ユミルさんの固ツイの積分

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。

$\begin{aligned} (1) & \int_{0}^{1} \frac{\ln x \ln (1 + x)}{1 + x} d x & = - \frac{1}{8} ζ (3) \\ (2) & \int_{0}^{1} \frac{\ln x \ln (1 - x)}{1 - x} d x & = ζ (3) \\ (3) & \int_{0}^{1} \frac{\ln x \ln (1 - x)}{1 + x} d x & = \frac{13}{8} ζ (3) - \frac{3}{2} ζ (2) \ln 2 \\ (4) & \int_{0}^{1} \frac{\ln x \ln (1 + x)}{1 - x} d x & = ζ (3) - \frac{3}{2} ζ (2) \ln 2 \\ (5) & \int_{0}^{1} \frac{\ln^{2} x \ln (1 - x)}{1 - x} d x & = - \frac{1}{2} ζ (4) \\ (6) & \int_{0}^{1} \frac{\ln^{2} x \ln (1 - x)}{1 + x} d x & = 3 ζ (4) + \frac{7}{2} ζ (3) \ln 2 - ζ (2) \ln^{2} 2 + \frac{1}{6} \ln^{4} 2 + 4 {L i}_{4} (\frac{1}{2}) \\ (7) & \int_{0}^{1} \frac{\ln^{2} x \ln (1 + x)}{1 - x} d x & = ζ (4) + ζ (2) \ln^{2} 2 - \frac{1}{6} \ln^{4} 2 - 4 {L i}_{4} (\frac{1}{2}) \\ (8) & \int_{0}^{1} \frac{\ln^{2} x \ln (1 + x)}{1 + x} d x & = - \frac{19}{8} ζ (4) + \frac{7}{2} ζ (3) \ln 2 \end{aligned}$
$\begin{aligned} (9) & \int_{0}^{1} \frac{\ln (1 - x) \ln x \ln (1 + x)}{1 - x} d x & = \frac{17}{16} ζ (4) + \frac{7}{8} ζ (3) \ln 2 - \frac{1}{4} ζ (2) \ln^{2} 2 - \frac{1}{12} \ln^{4} 2 - 2 {L i}_{4} (\frac{1}{2}) \\ (10) & \int_{0}^{1} \frac{\ln (1 - x) \ln x \ln (1 + x)}{1 + x} d x & = - 2 ζ (4) + \frac{21}{8} ζ (3) \ln 2 - \frac{5}{4} ζ (2) \ln^{2} 2 + \frac{1}{12} \ln^{4} 2 + 2 {L i}_{4} (\frac{1}{2}) \end{aligned}$

(1)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln x \ln (1 + x)}{1 + x} d x & = - \int_{0}^{1} \frac{d x}{- 1 - x} \int_{x}^{1} \frac{- d s}{s} \int_{0}^{x} \frac{d t}{1 + t} \\ = - \int_{0}^{1} \frac{d x}{- 1 - x} \int_{x}^{1} \frac{d s}{s} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = - \int_{0 < t < x < s < 1} \frac{d t}{- 1 - t} \frac{d x}{- 1 - x} \frac{d s}{s} \\ = - \int_{0}^{1} \frac{d s}{s} \int_{0}^{s} \frac{d x}{- 1 - x} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = - \int_{0}^{1} z_{2, 1} z_{1, 1} \\ = - ζ (1, \overset{―}{2}) \\ = - \frac{1}{8} ζ (3) \end{aligned}

(2)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln x \ln (1 - x)}{1 - x} d x & = \int_{0}^{1} \frac{d x}{1 - x} \int_{x}^{1} \frac{- d s}{s} \int_{0}^{x} \frac{- d t}{1 - t} \\ = \int_{0}^{1} \frac{d x}{1 - x} \int_{x}^{1} \frac{d s}{s} \int_{0}^{x} \frac{d t}{1 - t} \\ = \int_{0 < t < x < s < 1} \frac{d t}{1 - t} \frac{d x}{1 - x} \frac{d s}{s} \\ = \int_{0}^{1} \frac{d s}{s} \int_{0}^{s} \frac{d x}{1 - x} \int_{0}^{x} \frac{d t}{1 - t} \\ = \int_{0}^{1} z_{2, 0} z_{1, 0} \\ = ζ (1, 2) \\ = ζ (3) \end{aligned}

(3)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln x \ln (1 - x)}{1 + x} d x & = \int_{0}^{1} \frac{d x}{1 + x} \int_{x}^{1} \frac{- d s}{s} \int_{0}^{x} \frac{- d t}{1 - t} \\ = - \int_{0}^{1} \frac{d x}{- 1 - x} \int_{x}^{1} \frac{d s}{s} \int_{0}^{x} \frac{d t}{1 - t} \\ = - \int_{0 < t < x < s < 1} \frac{d t}{1 - t} \frac{d x}{- 1 - x} \frac{d s}{s} \\ = - \int_{0}^{1} \frac{d s}{s} \int_{0}^{s} \frac{d x}{- 1 - x} \int_{0}^{x} \frac{d t}{1 - t} \\ = - \int_{0}^{1} z_{2, 1} z_{1, 0} \\ = - ζ (\overset{―}{1}, \overset{―}{2}) \\ = \frac{13}{8} ζ (3) - \frac{3}{2} ζ (2) \ln 2 \end{aligned}

(4)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln x \ln (1 + x)}{1 - x} d x & = \int_{0}^{1} \frac{d x}{1 - x} \int_{x}^{1} \frac{- d s}{s} \int_{0}^{x} \frac{d t}{1 + t} \\ = \int_{0}^{1} \frac{d x}{1 - x} \int_{x}^{1} \frac{d s}{s} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = \int_{0 < t < x < s < 1} \frac{d t}{- 1 - t} \frac{d x}{1 - x} \frac{d s}{s} \\ = \int_{0}^{1} \frac{d s}{s} \int_{0}^{s} \frac{d x}{1 - x} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = \int_{0}^{1} z_{2, 0} z_{1, 1} \\ = ζ (\overset{―}{1}, 2) \\ = ζ (3) - \frac{3}{2} ζ (2) \ln 2 \end{aligned}

(5)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln^{2} x \ln (1 - x)}{1 - x} d x & = \int_{0}^{1} \frac{d x}{1 - x} \int_{x}^{1} \frac{- d s}{s} \int_{x}^{1} \frac{- d u}{u} \int_{0}^{x} \frac{- d t}{1 - t} \\ = - \int_{0}^{1} \frac{d x}{1 - x} \int_{x}^{1} \frac{d s}{s} \int_{x}^{1} \frac{d u}{u} \int_{0}^{x} \frac{d t}{1 - t} \\ = - \int_{0 < t < x < s, u < 1} \frac{d t}{1 - t} \frac{d x}{1 - x} \frac{d s}{s} \frac{d u}{u} \\ = - 2 \int_{0}^{1} \frac{d s}{s} \int_{0}^{s} \frac{d u}{u} \int_{0}^{u} \frac{d x}{1 - x} \int_{0}^{x} \frac{d t}{1 - t} \\ = - 2 \int_{0}^{1} z_{3, 0} z_{1, 0} \\ = - 2 ζ (1, 3) \\ = - \frac{1}{2} ζ (4) \end{aligned}

(6)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln^{2} x \ln (1 - x)}{1 + x} d x & = \int_{0}^{1} \frac{d x}{1 + x} \int_{x}^{1} \frac{- d s}{s} \int_{x}^{1} \frac{- d u}{u} \int_{0}^{x} \frac{- d t}{1 - t} \\ = - \int_{0}^{1} \frac{d x}{1 + x} \int_{x}^{1} \frac{d s}{s} \int_{x}^{1} \frac{d u}{u} \int_{0}^{x} \frac{d t}{1 - t} \\ = \int_{0 < t < x < s, u < 1} \frac{d t}{1 - t} \frac{d x}{- 1 - x} \frac{d s}{s} \frac{d u}{u} \\ = 2 \int_{0}^{1} \frac{d s}{s} \int_{0}^{s} \frac{d u}{u} \int_{0}^{u} \frac{d x}{- 1 - x} \int_{0}^{x} \frac{d t}{1 - t} \\ = 2 \int_{0}^{1} z_{3, 1} z_{1, 0} \\ = 2 ζ (\overset{―}{1}, \overset{―}{3}) \\ = 3 ζ (4) + \frac{7}{2} ζ (3) \ln 2 - ζ (2) \ln^{2} 2 + \frac{1}{6} \ln^{4} 2 + 4 {L i}_{4} (\frac{1}{2}) \end{aligned}

(7)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln^{2} x \ln (1 + x)}{1 - x} d x & = \int_{0}^{1} \frac{d x}{1 - x} \int_{x}^{1} \frac{- d s}{s} \int_{x}^{1} \frac{- d u}{u} \int_{0}^{x} \frac{d t}{1 + t} \\ = - \int_{0}^{1} \frac{d x}{1 - x} \int_{x}^{1} \frac{d s}{s} \int_{x}^{1} \frac{d u}{u} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = - \int_{0 < t < x < s, u < 1} \frac{d t}{- 1 - t} \frac{d x}{1 - x} \frac{d s}{s} \frac{d u}{u} \\ = - 2 \int_{0}^{1} \frac{d s}{s} \int_{0}^{s} \frac{d u}{u} \int_{0}^{u} \frac{d x}{1 - x} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = - 2 \int_{0}^{1} z_{3, 0} z_{1, 1} \\ = - 2 ζ (\overset{―}{1}, 3) \\ = ζ (4) + ζ (2) \ln^{2} 2 - \frac{1}{6} \ln^{4} 2 - 4 {L i}_{4} (\frac{1}{2}) \end{aligned}

(8)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln^{2} x \ln (1 + x)}{1 + x} d x & = \int_{0}^{1} \frac{d x}{1 + x} \int_{x}^{1} \frac{- d s}{s} \int_{x}^{1} \frac{- d u}{u} \int_{0}^{x} \frac{d t}{1 + t} \\ = \int_{0}^{1} \frac{d x}{- 1 - x} \int_{x}^{1} \frac{d s}{s} \int_{x}^{1} \frac{d u}{u} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = \int_{0 < t < x < s, u < 1} \frac{d t}{- 1 - t} \frac{d x}{- 1 - x} \frac{d s}{s} \frac{d u}{u} \\ = 2 \int_{0}^{1} \frac{d s}{s} \int_{0}^{s} \frac{d u}{u} \int_{0}^{u} \frac{d x}{- 1 - x} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = 2 \int_{0}^{1} z_{3, 1} z_{1, 1} \\ = 2 ζ (1, \overset{―}{3}) \\ = - \frac{19}{8} ζ (4) + \frac{7}{2} ζ (3) \ln 2 \end{aligned}

(9)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln (1 - x) \ln x \ln (1 + x)}{1 - x} d x = & \int_{0}^{1} \frac{d x}{1 - x} \int_{0}^{x} \frac{- d s}{1 - s} \int_{x}^{1} \frac{- d u}{u} \int_{0}^{x} \frac{d t}{1 + t} \\ = & - \int_{0}^{1} \frac{d x}{1 - x} \int_{0}^{x} \frac{d s}{1 - s} \int_{x}^{1} \frac{d u}{u} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = & - \int_{0 < s, t < x < u < 1} \frac{d t}{- 1 - t} \frac{d x}{1 - x} \frac{d s}{1 - s} \frac{d u}{u} \\ = & - \int_{0}^{1} \frac{d u}{u} \int_{0}^{u} \frac{d x}{1 - x} \int_{0}^{x} \frac{d t}{- 1 - t} \int_{0}^{t} \frac{d s}{1 - s} \\ - \int_{0}^{1} \frac{d u}{u} \int_{0}^{u} \frac{d x}{1 - x} \int_{0}^{x} \frac{d s}{1 - s} \int_{0}^{s} \frac{d t}{- 1 - t} \\ = & - \int_{0}^{1} z_{2, 0} z_{1, 1} z_{1, 0} + z_{2, 0} z_{1, 0} z_{1, 1} \\ = & - ζ (\overset{―}{1}, \overset{―}{1}, 2) - ζ (\overset{―}{1}, 1, 2) \\ = & \frac{17}{16} ζ (4) + \frac{7}{8} ζ (3) \ln 2 - \frac{1}{4} ζ (2) \ln^{2} 2 - \frac{1}{12} \ln^{4} 2 - 2 {L i}_{4} (\frac{1}{2}) \end{aligned}

(10)

の証明

\begin{aligned} \int_{0}^{1} \frac{\ln (1 - x) \ln x \ln (1 + x)}{1 + x} d x = & \int_{0}^{1} \frac{d x}{1 + x} \int_{0}^{x} \frac{- d s}{1 - s} \int_{x}^{1} \frac{- d u}{u} \int_{0}^{x} \frac{d t}{1 + t} \\ = & \int_{0}^{1} \frac{d x}{- 1 - x} \int_{0}^{x} \frac{d s}{1 - s} \int_{x}^{1} \frac{d u}{u} \int_{0}^{x} \frac{d t}{- 1 - t} \\ = & \int_{0 < s, t < x < u < 1} \frac{d t}{- 1 - t} \frac{d x}{- 1 - x} \frac{d s}{1 - s} \frac{d u}{u} \\ = & \int_{0}^{1} \frac{d u}{u} \int_{0}^{u} \frac{d x}{- 1 - x} \int_{0}^{x} \frac{d t}{- 1 - t} \int_{0}^{t} \frac{d s}{1 - s} \\ + \int_{0}^{1} \frac{d u}{u} \int_{0}^{u} \frac{d x}{- 1 - x} \int_{0}^{x} \frac{d s}{1 - s} \int_{0}^{s} \frac{d t}{- 1 - t} \\ = & \int_{0}^{1} z_{2, 1} z_{1, 1} z_{1, 0} + z_{2, 1} z_{1, 0} z_{1, 1} \\ = & ζ (\overset{―}{1}, 1, \overset{―}{2}) + ζ (\overset{―}{1}, \overset{―}{1}, \overset{―}{2}) \\ = & - 2 ζ (4) + \frac{21}{8} ζ (3) \ln 2 - \frac{5}{4} ζ (2) \ln^{2} 2 + \frac{1}{12} \ln^{4} 2 + 2 {L i}_{4} (\frac{1}{2}) \end{aligned}

投稿日：2022年1月13日

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ユミルさんの固ツイの積分