-1}\ln(1-t)dt \\&=& -2\sum_{n=1}^\infty\frac{1}{n}\left.\dd{B}{y}(x,y)\right|_{x=2n,y=1}\\&=&-2\sum_{n=1}^\infty\frac{1}{n}B(2n,1)\bigl(\psi(1)-\psi(2n+1)\bigr)\\&=&\sum_{n=1}^\infty\frac{H_{2n}}{n^2}\\&=&\frac{11}{4}\zeta(3)\tag{2}\end{eqnarray}where we used the formula related to Euler sum which we obtained in the previous post:

The second can be rewritten as\begin{eqnarray}\int_0^1\frac{\ln(1-t^2)\ln(1+t^2)}{t}dt &=& \frac{1}{2}\int_0^1\frac{\ln(1-u)\ln(1+u)}{u}du\quad(t^2=u)\\&=& -\frac{5}{16}\zeta(3)\tag{3}\end{eqnarray}where we used the formula deduced in "2022/7/10 C".

The third integral can be evaluated in the same manner as the first\begin{eqnarray}2\int_0^1\frac{\ln(1+t^2)\ln(1-t)}{t}dt &=& \sum_{n=1}^\infty(-1)^n\frac{H_{2n}}{n^2} \\&=& \frac{23}{16}\zeta(3)-\pi G\tag{4}\end{eqnarray}where we used another formula related to Euler sum which we obtained in the previous post.

In the last integral, putting $t^2=u$ gives\begin{eqnarray}\int_0^1\frac{\ln^2(1+t^2)}{t}dt &=& \frac{1}{2}\int_0^1\frac{\ln^2(1+u^2)}{u}du \\&=& -\Li_3\left(\frac{1}{2}\right)-\ln2\Li_2\left(\frac{1}{2}\right)-\frac{\ln^32}{3}+\zeta(3) \\&=& \frac{1}{8}\zeta(3)\tag{5}\end{eqnarray}where we used the formula obtained here. And the special values of polylogarithms are deduced here.

Hence, together with (1)(2)(3)(4)(5) we have$$\int_0^\frac{\pi}{2}\frac{\ln\cos x \ln(1-\sin x)}{\sin x}dx=\frac{7}{4}\zeta(3)+\pi G\approx 4.98$$

$$I(s):=\int_0^\infty\frac{dx}{1+x^s}=\frac{\pi}{s\sin\frac{\pi}{s}}\quad(s>0)$$PROOF.
Substitute $y=\frac{x^s}{1+x^s}$ to obtain\begin{eqnarray*}I(s) &=& \frac{1}{s}\int_0^1(1-y)\left[y^{\frac{1}{s}-1}(1-y)^{-\frac{1}{s}}+y^{\frac{1}{s}}(1-y)^{-\frac{1}{s}-1}\right]dy \\&=& \frac{1}{s}\int_0^1\left[y^{\frac{1}{s}-1}(1-y)^{1-\frac{1}{s}}+y^{\frac{1}{s}}(1-y)^{-\frac{1}{s}}\right]dy \\&=&\frac{1}{s}\left[B\left(\frac{1}{s},2-\frac{1}{s}\right)+B\left(\frac{1}{s}+1,1-\frac{1}{s}\right)\right]\\&=&\frac{1}{s}\left[\G\left(\frac{1}{s}\right)\G\left(2-\frac{1}{s}\right)+\G\left(\frac{1}{s}+1\right)\G\left(1-\frac{1}{s}\right)\right] \\&=& \frac{1}{s}\left[\left(1-\frac{1}{s}\right)\frac{\pi}{\sin\frac{\pi}{s}}+\frac{1}{s}\frac{\pi}{\sin\frac{\pi}{s}}\right]\end{eqnarray*}Hence,$$\int_0^\infty\frac{dx}{1+x^s}=\frac{\pi}{s\sin\frac{\pi}{s}}$$

From @SrinivasR1729 $$I:=\int_0^1\arcsin(-1)^x\cdot\arccos(-1)^xdx=\frac{i\pi\ln2}{2}$$PROOF.\begin{eqnarray*}I &=& \int_0^1\arcsin e^{i\pi x}\cdot\arccos e^{i\pi x}dx \\&=&\frac{1}{\pi} \int_0^\pi\arcsin e^{ix}\cdot\arccos e^{ix}dx\end{eqnarray*}Set $e^{ix}=z$ to rewrite this integral as$$I=\frac{1}{\pi i}\int_\G \frac{\arcsin z\arccos z}{z}dz$$where $\G$ is upper semi-circle of radius $1$ and we see the integrand holomorphic at $z=0$. Now we add the line $[-1,1]$ on the real axis in order to make the contour and find the contour integral equals to zero due to Cauchy's integral theorem. Therefore,\begin{eqnarray*}\int_\G \frac{\arcsin z\arccos z}{z}dz &=& -\int_{-1}^1\frac{\arcsin x\arccos x}{x}dx\\ &=& -\int_{-1}^1\frac{\arcsin x(\frac{\pi}{2}-\arcsin x)}{x}dx \\&=&-\frac{\pi}{2}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\t\cot\t d\t+\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\t^2\cot\t d\t\quad(x=\sin\t)\end{eqnarray*}The second integrand is odd function then \begin{eqnarray*}\int_\G \frac{\arcsin z\arccos z}{z}dz &=& -\pi\int_0^\frac{\pi}{2}\t\cot\t d\t \\&=&-\frac{\pi^2}{2}\ln2\end{eqnarray*}where we used the result obtained in "2022/5/7". Hence,$$\int_0^1\arcsin(-1)^x\cdot\arccos(-1)^xdx=\frac{i\pi\ln2}{2}$$