Integrals and Miscellaneous 13

From @SrinivasR1729$$I:=\int_0^1\arcsin\frac{1-x^2}{1+x^2}\arccos\frac{2x}{1+x^2}dx=2\pi\ln2-4G$$PROOF.
We substitute $x=\tan\t$ to get\begin{eqnarray*}I &=& \int_0^\frac{\pi}{4}\arcsin(\cos2\t)\arccos(\sin2\t)\frac{d\t}{\cos^2\t} \\&=& \int_0^\frac{\pi}{4}\left(\frac{\pi}{2}-2\t\right)^2\frac{d\t}{\cos^2\t} \\&=& 4\int_0^\frac{\pi}{4}\left(\frac{\pi}{2}-2\t\right)\tan\t d\t \\&=& -8\int_0^\frac{\pi}{4}\ln\cos\t d\t \\&=& 2\pi \ln2 -8\int_0^\frac{\pi}{4}\ln(2\cos \t) d\t \\&=& 2\pi \ln2 +8\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^\frac{\pi}{4}\cos 2n\t d\t \\&=& 2\pi \ln2 +4\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\sin\frac{n\pi}{2} \\&=& 2\pi \ln2 -4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2} \\&=& 2\pi\ln2-4G\end{eqnarray*}

From @intgralsbot$$I:=\int_{-\infty}^\infty \frac{\arctan{e^x}}{e^{ax}+e^{-ax}}dx=\frac{\pi^2}{8a}$$PROOF:
After setting $y=e^x$, we separate the interval into $[0,1]$ and $[1,\infty)$. Then, substitute $y\to1/y$ in latter integral.\begin{eqnarray*}I&=&\int_0^1\frac{\arctan y}{y^a+y^{-a}}\frac{dy}{y}+\int_1^\infty\frac{\arctan y}{y^a+y^{-a}}\frac{dy}{y}\\&=&\int_0^1\frac{\arctan y+\arctan\frac{1}{y}}{y^a+y^{-a}}\frac{dy}{y}\\&=&\frac{\pi}{2}\int_0^1\frac{1}{y^a+y^{-a}}\frac{dy}{y}\end{eqnarray*}We substitute $y^a=u$ to get\begin{eqnarray*} I &=& \frac{\pi}{2a}\int_0^1\frac{1}{u+u^{-1}}\frac{du}{u}\\&=&\frac{\pi^2}{8a}\end{eqnarray*}

From @SrinivasR1729$$S:=\sum_{n=1}^\infty\frac{(4n-1)!!(2n-1)!!}{(4n-2)!!(2n-2)!!}\frac{(\frac{1}{2})^n}{n}=\frac{16\G(\frac{1}{4})}{\sqrt{\sqrt{2}-1}\G(\frac{1}{8})^2}$$PROOF.
\begin{eqnarray*}(4n-1)!! = 4^{2n}(1/4)_n(3/4)_n \;&,&\; (4n-2)!!=4^{2n-1} (n-1)! (1/2)_n\\(2n-1)!!=2^n(1/2)_n \;&,&\; (2n)!!= 2^n n!\end{eqnarray*}yield that \begin{eqnarray*}S &=& 8\sum_{n=1}^\infty\frac{(\frac{1}{4})_n(\frac{3}{4})_n}{n!(n-1)!}\left(\frac{1}{2}\right)^n\\&=& \frac{3}{4}\sum_{n=0}^\infty\frac{(\frac{5}{4})_n(\frac{7}{4})_n}{(2)_nn!}\left(\frac{1}{2}\right)^n \\&=& \frac{3}{4}F\left(\begin{matrix}\frac{5}{4},\frac{7}{4}\\2\end{matrix};\frac{1}{2}\right)\end{eqnarray*}Pfaff transformation gives\begin{eqnarray*}S &=& \frac{3}{4}2^{\frac{5}{4}}F\left(\begin{matrix}\frac{5}{4},\frac{1}{4}\\2\end{matrix};-1\right) \\&=& \frac{3\cdot 2^\frac{5}{4}}{\G(\frac{1}{4})\G(\frac{3}{4})}\int_0^1t^\frac{1}{4}(1-t^2)^{-\frac{1}{4}}dt \\&=& \frac{3}{2^\frac{1}{4}\pi}\int_0^1u^{-\frac{3}{8}}(1-u)^{-\frac{1}{4}}du \\&=& \frac{4\sqrt{2}}{\pi\sqrt{\pi}}\G\left(\frac{5}{8}\right)\G\left(\frac{7}{8}\right) \\&=& \frac{8\sqrt{2}}{\sqrt{\pi}\sqrt{2-\sqrt{2}}}\frac{\G\left(\frac{5}{8}\right)}{\G\left(\frac{1}{8}\right)}\\&=&\frac{16\G(\frac{1}{4})}{\sqrt{\sqrt{2}-1}\G(\frac{1}{8})^2}\end{eqnarray*}

From @JMNVDVRN$$I:=\int_0^1\ln^2\frac{1+x}{1-x}dx=\frac{\pi^2}{3}$$PROOF.
Substitute $\frac{1-x}{1+x}=t$ gives\begin{eqnarray*}I &=& 2\int_0^1\frac{\ln^2t}{(1+t)^2}dt\\&=& 2\left[-\frac{\ln^2t}{1+t}\right]_\epsilon^1+4\int_0^1\frac{\ln t}{t(1+t)}dt \\&=& 2\frac{\ln^2\epsilon}{1+\epsilon}+4\int_\epsilon^1\left(\frac{\ln t}{t}-\frac{\ln t}{1+t}\right)dt \\&=& \frac{2\epsilon\ln^2\epsilon}{1+\epsilon}-4\int_0^1\frac{\ln t}{1+t}dt\ \\&=& 4\int_0^1\frac{\ln(1+t)}{t}dt \\&=& 4\int_0^{-1}\frac{\ln(1-t)}{t}dt \\&=& -4\Li_2(-1)\\&=&\frac{\pi^2}{3}\end{eqnarray*}

From @JMNVDVRN$$S:=\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}=\frac{7}{4}\zeta(3)-\frac{\pi^2}{4}\ln2$$PROOF.
Recall below formulae:\begin{eqnarray}\sum_{n=1}^\infty\frac{H_{2n}}{(2n+1)^2}&=&\frac{7}{16}\zeta(3)\tag{1}\\\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\tag{2}\end{eqnarray}where $H_n$ denotes harmonic numbers, $h_n$ denotes alternative harmonic numbers. We deduced (1) here, (2) here. From $h_{2n-1}=H_{2n}-H_n+\frac{1}{2n}$, it follows that$$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}=\frac{7}{4}\zeta(3)-\frac{\pi^2}{4}\ln2+\frac{\pi^2}{8}+\ln2-2+\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n(2n+1)^2}$$Now we calculate the last term of RHS\begin{eqnarray*}\sum_{n=1}^\infty\frac{1}{n(2n+1)^2} &=& \sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+\frac{1}{2}}\right)-2\sum_{n=1}^\infty\frac{1}{(2n+1)^2} \\&=& 2-2\ln2-\left(\frac{\pi^2}{4}-2\right) \\&=& 4-\frac{\pi^2}{4}-2\ln2\end{eqnarray*}Hence,$$S=\frac{7}{4}\zeta(3)-\frac{\pi^2}{4}\ln2$$

From @Ali39342137$$I:=\int_0^1\frac{x\sqrt{x}\ln x}{(1+x)(1+x^2)}dx=\frac{\psi'\left(\frac{7}{8}\right)-\psi'\left(\frac{5}{8}\right)}{16}$$PROOF.\begin{eqnarray*}I &=& \frac{1}{2}\int_0^1 x\sqrt{x}\ln x\left(\frac{1}{1+x}+\frac{1}{1+x^2}-\frac{x}{1+x^2}\right)dx \\&=& \frac{1}{2}\sum_{n=0}^\infty(-1)^n\left(\int_0^1x^{n+\frac{3}{2}}\ln xdx+\int_0^1x^{2n+\frac{3}{2}}\ln xdx-\int_0^1x^{2n+\frac{5}{2}}\ln xdx\right) \\&=& \frac{1}{2}\left[-\sum_{n=0}^\infty\frac{(-1)^n}{(n+\frac{5}{2})^2}-\sum_{n=0}^\infty\frac{(-1)^n}{(2n+\frac{5}{2})^2}+\sum_{n=0}^\infty\frac{(-1)^n}{(2n+\frac{7}{2})^2}\right] \\&=& \frac{1}{2}\left[-\sum_{n=0}^\infty\frac{1+(-1)^n}{(2n+\frac{5}{2})^2}+\sum_{n=0}^\infty\frac{1+(-1)^n}{(2n+\frac{7}{2})^2}\right] \\&=&-\sum_{n=0}^\infty\frac{1}{(4n+\frac{5}{2})^2}+\sum_{n=0}^\infty\frac{1}{(4n+\frac{7}{2})^2}\\&=& \frac{\psi'\left(\frac{7}{8}\right)-\psi'\left(\frac{5}{8}\right)}{16}\end{eqnarray*}

From @infseriesbot\begin{eqnarray*}{}_6F_5\left[\begin{matrix}1,1,1,\frac{1}{4},\frac{3}{4},\frac{13}{8}\\2,\frac{3}{2},\frac{5}{4},\frac{5}{4},\frac{5}{8}\end{matrix};1\right]=\frac{\pi^2}{20}+\frac{4}{5}G\end{eqnarray*}PROOF.
Recall a 6F5 formula:\begin{eqnarray*}&&{}_6F_5\left[\begin{matrix}1,1,a,b,c,\frac{c+3}{2}\\2,\frac{c+1}{2},c+1,c-a+2,c-b+2\end{matrix};1\right]\\&=&\frac{c(c-a+1)(c-b+1)}{(a-1)(b-1)(c+1)}\left[-\psi(c)+\psi(c-a+1)+\psi(c-b+1)-\psi(c-a-b+2)\right]\end{eqnarray*}which we obtained here. Substituting $a=1-\epsilon$ , $b=\frac{3}{4}$ , $c=\frac{1}{4}$ yields\begin{eqnarray*}&&{}_6F_5\left[\begin{matrix}1,1,1,\frac{1}{4},\frac{3}{4},\frac{13}{8}\\2,\frac{3}{2},\frac{5}{4},\frac{5}{4},\frac{5}{8}\end{matrix};1\right]\\&=&\frac{1}{10\epsilon}\left[\psi\left(\frac{1}{4}+\epsilon\right)-\psi\left(\frac{1}{4}\right)-\psi\left(\frac{1}{2}+\epsilon\right)+\psi\left(\frac{1}{2}\right)\right]\\&=&\frac{1}{10}\left[\psi'\left(\frac{1}{4}\right)-\psi'\left(\frac{1}{2}\right)\right]\quad(\mathrm{as}\:\epsilon\to0)\\&=&\frac{1}{10}\left(8G+\pi^2-\frac{\pi^2}{2}\right)\end{eqnarray*}Hence,$${}_6F_5\left[\begin{matrix}1,1,1,\frac{1}{4},\frac{3}{4},\frac{13}{8}\\2,\frac{3}{2},\frac{5}{4},\frac{5}{4},\frac{5}{8}\end{matrix};1\right]=\frac{\pi^2}{20}+\frac{4}{5}G$$

From @Ali39342137\begin{eqnarray*}\Omega &:=& \lim_{n\to\infty}\frac{1}{n^5}\left[4!+\frac{5!}{1!}+\frac{6!}{2!}+\cdots\frac{(n+4)!}{n!}\right] \\&=& \lim_{n\to\infty}\frac{1}{n^5}\sum_{k=0}^n(k+1)(k+2)(k+3)(k+4)\\&=&\frac{1}{5}\lim_{n\to\infty}\frac{1}{n^5}\sum_{k=0}^n(k+1)(k+2)(k+3)(k+4)(k+5-k)\\&=&\frac{1}{5}\lim_{n\to\infty}\frac{1}{n^5}\sum_{k=0}^n\left[(k+1)(k+2)(k+3)(k+4)(k+5)-k(k+1)(k+2)(k+3)(k+4)\right]\\&=&\frac{1}{5}\lim_{n\to\infty}\frac{1}{n^5}(n+1)(n+2)(n+3)(n+4)(n+5)\\&=&\frac{1}{5}\end{eqnarray*}

From @integralsbot$$\int_0^\infty\frac{\ln(\cos^2x)}{x^2}dx=-\pi$$PROOF.
Lobachevsky integral formula gives$$\int_0^\infty\frac{\ln(\cos^2x)}{x^2}dx=\int_0^\infty\frac{\sin^2x}{x^2}\frac{\ln(\cos^2x)}{\sin^2x}dx=\int_0^\frac{\pi}{2}\frac{\ln(\cos^2x)}{\sin^2x}dx$$Therefore,\begin{eqnarray*}\int_0^\infty\frac{\ln(\cos^2x)}{x^2}dx &=& \int_0^\frac{\pi}{2}\frac{\ln(\cos^2x)}{\sin^2x}dx\\&=&-\left[\cot x\ln\cos^2x\right]_0^\frac{\pi}{2}-2\int_0^\frac{\pi}{2}dx\\&=&-\pi\end{eqnarray*}

$$\int_0^\infty\frac{dx}{\sqrt{x^4+21x^2+112}}=\frac{\G(\frac{1}{7})\G(\frac{2}{7})\G(\frac{4}{7})}{8\sqrt{7}\pi}$$$$\int_0^\infty\frac{dx}{\sqrt{x^4+3x^2+3}}=\frac{\G^3(\frac{1}{3})}{2^\frac{7}{3}\pi}$$

複２次式の平方根を含む積分（楕円積分の応用）

From @JMNVDVRN$$S:=\sum_{n=1}^\infty\frac{4^n(n!)^2}{n^3(2n)!}=\pi^2\ln2-\frac{7}{2}\zeta(3)$$PROOF.
We easily see that $$S=\sum_{n=1}^\infty\frac{(n-1)!}{n^2(\frac{1}{2})_n}$$ and using $\frac{1}{n}=\int_0^1x^{n-1}dx$ we have\begin{eqnarray*}S &=& \sum_{n=1}^\infty\frac{(n-1)!}{(\frac{1}{2})_n}\int_0^1x^{n-1}dx\int_0^1 y^{n-1}dy \\&=& \int_0^1\int_0^1\sum_{n=1}^\infty\frac{(n-1)!}{(\frac{1}{2})_n} (xy)^{n-1}dxdy \\&=& 2\int_0^1\int_0^1\sum_{n=0}^\infty\frac{n!}{(\frac{3}{2})_n} (xy)^ndxdy \\&=& 2\int_0^1\int_0^1 F\left(\begin{matrix}1,1\\\frac{3}{2}\end{matrix};xy\right) dxdy\end{eqnarray*}We deduced in "2022/9/27" here that$${}_2F_1\left[\begin{matrix}1,1\\\frac{3}{2}\end{matrix};x\right]=\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}$$So we have\begin{eqnarray*}S &=& 2\int_0^1\left(\int_0^1 \frac{\arcsin\sqrt{xy}}{\sqrt{xy(1-xy)}}dy\right)dx \\&=& 2\int_0^1\left(\int_0^x \frac{\arcsin\sqrt{u}}{\sqrt{u(1-u)}}du\right)\frac{dx}{x} \\&=& 4\int_0^1\left(\int_0^\sqrt{x} \frac{\arcsin t}{\sqrt{1-t^2}}dt\right)\frac{dx}{x} \\&=&2\int_0^1\frac{\arcsin^2\sqrt{x}}{x}dx \\&=&2\int_0^1\frac{\arcsin^2 s}{s}ds \\&=& 4\int_0^\frac{\pi}{2}\t^2\cot\t d\t\quad(\sin\t=s) \\&=& \frac{4}{3}\int_0^\frac{\pi}{2}\frac{\t^3}{\sin^2\t}d\t\end{eqnarray*}We calculated the remaining integral here, namely$$\int_0^\frac{\pi}{2}\frac{x^3}{\sin^2 x}dx=\frac{3}{4}\pi^2\ln2-\frac{21}{8}\zeta(3)$$Hence,$$S=\pi^2\ln2-\frac{7}{2}\zeta(3)$$

From @JMNVDVRN$$S:=\sum_{n=1}^\infty\frac{4^n(n!)^2}{2n(2n+1)!}=1$$PROOF.
We easily see that $$S=\frac{1}{2}\sum_{n=1}^\infty\frac{n!}{n(\frac{3}{2})_n}$$ and using $\frac{1}{n}=\int_0^1x^{n-1}dx$ we have\begin{eqnarray*}S &=& \frac{1}{2}\int_0^1\sum_{n=1}^\infty\frac{n!}{(\frac{3}{2})_n} x^{n-1}dx \\&=& \frac{1}{3}\int_0^1 F\left(\begin{matrix}1,2\\\frac{5}{2}\end{matrix};x\right) dx\end{eqnarray*}We deduced in "2022/9/27" here that$${}_2F_1\left[\begin{matrix}1,2\\\frac{5}{2}\end{matrix};x\right]=\frac{3}{2x}\left(\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}-1\right)$$So we have\begin{eqnarray*}S &=& \frac{1}{2}\int_0^1\left(\frac{\arcsin\sqrt{x}}{\sqrt{x(1-x)}}-1\right)\frac{dx}{x} \\&=& \int_0^\frac{\pi}{2}\left(\frac{\t}{\sin^2\t}-\cot\t\right)d\t\quad(\sin\t=\sqrt{x}) \\&=& \left[-\t\cot\t\right]_0^\frac{\pi}{2}\\&=& 1\end{eqnarray*}

About Fuchsian differential equation, Riemann P-equation, regular singularity and exponent.

フックス型微分方程式と確定特異点１（基本と例題）

フックス型微分方程式と確定特異点2 (RiemannのP方程式)

リーマンのP微分方程式の指数と解の関係