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ユミルさんの固ツイの積分

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$$\newcommand{ds}[0]{\displaystyle} \newcommand{Li}[0]{{\rm Li}} \newcommand{ol}[1]{\overline{#1}} $$

ユミルさんの固ツイの積分 を全部求めます。NKSさんによる特殊値の表を利用しています。許可が取れ次第その表を投稿しようと思います。

\begin{align} (1)&&\int_0^1\frac{\ln x\ln(1+x)}{1+x}dx &=-\frac18\zeta(3) \\ (2)&&\int_0^1\frac{\ln x\ln(1-x)}{1-x}dx &=\zeta(3) \\ (3)&&\int_0^1\frac{\ln x\ln(1-x)}{1+x}dx &=\frac{13}{8}\zeta(3)-\frac32\zeta(2)\ln2 \\ (4)&&\int_0^1\frac{\ln x\ln(1+x)}{1-x}dx &=\zeta(3)-\frac32\zeta(2)\ln2 \\ (5)&&\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx&=-\frac12\zeta(4) \\ (6)&&\int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx&=3\zeta(4)+\frac72\zeta(3)\ln2-\zeta(2)\ln^22+\frac16\ln^42+4\Li_4\left(\frac12\right) \\ (7)&&\int_0^1\frac{\ln^2x\ln(1+x)}{1-x}dx&=\zeta(4)+\zeta(2)\ln^22-\frac16\ln^42-4\Li_4\left(\frac12\right) \\ (8)&&\int_0^1\frac{\ln^2x\ln(1+x)}{1+x}dx&=-\frac{19}{8}\zeta(4)+\frac{7}{2}\zeta(3)\ln2 \\ \end{align}
\begin{align} (9)&&\int_0^1\frac{\ln(1-x)\ln x\ln(1+x)}{1-x}dx&=\frac{17}{16}\zeta(4)+\frac78\zeta(3)\ln2-\frac14\zeta(2)\ln^22-\frac{1}{12}\ln^42-2\Li_4\left(\frac12\right) \\ (10)&&\int_0^1\frac{\ln(1-x)\ln x\ln(1+x)}{1+x}dx&=-2\zeta(4)+\frac{21}{8}\zeta(3)\ln2-\frac54\zeta(2)\ln^22+\frac{1}{12}\ln^42+2\Li_4\left(\frac12\right) \end{align}

$(1)$の証明
\begin{align} \int_0^1\frac{\ln x\ln(1+x)}{1+x}dx&=-\int_0^1\frac{dx}{-1-x}\int_x^1\frac{-ds}{s}\int_0^x\frac{dt}{1+t} \\ &=-\int_0^1\frac{dx}{-1-x}\int_x^1\frac{ds}{s}\int_0^x\frac{dt}{-1-t} \\ &=-\int_{0< t< x< s<1}\frac{dt}{-1-t}\frac{dx}{-1-x}\frac{ds}{s} \\ &=-\int_0^1\frac{ds}{s}\int_0^s\frac{dx}{-1-x}\int_0^x\frac{dt}{-1-t} \\ &=-\int_0^1z_{2,1}z_{1,1} \\ &=-\zeta(1,\ol2) \\ &=-\frac18\zeta(3) \end{align}
$(2)$の証明
\begin{align} \int_0^1\frac{\ln x\ln(1-x)}{1-x}dx&=\int_0^1\frac{dx}{1-x}\int_x^1\frac{-ds}{s}\int_0^x\frac{-dt}{1-t} \\ &=\int_0^1\frac{dx}{1-x}\int_x^1\frac{ds}{s}\int_0^x\frac{dt}{1-t} \\ &=\int_{0< t< x< s<1}\frac{dt}{1-t}\frac{dx}{1-x}\frac{ds}{s} \\ &=\int_0^1\frac{ds}{s}\int_0^s\frac{dx}{1-x}\int_0^x\frac{dt}{1-t} \\ &=\int_0^1z_{2,0}z_{1,0} \\ &=\zeta(1,2) \\ &=\zeta(3) \end{align}
$(3)$の証明
\begin{align} \int_0^1\frac{\ln x\ln(1-x)}{1+x}dx&=\int_0^1\frac{dx}{1+x}\int_x^1\frac{-ds}{s}\int_0^x\frac{-dt}{1-t} \\ &=-\int_0^1\frac{dx}{-1-x}\int_x^1\frac{ds}{s}\int_0^x\frac{dt}{1-t} \\ &=-\int_{0< t< x< s<1}\frac{dt}{1-t}\frac{dx}{-1-x}\frac{ds}{s} \\ &=-\int_0^1\frac{ds}{s}\int_0^s\frac{dx}{-1-x}\int_0^x\frac{dt}{1-t} \\ &=-\int_0^1z_{2,1}z_{1,0} \\ &=-\zeta(\ol1,\ol2) \\ &=\frac{13}{8}\zeta(3)-\frac32\zeta(2)\ln2 \end{align}
$(4)$の証明
\begin{align} \int_0^1\frac{\ln x\ln(1+x)}{1-x}dx&=\int_0^1\frac{dx}{1-x}\int_x^1\frac{-ds}{s}\int_0^x\frac{dt}{1+t} \\ &=\int_0^1\frac{dx}{1-x}\int_x^1\frac{ds}{s}\int_0^x\frac{dt}{-1-t} \\ &=\int_{0< t< x< s<1}\frac{dt}{-1-t}\frac{dx}{1-x}\frac{ds}{s} \\ &=\int_0^1\frac{ds}{s}\int_0^s\frac{dx}{1-x}\int_0^x\frac{dt}{-1-t} \\ &=\int_0^1z_{2,0}z_{1,1} \\ &=\zeta(\ol1,2) \\ &=\zeta(3)-\frac32\zeta(2)\ln2 \end{align}
$(5)$の証明
\begin{align} \int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx&=\int_0^1\frac{dx}{1-x}\int_x^1\frac{-ds}{s}\int_x^1\frac{-du}{u}\int_0^x\frac{-dt}{1-t} \\ &=-\int_0^1\frac{dx}{1-x}\int_x^1\frac{ds}{s}\int_x^1\frac{du}{u}\int_0^x\frac{dt}{1-t} \\ &=-\int_{0< t< x< s,u<1}\frac{dt}{1-t}\frac{dx}{1-x}\frac{ds}{s}\frac{du}{u} \\ &=-2\int_0^1\frac{ds}{s}\int_0^s\frac{du}{u}\int_0^u\frac{dx}{1-x}\int_0^x\frac{dt}{1-t} \\ &=-2\int_0^1z_{3,0}z_{1,0} \\ &=-2\zeta(1,3) \\ &=-\frac12\zeta(4) \end{align}
$(6)$の証明
\begin{align} \int_0^1\frac{\ln^2x\ln(1-x)}{1+x}dx&=\int_0^1\frac{dx}{1+x}\int_x^1\frac{-ds}{s}\int_x^1\frac{-du}{u}\int_0^x\frac{-dt}{1-t} \\ &=-\int_0^1\frac{dx}{1+x}\int_x^1\frac{ds}{s}\int_x^1\frac{du}{u}\int_0^x\frac{dt}{1-t} \\ &=\int_{0< t< x< s,u<1}\frac{dt}{1-t}\frac{dx}{-1-x}\frac{ds}{s}\frac{du}{u} \\ &=2\int_0^1\frac{ds}{s}\int_0^s\frac{du}{u}\int_0^u\frac{dx}{-1-x}\int_0^x\frac{dt}{1-t} \\ &=2\int_0^1z_{3,1}z_{1,0} \\ &=2\zeta(\ol1,\ol3) \\ &=3\zeta(4)+\frac72\zeta(3)\ln2-\zeta(2)\ln^22+\frac16\ln^42+4\Li_4\left(\frac12\right) \end{align}
$(7)$の証明
\begin{align} \int_0^1\frac{\ln^2x\ln(1+x)}{1-x}dx&=\int_0^1\frac{dx}{1-x}\int_x^1\frac{-ds}{s}\int_x^1\frac{-du}{u}\int_0^x\frac{dt}{1+t} \\ &=-\int_0^1\frac{dx}{1-x}\int_x^1\frac{ds}{s}\int_x^1\frac{du}{u}\int_0^x\frac{dt}{-1-t} \\ &=-\int_{0< t< x< s,u<1}\frac{dt}{-1-t}\frac{dx}{1-x}\frac{ds}{s}\frac{du}{u} \\ &=-2\int_0^1\frac{ds}{s}\int_0^s\frac{du}{u}\int_0^u\frac{dx}{1-x}\int_0^x\frac{dt}{-1-t} \\ &=-2\int_0^1z_{3,0}z_{1,1} \\ &=-2\zeta(\ol1,3) \\ &=\zeta(4)+\zeta(2)\ln^22-\frac16\ln^42-4\Li_4\left(\frac12\right) \end{align}
$(8)$の証明
\begin{align} \int_0^1\frac{\ln^2x\ln(1+x)}{1+x}dx&=\int_0^1\frac{dx}{1+x}\int_x^1\frac{-ds}{s}\int_x^1\frac{-du}{u}\int_0^x\frac{dt}{1+t} \\ &=\int_0^1\frac{dx}{-1-x}\int_x^1\frac{ds}{s}\int_x^1\frac{du}{u}\int_0^x\frac{dt}{-1-t} \\ &=\int_{0< t< x< s,u<1}\frac{dt}{-1-t}\frac{dx}{-1-x}\frac{ds}{s}\frac{du}{u} \\ &=2\int_0^1\frac{ds}{s}\int_0^s\frac{du}{u}\int_0^u\frac{dx}{-1-x}\int_0^x\frac{dt}{-1-t} \\ &=2\int_0^1z_{3,1}z_{1,1} \\ &=2\zeta(1,\ol3) \\ &=-\frac{19}{8}\zeta(4)+\frac{7}{2}\zeta(3)\ln2 \end{align}
$(9)$の証明
\begin{align} \int_0^1\frac{\ln(1-x)\ln x\ln(1+x)}{1-x}dx=&\int_0^1\frac{dx}{1-x}\int_0^x\frac{-ds}{1-s}\int_x^1\frac{-du}{u}\int_0^x\frac{dt}{1+t} \\ =&-\int_0^1\frac{dx}{1-x}\int_0^x\frac{ds}{1-s}\int_x^1\frac{du}{u}\int_0^x\frac{dt}{-1-t} \\ =&-\int_{0< s,t< x< u<1}\frac{dt}{-1-t}\frac{dx}{1-x}\frac{ds}{1-s}\frac{du}{u} \\ =&-\int_0^1\frac{du}{u}\int_0^u\frac{dx}{1-x}\int_0^x\frac{dt}{-1-t}\int_0^t\frac{ds}{1-s} \\ &-\int_0^1\frac{du}{u}\int_0^u\frac{dx}{1-x}\int_0^x\frac{ds}{1-s}\int_0^s\frac{dt}{-1-t} \\ =&-\int_0^1z_{2,0}z_{1,1}z_{1,0}+z_{2,0}z_{1,0}z_{1,1} \\ =&-\zeta(\ol1,\ol1,2)-\zeta(\ol1,1,2) \\ =&\frac{17}{16}\zeta(4)+\frac78\zeta(3)\ln2-\frac14\zeta(2)\ln^22-\frac{1}{12}\ln^42-2\Li_4\left(\frac12\right) \end{align}
$(10)$の証明
\begin{align} \int_0^1\frac{\ln(1-x)\ln x\ln(1+x)}{1+x}dx=&\int_0^1\frac{dx}{1+x}\int_0^x\frac{-ds}{1-s}\int_x^1\frac{-du}{u}\int_0^x\frac{dt}{1+t} \\ =&\int_0^1\frac{dx}{-1-x}\int_0^x\frac{ds}{1-s}\int_x^1\frac{du}{u}\int_0^x\frac{dt}{-1-t} \\ =&\int_{0< s,t< x< u<1}\frac{dt}{-1-t}\frac{dx}{-1-x}\frac{ds}{1-s}\frac{du}{u} \\ =&\int_0^1\frac{du}{u}\int_0^u\frac{dx}{-1-x}\int_0^x\frac{dt}{-1-t}\int_0^t\frac{ds}{1-s} \\ &+\int_0^1\frac{du}{u}\int_0^u\frac{dx}{-1-x}\int_0^x\frac{ds}{1-s}\int_0^s\frac{dt}{-1-t} \\ =&\int_0^1z_{2,1}z_{1,1}z_{1,0}+z_{2,1}z_{1,0}z_{1,1} \\ =&\zeta(\ol1,1,\ol2)+\zeta(\ol1,\ol1,\ol2) \\ =&-2\zeta(4)+\frac{21}{8}\zeta(3)\ln2-\frac54\zeta(2)\ln^22+\frac{1}{12}\ln^42+2\Li_4\left(\frac12\right) \end{align}
投稿日:2022113

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