Integrals and Miscellaneous 15

\begin{eqnarray}I&:=&\int_0^1\frac{\ln^nx\Li_{n+1}(-x)}{1+x^2}dx\\&=&\frac{(-1)^{n-1}n!}{2^{n+1}}\left(1-\frac{1}{2^n}\right)\zeta(n+1)\beta(n+1)\end{eqnarray}\begin{eqnarray}J&:=&\int_0^1\frac{x\ln^nx\Li_{n+1}(x)}{1+x^2}dx\\&=&\frac{(-1)^nn!}{2}\left[\zeta(2n+2)-\b^2(n+1)-\frac{1}{2^{2n+2}}\left(1-\frac{1}{2^n}\right)^2\zeta^2(n+1)\right]\end{eqnarray}\begin{eqnarray*}\int_0^1\frac{\ln x\Li_{2}(-x)}{1+x^2}dx&=&\frac{\pi^2 G}{48}\\\int_0^1\frac{\ln^2 x\Li_{3}(-x)}{1+x^2}dx&=& -\frac{3\pi^3}{512}\zeta(3)\\\int_0^1\frac{x\ln x\Li_{2}(x)}{1+x^2}dx&=&\frac{G^2}{2}-\frac{123}{256}\zeta(4)\end{eqnarray*}$$\int_0^1\frac{x\ln^nx}{1+x^2}dx=\frac{(-1)^{n-1}n!}{2^{n+1}}\left(\frac{1}{2^{n}}-1\right)\zeta(n+1)$$$$\int_0^1\frac{\ln^nx}{1+x^2}dx=(-1)^n n!\b(n+1)$$

Proof:

ポリログを含む積分2（重積分・級数展開）

From @integralsbot\begin{eqnarray*}\int_0^1\frac{dx}{(2-x)\sqrt[5]{x^2(1-x)^3}} &=& \frac{1}{2}\G\left(\frac{2}{5}\right)\G\left(\frac{3}{5}\right) F\left[\begin{matrix}\frac{3}{5},1\\1\end{matrix};\frac{1}{2}\right]\\&=&\frac{\pi}{2\sin\frac{2}{5}\pi}\sum_{n=0}^\infty\frac{(\frac{3}{5})_n}{n!}\frac{1}{2^n}\\&=&\frac{\pi}{2\sin\frac{2}{5}\pi}\left(1-\frac{1}{2}\right)^{-\frac{3}{5}}\\&=&\frac{2^\frac{11}{10}\pi}{\sqrt{5+\sqrt{5}}}\end{eqnarray*}

$${}_3F_2\left[\begin{matrix}a,b,c\\1+a-b,1+a-c\end{matrix};1\right]=\frac{\G(1+\frac{a}{2})\G(1+a-b)\G(1+a-c)\G(1+\frac{a}{2}-b-c)}{\G(1+a)\G(1+\frac{a}{2}-b)\G(1+\frac{a}{2}-c)\G(1+a-b-c)}$$

Proof

Dixonの定理の導出（一般化超幾何級数3F2）

Dixonの定理の導出2・オリジナル論文より（未完）

$${}_3F_2\left[\begin{matrix}a,b,c\\\frac{a+b+1}{2},2c\end{matrix};1\right]=\frac{\sqrt{\pi}\G(\frac{a+b+1}{2})\G(c+\frac{1}{2})\G(\frac{1-a-b}{2}+c)}{\G(\frac{a+1}{2})\G(\frac{b+1}{2})\G(\frac{1-a}{2}+c)\G(\frac{1-b}{2}+c)}$$

Watsonの定理（一般化超幾何級数3F2）

\begin{equation}{}_3F_2\left[\begin{matrix}a,1-a,c\\e,1+2c-e\end{matrix};1\right]=\frac{2^{1-2c}\pi\G(e)\G(1+2c-e)}{\G(\frac{a+e}{2})\G(\frac{1-a+e}{2})\G(\frac{1+a-e}{2}+c)\G(1+c-\frac{a+e}{2})}\end{equation}

Whippleの和定理（一般化超幾何級数3F2）

\begin{eqnarray*}&&{}_{s+4}F_{s+3}\left[\begin{matrix}a,&b,&c,&a_1,&a_2,\cdots,&a_s,&-m\\ &1+a-b,&1+a-c,&b_1,&b_2,\cdots,&b_s,&b_{s+1}\end{matrix};z\right]\\&=&\sum_{k=0}^m \frac{(\frac{a}{2})_k(\frac{a+1}{2})_k(a-b-c+1)_k }{(1+a-b)_k(1+a-c)_k k!}\frac{(a_1)_k\cdots(a_s)_k(-m)_k}{(b_1)_k\cdots(b_s)_k(b_{s+1})_k}(-4z)^k\\&&\times {}_{s+2}F_{s+1}\left[\begin{matrix}a+2k,&a_1+k,&a_2+k,\cdots,&a_s+k,&-m+k\\ &b_1+k,&b_2+k,\cdots,&b_s+k,&b_{s+1}+k\end{matrix};z\right]\end{eqnarray*}

Proof:

Well-poisedな一般化超幾何関数の変換公式とDougallの5F4-和公式

\begin{eqnarray}{}_5F_{4}&&\left[\begin{matrix}a,&1+\frac{a}{2},&c,&d,&-m\\ &\frac{a}{2},&1+a-c,&1+a-d,&1+a+m\end{matrix};1\right]\\&&=\frac{(1+a)_m (1+a-c-d)_m}{(1+a-c)_m(1+a-d)_m}\end{eqnarray}and its corollary:\begin{eqnarray}{}_4F_{3}&&\left[\begin{matrix}a,&1+\frac{a}{2},&c,&d\\ &\frac{a}{2},&1+a-c,&1+a-d\end{matrix};-1\right]\\&&=\frac{\G(1+a-c)\G(1+a-d)}{\G(1+a)\G(1+a-c-d)}\end{eqnarray}
proof:

Well-poisedな一般化超幾何関数の変換公式とDougallの5F4-和公式

Whipple's 7F6 transformation\begin{eqnarray}&&{}_7F_6\left[\begin{matrix}a,&1+\frac{a}{2},&b,&c,&d,&e,&-m\\ &\frac{a}{2},&1+a-b,&1+a-c,&1+a-d,&1+a-e,&1+a+m\end{matrix};1\right]\\&=&\frac{(1+a)_m(1+a-d-e)_m}{(1+a-d)_m(1+a-e)_m}{}_4F_3\left[\begin{matrix}1+a-b-c\:,\:d\:,\:e\:,\:-m\\ 1+a-b\:,\:1+a-c\:,\:d+e-a-m\end{matrix};1\right]\end{eqnarray}

Dougall's 7F6 summation formula
$1+2a=b+c+d+e-m$ のとき\begin{eqnarray*}&&{}_7F_6\left[\begin{matrix}a,&1+\frac{a}{2},&b,&c,&d,&e,&-m\\&\frac{a}{2},&1+a-b,&1+a-c,&1+a-d,&1+a-e,&1+a+m\end{matrix};1\right]\\&=&\frac{(1+a)_m(1+a-b-c)_m(1+a-b-d)_m(1+a-c-d)_m}{(1+a-b)_m(1+a-c)_m(1+a-d)_m(1+a-b-c-d)_m}\end{eqnarray*}

Corollaries\begin{eqnarray}&&{}_5F_4\left[\begin{matrix}a,&1+\frac{a}{2},&c,&d,&e\\ &\frac{a}{2},&1+a-c,&1+a-d,&1+a-e\end{matrix};1\right]\\&=& \frac{\G(1+a-c)\G(1+a-d)\G(1+a-e)\G(1+a-c-d-e)}{\G(1+a)\G(1+a-d-e)\G(1+a-c-e)\G(1+a-c-d)}\end{eqnarray}\begin{eqnarray}&&{}_6F_5\left[\begin{matrix}a,&1+\frac{a}{2},&b,&c,&d,&e\\ &\frac{a}{2},&1+a-b,&1+a-c,&1+a-d,&1+a-e\end{matrix};-1\right]\\&=&\frac{\G(1+a-d)\G(1+a-e)}{\G(1+a)\G(1+a-d-e)}{}_3F_2\left[\begin{matrix}1+a-b-c\:,\:d\:,\:e\\ 1+a-b\:,\:1+a-c\end{matrix};1\right]\end{eqnarray}$${}_4F_3\left[\begin{matrix}a,&1+\frac{a}{2},&d,&e\\ &\frac{a}{2},&1+a-d,&1+a-e\end{matrix};-1\right]=\frac{\G(1+a-d)\G(1+a-e)}{\G(1+a)\G(1+a-d-e)}$$

PROOF:

Whippleの7F6変換公式とDougallの7F6-和公式・6F5・5F4・4F3への応用（一般化超幾何関数）

$$\int_0^1\int_0^1\frac{\arctan xy}{1+x^2y^2}dxdy = \frac{\pi G}{4}-\frac{7}{16}\zeta(3)$$PROOF.\begin{eqnarray*}LHS &=& \int_0^1\frac{dx}{x}\int_0^x du\frac{\arctan u}{1+u^2}\quad(u=xy)\\&=& \frac{1}{2}\int_0^1\frac{\arctan^2 x}{x}dx \\&=& \int_0^\frac{\pi}{4}\frac{s^2ds}{\sin 2s}\quad(\tan s=x)\\&=&\frac{1}{8}\int_0^\frac{\pi}{2}\frac{s^2ds}{\sin s}\end{eqnarray*}We obtaind here that$$\int_0^\frac{\pi}{2}\frac{x^2}{\sin x}dx=2\pi G-\frac{7}{2}\zeta(3)$$Hence,$$\int_0^1\int_0^1\frac{\arctan xy}{1+x^2y^2}dxdy = \frac{\pi G}{4}-\frac{7}{16}\zeta(3)$$

$$\int_0^\infty\frac{\arctan(bx)-\arctan(ax)}{x}dx=\frac{\pi}{2}\ln\frac{b}{a}\quad(0<a<b)$$PROOF.$$\int_0^\infty\frac{\arctan(bx)-\arctan(ax)}{x}dx=\lim_{R\to\infty}\int_0^R\left(\int_a^b\frac{ds}{1+x^2s^2}\right)dx$$From Tonelli's theorem we may switch the order of integral then\begin{eqnarray*}\lim_{R\to\infty}\int_0^R\left(\int_a^b\frac{ds}{1+x^2s^2}\right)dx &=& \lim_{R\to\infty}\int_a^b\left(\int_0^R\frac{dx}{1+x^2s^2}\right)ds \\&=& \lim_{R\to\infty}\int_a^b\frac{\arctan sR}{s}ds\end{eqnarray*}Since the integrand is bounded, we find by dominated convergence theorem that\begin{eqnarray*}\lim_{R\to\infty}\int_a^b\frac{\arctan sR}{s}ds &=& \int_a^b\lim_{R\to\infty}\frac{\arctan sR}{s}ds \\&=& \int_a^b\frac{\pi}{2s}ds \\&=& \frac{\pi}{2}\ln\frac{b}{a}\end{eqnarray*}

From @infseriesbot$$\int_0^\frac{\pi}{2}\frac{\tan x}{e^{\tan x}-1}dx=-\frac{1}{2}\left[\pi+\ln2\pi+\psi\left(\frac{1}{2\pi}\right)\right]$$PROOF.
Substituting $2\pi t=\tan x$ yields$$\int_0^\frac{\pi}{2}\frac{\tan x}{e^{\tan x}-1}dx=4\pi^2\int_0^\infty\frac{t}{(e^{2\pi t}-1)(1+4\pi^2t^2)}dt$$From Abel-Plana summation formula, we deduced here that$$\frac{d}{dz}\log\G(z)=-\frac{1}{2z}+\log z-2\int_0^\infty \frac{1}{z^2+t^2}\frac{tdt}{e^{2\pi t}-1}$$We set $z=1/2\pi$ to find$$\psi\left(\frac{1}{2\pi}\right)=-\pi-\ln2\pi-8\pi^2\int_0^\infty\frac{t}{(e^{2\pi t}-1)(1+4\pi^2t^2)}dt$$Hence,$$\int_0^\frac{\pi}{2}\frac{\tan x}{e^{\tan x}-1}dx=-\frac{1}{2}\left[\pi+\ln2\pi+\psi\left(\frac{1}{2\pi}\right)\right]$$

From @SrinivasR1729\begin{eqnarray*}\int_0^\frac{\pi}{2}\frac{\tan x}{e^{2\pi \tan x}-1}dx &=& \frac{2\g-1}{4} \\\int_0^\frac{\pi}{2}\frac{\tan x}{e^{2\pi \tan x}+1}dx &=& 1-\frac{\g}{2}-\ln2\end{eqnarray*}PROOF.
Substituting $t=\tan x$ yields$$\int_0^\frac{\pi}{2}\frac{\tan x}{e^{2\pi \tan x}\mp 1}dx=\int_0^\infty\frac{t}{(e^{2\pi t}\mp 1)(1+t^2)}dt$$From Abel-Plana summation formula and its variation, we deduced here and here that$$\psi(z)=-\frac{1}{2z}+\log z-2\int_0^\infty \frac{t}{(t^2+z^2)(e^{2\pi t}-1)}dt$$$$\psi\left(z+\frac{1}{2}\right)=\ln z+2\int_0^\infty\frac{t}{(t^2+z^2)(e^{2\pi t}+1)}dt$$Setting $z=1$ gives\begin{eqnarray*}\int_0^\frac{\pi}{2}\frac{\tan x}{e^{2\pi \tan x}-1}dx &=& \frac{2\g-1}{4} \\\int_0^\frac{\pi}{2}\frac{\tan x}{e^{2\pi \tan x}+1}dx &=& 1-\frac{\g}{2}-\ln2\end{eqnarray*}

$$\int_0^\infty\frac{\cos bx-\cos ax}{x^2}dx=\frac{\pi}{2}(a-b)$$PROOF.\begin{eqnarray*}LHS &=& -\int_0^\infty\left(\int_a^b \frac{\sin sx}{x}ds\right)dx \\&=& -\int_a^b\left(\int_0^\infty \frac{\sin sx}{x}dx\right)ds \\&=& -\int_a^b\left(\int_0^\infty \frac{\sin u}{u}du\right)ds\quad(sx=u)\\&=&-\int_a^b\frac{\pi}{2}ds\\&=&\frac{\pi}{2}(a-b)\end{eqnarray*}

Suppose $s>0$ then$$I:=\int_0^1\frac{x^{s-1}}{1+x}dx=\psi(s)-\psi\left(\frac{s}{2}\right)-\ln2$$PROOF.\begin{eqnarray*} I &=& \sum_{n=0}^\infty (-1)^n\int_0^1x^{s+n-1}dx \\&=& \sum_{n=0}^\infty \frac{(-1)^n}{n+s} \\&=& \sum_{n=0}^\infty \left(\frac{1}{2n+s}-\frac{1}{2n+s+1}\right) \\&=& \frac{1}{2} \sum_{n=0}^\infty\left(\frac{1}{n+\frac{s}{2}}-\frac{1}{n+\frac{s+1}{2}}\right) \\&=& \frac{1}{2}\left[\psi\left(\frac{s+1}{2}\right)-\psi\left(\frac{s}{2}\right)\right]\end{eqnarray*}We obtained here that$$\psi(2z)=\frac{1}{2}\left[\psi(z)+\psi\left(z+\frac{1}{2}\right)\right]+\log2$$Setting $z=s/2$ yields$$\int_0^1\frac{x^{s-1}}{1+x}dx=\psi(s)-\psi\left(\frac{s}{2}\right)-\ln2$$

Suppose $0<s<1$ then$$J:=\int_0^\infty\frac{x^{s-1}}{1+x}dx=\frac{\pi}{\sin\pi s}$$PROOF.
We define $f(z):=\dfrac{z^{s-1}}{1+z}$ and the contour $C$ as the below figure.

The integrals of the big and small arcs vanish since $s<1$ and $s>0$ respectively. $f(z)$ has the only pole at $z=-1$ with order 1 then$$(1-e^{2\pi si})J=-2\pi i e^{\pi s i}$$Therefore,$$J:=\frac{\pi}{\sin\pi s}$$

Detailed proof here:

複素積分演習（logの分岐点と切断）

From (Almost) Impossible Integrals, Sums, and Series(Amazon)\begin{eqnarray*}I&:=&\int_0^1dx\int_0^1dy\frac{\ln(1-x)\ln(1-y)\ln(1+xy)}{xy(1+xy)} \\&=& \sum_{n=1}^\infty(-1)^{n-1}H_n\int_0^1dx\int_0^1dy\; x^{n-1}y^{n-1}\ln(1-x)\ln(1-y) \\&=& \sum_{n=1}^\infty(-1)^{n-1}H_n\left(\int_0^1x^{n-1}\ln(1-x)dx\right)^2 \\&=& \sum_{n=1}^\infty(-1)^{n-1}H_n\left[\left.\dd{B(p,q)}{q}\right|_{p=n,q=1}\right]^2 \\&=& \sum_{n=1}^\infty(-1)^{n-1}H_n\left[-\frac{H_n}{n}\right]^2 \\&=&\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^3}{n^2}\\&=& 6\Li_5\left(\frac{1}{2}\right)+6\Li_4\left(\frac{1}{2}\right)\ln2+\frac{\ln^52}{2}+\frac{21}{8}\zeta(3)\ln^22\\&&-\frac{9}{4}\zeta(5)-\zeta(2)\ln^32-\frac{27}{16}\zeta(2)\zeta(3)\end{eqnarray*}*The last Iine we used the formula in [Xu & Cai(2017),§3.3] .

From @CerereBaccho$$I_n:=\int_0^\pi(1-\sin x)^n\sqrt{\sin x(1-\cos x)}dx=?$$\begin{eqnarray}I_n &=& \sqrt{2}\sum_{r=0}^n(-1)^r\binom{n}{r}\int_0^\pi\sin^{r+\frac{1}{2}}x\sin\frac{x}{2}dx \\&=& \sqrt{2}\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{\pi \sin\frac{\pi}{4}}{2^{r+\frac{1}{2}}(r+\frac{3}{2})B(\frac{r+3}{2},\frac{r+2}{2})}\tag{1}\\&=&\sqrt{2\pi}\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{\G(r+\frac{3}{2})}{(r+1)!} \\&=& \frac{\pi}{\sqrt{2}}\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{(\frac{3}{2})_r}{(2)_r} \\&=& \frac{\pi}{\sqrt{2}}\sum_{r=0}^n \frac{(\frac{3}{2})_r(-n)_r}{(2)_r r!}\tag{2} \\&=& \frac{\pi}{\sqrt{2}} F \left[\begin{matrix}\frac{3}{2}, -n\\2\end{matrix};1\right] \\&=& \frac{\pi(\frac{1}{2})_n}{\sqrt{2} (n+1)!}\tag{3}\end{eqnarray}Hence,$$I_n=\frac{\pi(\frac{1}{2})_n}{\sqrt{2} (n+1)!}$$

We used for (1) $$\int_0^\pi\sin^{s-1}\t\sin a\t d\t=\frac{\pi\sin\frac{\pi a}{2}}{2^{s-1}sB(\frac{s+a+1}{2},\frac{s-a+1}{2})}$$which we obtained here. From $\binom{n}{r}=\frac{(-1)^r(-n)_r}{r!}$ , (2) follows. And (3) is deduced with Vandermonde's theorem:$$F \left[\begin{matrix}a, -n\\c\end{matrix};1\right]=\frac{(c-a)_n}{(c)_n}$$

\begin{eqnarray}\sum_{n=1}^\infty H_n^3x^n &=& \frac{1}{1-x}\biggl[\Li_3(x)+3\Li_3(1-x)+\frac{3}{2}\ln x\ln^2(1-x)\\&&\quad-\ln^3(1-x)-3\zeta(2)\ln(1-x)-3\zeta(3)\biggr]\end{eqnarray}\begin{eqnarray*}\sum_{n=1}^\infty \frac{H_n^3}{n+1}x^{n+1} &=& -\ln(1-x)\Li_3(x)-3\ln(1-x)\Li_3(1-x)\\&&-\frac{1}{2}\Li_2(x)^2+\frac{3}{2}\ln^2(1-x)\Li_2(1-x)\\&&+\frac{3}{2}\zeta(2)\ln^2(1-x)+3\zeta(3)\ln(1-x)+\frac{1}{4}\ln^4(1-x)\end{eqnarray*}\begin{eqnarray*}\sum_{n=1}^\infty \frac{H_n^3}{n}x^n &=& -2\Li_4(x)-3\Li_4\left(\frac{x}{x-1}\right)-3\Li_4(1-x)\\&&+2\ln(1-x)\Li_3(x)+3\ln(1-x)\Li_3(1-x)\\&&+\Li_2(x)^2-\frac{3}{2}\ln^2(1-x)\Li_2(1-x)\\&&+\frac{1}{8}\ln^4(1-x)-\frac{1}{2}\ln x\ln^3(1-x)+3\zeta(4)\end{eqnarray*}PROOF here:

調和数を含んだ級数(Euler-sum)とゼータ関数 part13

Suppose $r>s$,$$\int_0^\infty \frac{e^{px}-e^{qx}}{e^{rx}-e^{sx}}dx=\frac{1}{r-s}\left\{\psi\left(\frac{r-q}{r-s}\right)-\psi\left(\frac{r-p}{r-s}\right)\right\}$$PROOF.\begin{eqnarray} LHS &=& \int_0^\infty\frac{e^{(p-r)x}-e^{(q-r)x}}{1-e^{-(r-s)x}}dx \\&=& \int_0^\infty\frac{x}{1-e^{-(r-s)x}}\left(\int_{q-r}^{p-r} e^{xu}du\right)dx \\&=& \frac{1}{(r-s)^2}\int_{q-r}^{p-r}\left(\int_0^\infty\frac{ye^{\frac{u}{r-s}y}}{1-e^{-y}}dy\right)du \\&=& \frac{1}{(r-s)^2}\int_{q-r}^{p-r}\zeta\left(2,\frac{-u}{r-s}\right)du\tag{1} \\&=& -\frac{1}{r-s}\left[\psi\left(\frac{-u}{r-s}\right)\right]_{q-r}^{p-r} \\&=& \frac{1}{r-s}\left\{\psi\left(\frac{r-q}{r-s}\right)-\psi\left(\frac{r-p}{r-s}\right)\right\}\end{eqnarray}(1) is proved here.