Integrals and Miscellaneous 4

$$I\equiv \int_0^\infty\frac{\sqrt[3]{x+1}-\sqrt[3]{x}}{\sqrt{x}}dx$$Let $x=\frac{1-t}{t}$ then,\begin{eqnarray*}I&&=\int_0^1t^{-\frac{11}{6}}(1-t)^{-\frac{1}{2}}\left[1-(1-t)^{\frac{1}{3}}\right]dt\\&&=-\int_0^1t^{-\frac{11}{6}}(1-t)^{-\frac{1}{2}}\sum_{n=1}^\infty\frac{(-\frac{1}{3})_n}{n!}t^{n}dt\\&&=-\sum_{n=1}^\infty\frac{(-\frac{1}{3})_n}{n!}B\left(n-\frac{5}{6},\frac{1}{2}\right)\\&&=-\sqrt{\pi}\sum_{n=1}^\infty\frac{(-\frac{1}{3})_n\G(n-\frac{5}{6})}{n!\G(n-\frac{1}{3})}\\&&=-\sqrt{\pi}\frac{\G(-\frac{5}{6})}{\G(-\frac{1}{3})}\sum_{n=1}^\infty\frac{(-\frac{5}{6})_n}{n!}\\&&=-\frac{2\sqrt{\pi}\G(\frac{1}{6})}{5\G(\frac{2}{3})}\sum_{n=1}^\infty\frac{(-\frac{5}{6})_n}{n!}\\&&=-\frac{2\sqrt{\pi}\G(\frac{1}{6})}{5\G(\frac{2}{3})}\left[(1-x)^\frac{5}{6}-1\right]_{x=1}\end{eqnarray*}Hence,$$\int_0^\infty\frac{\sqrt[3]{x+1}-\sqrt[3]{x}}{\sqrt{x}}dx=\frac{2\sqrt{\pi}\G(\frac{1}{6})}{5\G(\frac{2}{3})}$$

$$I\equiv \int_0^\infty\frac{x\log(1+\frac{1}{x^3})}{1+x^3}dx$$Let $t=\frac{x^3}{1+x^3}$ then,\begin{eqnarray*}I&&=\int_0^1\left(\frac{t}{1-t}\right)^{\frac{1}{3}}(1-t)\left(\log\frac{1}{t}\right)\frac{1}{3}\left(\frac{t}{1-t}\right)^{-\frac{2}{3}}\frac{dt}{(1-t)^2}\\&&=-\frac{1}{3}\int_0^1t^{-\frac{1}{3}}(1-t)^{-\frac{2}{3}}\log t dt\\&&=-\frac{1}{3}\left.\dd{B(x,y)}{x}\right|_{x=\frac{2}{3},y=\frac{1}{3}}\\&&=-\frac{1}{3}B\left(\frac{2}{3},\frac{1}{3}\right)\left[\psi\left(\frac{2}{3}\right)-\psi(1)\right]\\&&=-\frac{1}{3}\G\left(\frac{2}{3}\right)\G\left(\frac{1}{3}\right)\left[-\frac{3}{2}\log3+\frac{\pi}{2\sqrt{3}}\right]\\&&=\frac{\pi}{\sqrt{3}}\log3-\frac{\pi^2}{9}\end{eqnarray*}Hence,$$ \int_0^\infty\frac{x\log(1+\frac{1}{x^3})}{1+x^3}dx=\frac{\pi}{\sqrt{3}}\log3-\frac{\pi^2}{9}$$

About derivative of the beta function:

2021年11月29日 ∫logsin xdx 対数正弦積分その1

About $\psi(2/3)$ :

2022年2月17日 【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）

\begin{eqnarray*}\left(\left(\frac{1}{2}\right)^2\right)^{-2}&&=\left(\frac{1}{4}\right)^{-2}\\&&=\sum_{n=0}^\infty\frac{(-1)^n(2)_n}{n!}\left(-\frac{3}{4}\right)^n\\&&=\sum_{n=0}^\infty\frac{(2)_n}{n!}\left(\frac{3}{4}\right)^n\\&&=\sum_{n=0}^\infty(n+1)\left(\frac{3}{4}\right)^n\\&&=4\cdot\frac{1}{1-\frac{3}{4}}\\&&=16\end{eqnarray*}

\begin{eqnarray*}I&&=\int_0^\pi\frac{dx}{\sqrt{5}-2\cos x}\\&&=\int_0^\frac{\pi}{2}\frac{dx}{\sqrt{5}-2\cos x}+\int_0^\frac{\pi}{2}\frac{dx}{\sqrt{5}+2\cos x}\\&&=2\sqrt{5}\int_0^\frac{\pi}{2}\frac{dx}{5-4\cos^2x}\\&&=2\sqrt{5}\int_0^\infty\frac{dy}{5+y^2}\quad(y=\cot x)\\&&=2\left[\arctan x\right]_0^\infty\\&&=\pi\end{eqnarray*}

$$I=\int_0^\infty\left(1-x\arctan\frac{1}{x}\right)dx$$\begin{eqnarray*}\int\left(1-x\arctan\frac{1}{x}\right)dx&&=x-\frac{x^2}{2}\arctan\frac{1}{x}+\int\frac{x^2}{2}\frac{-\frac{1}{x^2}}{1+\frac{1}{x^2}}dx\\&&=\frac{x}{2}-\frac{x^2}{2}\arctan\frac{1}{x}+\arctan x+C\end{eqnarray*}Therefore,\begin{eqnarray*}\displaystyle\lim_{R\to\infty}\int_0^R\left(1-x\arctan\frac{1}{x}\right)dx&&=\displaystyle\lim_{R\to\infty}\left(\frac{R}{2}-\frac{R^2}{2}\arctan\frac{1}{R}+\frac{\pi}{4}\right)\\&&=\displaystyle\lim_{R\to\infty}\left(\frac{R}{2}-\frac{R^2}{2}\left(\frac{1}{R}+O(R^{-3})\right)+\frac{\pi}{4}\right)\\&&=\frac{\pi}{4}\end{eqnarray*}Hence,$$\int_0^\infty\left(1-x\arctan\frac{1}{x}\right)dx=\frac{\pi}{4}$$

\begin{eqnarray*}I&&=\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx=-2\int_0^\frac{\pi}{4}\ln\cos\t d\t\quad(x=\tan\t)\\&&=\frac{\pi}{2}\ln2-2\int_0^\frac{\pi}{4}\ln(2\cos\t)d\t\end{eqnarray*}\begin{eqnarray*}\ln(2\cos\t)&&=\ln(e^{i\t}+e^{-i\t})=i\t+\ln(1+e^{-2i\t})\\&&=i\t+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}e^{-2in\t}\end{eqnarray*}\begin{eqnarray*}\int_0^\frac{\pi}{4}\ln(2\cos\t)d\t&&=\int_0^\frac{\pi}{4}\left(i\t+\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}e^{-2in\t}\right)d\t\\&&=\frac{\pi^2}{32}i-\frac{i}{2}\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(e^{-i\frac{n\pi}{2}}-1\right)\\&&=\frac{\pi^2}{32}i-\frac{i}{2}\left[\sum_{n=0}^\infty\frac{i(-1)^n+1}{(2n+1)^2}+\sum_{n=1}^\infty\frac{(-1)^n-1}{4n^2}\right]\\&&=\frac{\pi^2}{32}i+\frac{G}{2}-\frac{i}{4}\sum_{n=0}^\infty\frac{1}{(2n+1)^2}\\&&=\frac{G}{2}\end{eqnarray*}Hence,$$\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx=\frac{\pi}{2}\ln2-G$$

$$S\equiv \sum_{n=0}^\infty\frac{1}{(3n+1)^3}=\frac{1}{27}\zeta\left(3,\frac{1}{3}\right)$$Now consider Fourier Expansion of Zeta Function:$$\zeta\left(1-s,\frac{m}{n}\right)=\frac{2\G(s)}{(2\pi n)^s}\sum_{k=1}^n\left[\cos\left(\frac{\pi s}{2}-\frac{2\pi km}{n}\right)\zeta\left(s,\frac{k}{n}\right)\right]$$Then we get\begin{equation}\zeta\left(-2,\frac{1}{3}\right)=-\frac{2\sqrt{3}}{216\pi^3}\left[\zeta\left(3,\frac{1}{3}\right)-\zeta\left(3,\frac{2}{3}\right)\right]\tag{1}\end{equation}By the way,\begin{eqnarray}\zeta\left(3,\frac{2}{3}\right)&&=27\left[\zeta(3)-S-\frac{1}{27}\zeta(3)\right]\\&&=26\zeta(3)-\zeta\left(3,\frac{1}{3}\right)\tag{2}\end{eqnarray}The relation between zeta and bernoulli function:\begin{equation}\zeta\left(-2,\frac{1}{3}\right)=-\frac{1}{3}B_3(1/3)=-\frac{1}{81}\tag{3}\end{equation} From (1)(2)(3),we obtain $$ \sum_{n=0}^\infty\frac{1}{(3n+1)^3}=\frac{13}{27}\zeta(3)+\frac{2\pi^3}{81\sqrt{3}}$$and$$ \sum_{n=0}^\infty\frac{1}{(3n+2)^3}=\frac{13}{27}\zeta(3)-\frac{2\pi^3}{81\sqrt{3}}$$

Fourier Expansion of Zeta Function:

2022年3月19日 【ζ4】フルヴィッツゼータ関数のフーリエ展開・複素積分・Critical Stripへの拡張(ゼータ関数の基礎4)

$$S\equiv\sum_{n=0}^\infty\frac{1}{(4n+1)^3}\quad,\quad S'\equiv\sum_{n=0}^\infty\frac{1}{(4n+3)^3}$$\begin{eqnarray}S+S'&&=\zeta(3)-\sum_{n=1}^\infty\frac{1}{(4n)^3}-\sum_{n=0}^\infty\frac{1}{(4n+2)^3}\\&&=\frac{63}{64}\zeta(3)-\frac{1}{8}\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\\&&=\frac{63}{64}\zeta(3)-\frac{7}{64}\zeta(3)\\&&=\frac{7}{8}\zeta(3)\tag{1}\end{eqnarray}\begin{eqnarray}S-S'&&=1+\frac{1}{5^3}+\frac{1}{9^3}+\cdots-\frac{1}{3^3}-\frac{1}{7^3}-\frac{1}{11^3}-\cdots\\&&=\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\&&=\beta(3)=\frac{\pi^3}{32}\tag{2}\end{eqnarray}From (1)(2), we get$$\sum_{n=0}^\infty\frac{1}{(4n+1)^3}=\frac{7}{16}\zeta(3)+\frac{\pi^3}{64}$$$$\sum_{n=0}^\infty\frac{1}{(4n+3)^3}=\frac{7}{16}\zeta(3)-\frac{\pi^3}{64}$$

$$S\equiv\sum_{n=0}^\infty\frac{1}{(6n+1)^3}\quad,\quad S'\equiv\sum_{n=0}^\infty\frac{1}{(6n+5)^3}$$\begin{eqnarray}S+S'&&=\zeta(3)-\frac{\zeta(3)}{216}-\sum_{n=0}^\infty\left(\frac{1}{(6n+2)^3}+\frac{1}{(6n+4)^3}+\frac{1}{(6n+3)^3}\right)\\&&=\frac{215}{216}\zeta(3)-\frac{1}{8}\left(\zeta(3)-\frac{1}{27}\zeta(3)\right)-\frac{1}{27}\left(1-\frac{1}{8}\right)\zeta(3)\\&&=\frac{91}{108}\zeta(3)\tag{1}\end{eqnarray}

Fourier Expansion of Zeta Function:$$\zeta\left(-2,\frac{1}{6}\right)=-\frac{1}{288\sqrt{3}\pi^3}\left[\zeta\left(3,\frac{1}{6}\right)-\zeta\left(3,\frac{5}{6}\right)+\zeta\left(3,\frac{1}{3}\right)-\zeta\left(3,\frac{2}{3}\right)\right]$$When calculating $\zeta(3,1/3)$ ,We obtained the relation:$$\zeta\left(3,\frac{1}{3}\right)-\zeta\left(3,\frac{2}{3}\right)=\frac{4\pi^3}{3\sqrt{3}}$$Therefore,$$\zeta\left(-2,\frac{1}{6}\right)=-\frac{1}{288\sqrt{3}\pi^3}\left[\zeta\left(3,\frac{1}{6}\right)-\zeta\left(3,\frac{5}{6}\right)\right]-\frac{1}{648}$$Using bernoulli function, $$\zeta\left(-2,\frac{1}{6}\right)=-\frac{B_3(1/6)}{3}=-\frac{1}{648}-1/72$$Since we get here $\zeta(3,1/6)-\zeta(3,5/6)=4\sqrt{3}\pi^3$, \begin{equation}S-S'=\frac{4\sqrt{3}\pi^3}{216}=\frac{\sqrt{3}}{54}\pi^3\tag{2}\end{equation}From (1)(2), finally we have$$\sum_{n=0}^\infty\frac{1}{(6n+1)^3}=\frac{91}{216}\zeta(3)+\frac{\sqrt{3}\pi^3}{108}$$$$\sum_{n=0}^\infty\frac{1}{(6n+5)^3}=\frac{91}{216}\zeta(3)-\frac{\sqrt{3}\pi^3}{108}$$

$$I\equiv\int_0^\frac{\pi}{2}\mathrm{arctanh}^2(\sin x)dx$$\begin{eqnarray*}I&&=\frac{1}{4}\int_0^\frac{\pi}{2}\ln^2\frac{1+\sin x}{1-\sin x}dx\\&&=\frac{1}{4}\int_0^\frac{\pi}{2}\ln^2\frac{1+\cos x}{1-\cos x}dx\\&&=\int_0^\frac{\pi}{2}\ln^2\left(\tan\frac{x}{2}\right)dx\\&&=2\int_0^1\frac{\ln^2t}{1+t^2}dt\quad(t=\tan\frac{x}{2})\\&&=2\int_0^1\ln^2t\sum_{n=0}^\infty(-t^2)^ndt\\&&=2\sum_{n=0}^\infty(-1)^n\int_0^1t^{2n}\ln^2tdt\end{eqnarray*}Define $I(s)\equiv\int_0^1t^sdt=(s+1)^{-1}$ and differentiate twice,then we have$$\frac{d^2I(s)}{ds^2}=\int_0^1t^s\ln^2tdt=\frac{2}{(s+1)^3}$$Now apply this equation to the above,$$I=2\sum_{n=0}^\infty(-1)^n\frac{2}{(2n+1)^3}=4\beta(3)=\frac{\pi^3}{8}$$Hence,$$\int_0^\frac{\pi}{2}\mathrm{arctanh}^2(\sin x)dx=\frac{\pi^3}{8}$$

$$I\equiv\int_0^\frac{\pi}{2}x^2\sqrt{\cos x}dx$$The beta function has the formula:$$\int_0^\frac{\pi}{2}\cos^{x-1}\t\cos y\t d\t=\frac{\pi}{2^xxB(\frac{x+y+1}{2},\frac{x-y+1}{2})}$$Define a function where we let $x=3/2$ above:$$A(y)=\int_0^\frac{\pi}{2}\cos y\t\sqrt{\cos\t}d\t=\frac{\pi\sqrt{\pi}}{4\sqrt{2}}\frac{1}{\G(\frac{5}{4}+\frac{y}{2})\G(\frac{5}{4}-\frac{y}{2})}$$Obviously,$$I=-A''(0)$$Differentiate twice\begin{multline}A''(y)=\frac{1}{4}A(y)\left[\psi\left(\frac{5}{4}+\frac{y}{2}\right)-\psi\left(\frac{5}{4}-\frac{y}{2}\right)\right]^2\\-\frac{1}{4}A(y)\left[\psi'\left(\frac{5}{4}+\frac{y}{2}\right)+\psi'\left(\frac{5}{4}-\frac{y}{2}\right)\right]\end{multline}Therefore, since $A(0)$ be evaluated easily,$$I=\frac{\pi\sqrt{2\pi}}{\G^2(\frac{1}{4})}\psi'\left(\frac{5}{4}\right)=\frac{\pi\sqrt{2\pi}}{\G^2(\frac{1}{4})}\left[\psi'\left(\frac{1}{4}\right)-16\right]$$$$\psi'\left(\frac{1}{4}\right)=\zeta\left(2,\frac{1}{4}\right)=\pi^2+8G$$Hence,$$\int_0^\frac{\pi}{2}x^2\sqrt{\cos x}dx=\frac{\pi\sqrt{2\pi}(\pi^2+8G-16)}{\G^2(\frac{1}{4})}$$

About poly-gamma function:

2022年2月19日 【γ10】ポリガンマ関数の値、極、級数表示、ゼータ関数との関係（ガンマ関数の基礎シリーズ10）

Reciprocal beta function formula:

2022年5月6日 ベータ関数の逆数の積分表示（複素積分演習）

\begin{eqnarray}I&&\equiv\int_0^1\int_0^1\frac{\arcsin xy}{xy}dxdy\\&&=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!(2n+1)}\int_0^1\int_0^1x^{2n}y^{2n}dxdy\\&&=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!(2n+1)^3}\\&&={}_4F_3\left[\begin{matrix}\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\\\frac{3}{2},\frac{3}{2},\frac{3}{2}\end{matrix};1\right]\\&&=\frac{1}{16}\left.\frac{d^2B(x,\frac{1}{2})}{dx^2}\right|_{x=1/2}\\&&=\frac{1}{16}B\left(\frac{1}{2},\frac{1}{2}\right)\left[\psi'\left(\frac{1}{2}\right)-\psi'(1)+\left\{\psi\left(\frac{1}{2}\right)-\psi(1)\right\}^2\right]\\&&=\frac{1}{16}\pi\left(\frac{\pi^2}{3}+4\ln^22\right)\\&&=\frac{\pi^3}{48}+\frac{\pi}{4}\ln^22\end{eqnarray}

2022年5月4日 一般化された超幾何関数とベータ関数の微分の関係