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$\int \frac{\ln^{3} x \ln (1 - x)}{1 - x} d x = - H_{1} (x) \ln^{3} x + {Li}_{2} (x) \ln^{3} x + 3 H_{2} (x) \ln^{2} x - 3 {Li}_{3} (x) \ln^{2} x - 6 H_{3} (x) \ln x + 6 {Li}_{4} (x) \ln x + 6 H_{4} (x) - 6 {Li}_{5} (x)$

$\begin{aligned} \int \frac{\ln^{3} x \ln (1 - x)}{1 - x} d x & = - \int \sum_{k \geq 1} H_{k} x^{k} \ln^{3} k d x \\ = - \sum_{k \geq 1} H_{k} \int x^{k} \ln^{3} k d x \\ = - \sum_{k \geq 1} H_{k} \frac{\partial^{3}}{\partial k^{3}} [\int x^{k} d x] \\ = - \sum_{k \geq 1} H_{k} \frac{\partial^{3}}{\partial k^{3}} [\frac{x^{k + 1}}{k + 1}] \\ = - \sum_{k \geq 1} H_{k} [\frac{x^{k + 1} \ln^{3} x}{k + 1} - \frac{3 x^{k + 1} \ln^{2} x}{(k + 1)^{2}} + \frac{6 x^{k + 1} \ln x}{(k + 1)^{3}} - \frac{6 x^{k + 1}}{(k + 1)^{4}}] \\ = - \ln^{3} x \sum_{k \geq 1} \frac{H_{k + 1} x^{k + 1}}{k + 1} + \ln^{3} x \sum_{k \geq 1} \frac{x^{k + 1}}{(k + 1)^{2}} + 3 \ln^{2} x \sum_{k \geq 1} \frac{H_{k + 1} x^{k + 1}}{(k + 1)^{2}} - 3 \ln^{2} x \sum_{k \geq 1} \frac{x^{k + 1}}{(k + 1)^{3}} - 6 \ln x \sum_{k \geq 1} \frac{H_{k + 1} x^{k + 1}}{(k + 1)^{3}} + 6 \ln x \sum_{k \geq 1} \frac{x^{k + 1}}{(k + 1)^{4}} + 6 \sum_{k \geq 1} \frac{H_{k + 1} x^{k + 1}}{(k + 1)^{4}} - 6 \ln x \sum_{k \geq 1} \frac{x^{k + 1}}{(k + 1)^{4}} \\ = - H_{1} (x) \ln^{3} x + {Li}_{2} (x) \ln^{3} x + 3 H_{2} (x) \ln^{2} x - 3 {Li}_{3} (x) \ln^{2} x - 6 H_{3} (x) \ln x + 6 {Li}_{4} (x) \ln x + 6 H_{4} (x) - 6 {Li}_{5} (x) \end{aligned}$

$\int_{0}^{1} \frac{\ln x \ln (1 + x)}{1 - x} d x = - \frac{3}{2} ζ (2) \ln 2 + ζ (3)$

$\begin{aligned} \int_{0}^{1} \frac{\ln x \ln (1 + x)}{1 - x} d x & = {[\int_{0}^{x} \frac{\ln y}{1 - y} d y \ln (1 + x)]}_{0}^{1} - \int_{0}^{1} \frac{J (x)}{1 + x} d x (J (x) = \int_{0}^{x} \frac{\ln y}{1 - y} d y x \in [0, 1]) \\ = - ζ (2) \ln 2 - \int_{0}^{1} \int_{0}^{1} \frac{x \ln x y}{(1 - x y) (1 + x)} d y d x \\ = - ζ (2) \ln 2 - \int_{0}^{1} (\int_{0}^{1} \frac{x \ln y}{(1 - x y) (1 + x)} d x) d y - \int_{0}^{1} (\int_{0}^{1} \frac{x \ln x}{(1 - x y) (1 + x)} d y) d x \\ = - ζ (2) \ln 2 + \int_{0}^{1} {[\frac{\ln (1 - x y)}{y (1 + y)} + \frac{\ln (1 + x)}{1 + y}]}_{x = 0}^{x = 1} \ln y d y + \int_{0}^{1} {[\frac{\ln (1 - x y)}{1 + x}]}_{y = 0}^{y = 1} \ln x d x \\ = - ζ (2) \ln 2 + \int_{0}^{1} \frac{\ln (1 - y) \ln y}{y (1 + y)} d y + \ln 2 \int_{0}^{1} \frac{\ln y}{1 + y} d y + \int_{0}^{1} \frac{\ln (1 - x) \ln x}{1 + x} d x \\ = - ζ (2) \ln 2 + \int_{0}^{1} \frac{\ln (1 - y) \ln y}{y} d y + \ln 2 \int_{0}^{1} \frac{\ln y}{1 + y} d y \\ = - ζ (2) \ln 2 + \frac{1}{2} (\ln^{2} x \ln (1 - x) + \int_{0}^{1} \frac{\ln^{2} y}{1 - y} d y) + \ln 2 (\int_{0}^{1} \frac{\ln y}{1 - y} d y - \int_{0}^{1} \frac{2 y \ln y}{1 - y^{2}} d y) \\ = - ζ (2) \ln 2 + \frac{1}{2} \int_{0}^{1} \frac{\ln^{2} y}{1 - y} d y + \ln 2 (\int_{0}^{1} \frac{\ln y}{1 - y} d y - \frac{1}{2} \int_{0}^{1} \frac{\ln y}{1 - y} d y) \\ = - \frac{3}{2} ζ (2) \ln 2 + ζ (3) \end{aligned}$

$\int_{0}^{1} (\frac{\ln^{2} (1 - x) \ln^{2} (1 + x)}{1 - x} - \frac{\ln^{2} 2 \ln^{2} (1 - x)}{1 - x}) d x = \frac{1}{8} ζ (5) - \frac{1}{2} \ln 2 ζ (4) + 2 \ln^{2} 2 ζ (3) - \frac{2}{3} \ln^{3} 2 ζ (2) - 2 ζ (2) ζ (3) + \frac{1}{10} \ln^{5} 2 + 4 {Li}_{5} (\frac{1}{2})$

$\begin{aligned} I = \int_{0}^{1} (\frac{\ln^{2} (1 - x) \ln^{2} (1 + x)}{1 - x} - \frac{\ln^{2} 2 \ln^{2} (1 - x)}{1 - x}) d x & = \int_{0}^{1} \frac{\ln^{2} (1 - x)}{1 - x} (\ln^{2} (1 + x) - \ln^{2} 2) d x \\ \overset{I B P}{=} \frac{2}{3} \int_{0}^{1} \frac{\ln^{3} (1 - x) \ln (1 + x)}{1 + x} d x \end{aligned}$
ここで
$\begin{array}{r} a^{3} b = \frac{(a + b)^{4}}{8} - \frac{(a - b)^{4}}{8} - a b^{3} \to_{b = \ln (1 + x)}^{a = \ln (1 - x)} (\ln (1 - x))^{3} \ln (1 + x) = \frac{(\ln (1 - x) + \ln (1 + x))^{4}}{8} - \frac{(\ln (1 - x) - \ln (1 + x))^{4}}{8} - (\ln (1 - x)) (\ln (1 + x))^{3} \end{array}$
であるから
$\begin{aligned} I & = \frac{1}{12} \int_{0}^{1} \frac{\ln^{4} (1 - x^{2})}{1 + x} d x - \frac{1}{12} \int_{0}^{1} \frac{\ln^{4} (1 - x) - \ln^{4} (1 + x)}{1 + x} d x - \frac{2}{3} \int_{0}^{1} \frac{\ln (1 - x) \ln^{3} (1 + x)}{1 + x} d x \\ = (\frac{16}{5} \ln^{5} 2 - 16 \ln^{3} 2 ζ (2) + 48 \ln^{2} 2 ζ (3) - 54 \ln 2 ζ (4) - 24 ζ (2) ζ (3) + 72 ζ (5)) - (\frac{45}{2} ζ (5)) - (- 6 {Li}_{5} (\frac{1}{2}) + 6 ζ (5) - 6 \ln 2 ζ (4) + 3 \ln^{2} 2 ζ (3) - \ln^{3} 2 ζ (2) + \frac{1}{4} \ln^{5} 2) \\ = \frac{1}{8} ζ (5) - \frac{1}{2} \ln 2 ζ (4) + 2 \ln^{2} 2 ζ (3) - \frac{2}{3} \ln^{3} 2 ζ (2) - 2 ζ (2) ζ (3) + \frac{1}{10} \ln^{5} 2 + 4 {Li}_{5} (\frac{1}{2}) \end{aligned}$

$\begin{aligned} \int_{0}^{1} \frac{\ln^{4} (1 - x^{2})}{1 + x} d x & = \int_{0}^{1} (1 - x) \frac{\ln^{4} (1 - x^{2})}{1 - x^{2}} d x \\ \overset{x^{2} = y}{=} \frac{1}{2} \int_{0}^{1} \frac{(1 - \sqrt{y}) (\ln^{4} (1 - y))}{(1 - y) \sqrt{y}} d y \\ \overset{I B P}{=} - \frac{1}{20} \int_{0}^{1} \frac{\ln^{5} (1 - y)}{y^{3 / 2}} d y = - \frac{1}{20} lim_{\begin{array}{c} x \mapsto - \frac{1}{2} \\ y \mapsto 1 \end{array}} \frac{\partial^{5}}{\partial y^{5}} B (x, y) \\ = \frac{16}{5} \ln^{5} 2 - 16 \ln^{3} 2 ζ (2) + 48 \ln^{2} 2 ζ (3) - 54 \ln 2 ζ (4) - 24 ζ (2) ζ (3) + 72 ζ (5) \end{aligned}$

$\begin{array}{r} \int_{0}^{1} \frac{\ln^{4} (1 - x) - \ln^{4} (1 + x)}{1 + x} d x \overset{\frac{1 - x}{1 + x} = y}{=} \int_{0}^{1} \frac{\ln^{4} x}{1 + x} d x = \frac{45}{2} ζ (5) \end{array}$

$\begin{aligned} \int_{0}^{1} \frac{\ln (1 - x) \ln^{3} (1 + x)}{1 + x} d x & \overset{\frac{1}{1 + x} = y}{=} - \int_{\frac{1}{2}}^{1} \frac{\ln (\frac{2 x - 1}{x}) \ln^{3} x}{x} d x \\ = \int_{\frac{1}{2}}^{1} \frac{\ln^{4} x}{x} d x - \int_{\frac{1}{2}}^{1} \frac{\ln (2 x - 1) \ln^{3} x}{x} d x \\ = \frac{1}{5} \ln^{5} 2 - \int_{\frac{1}{2}}^{1} \frac{\ln (1 - 2 x) \ln^{3} x}{x} d x - i \frac{π}{4} \ln^{4} 2 ∵ \ln (2 x - 1) = \ln (1 - 2 x) - i π \\ = \frac{1}{5} \ln^{5} 2 + \sum_{n \geq 1} \frac{2^{n}}{n} \int_{1 / 2}^{1} x^{n - 1} \ln^{3} x d x - i \frac{π}{4} \ln^{4} 2 \\ = \frac{1}{5} \ln^{5} 2 + \sum_{n \geq 1} \frac{2^{n}}{n} (\frac{\ln^{3} 2}{n 2^{n}} + \frac{3 \ln^{2} 2}{n^{2} 2^{n}} + \frac{6 \ln 2}{n^{3} 2^{n}} + \frac{6}{n^{4} 2^{n}} - \frac{6}{n^{4}}) - i \frac{π}{4} \ln^{4} 2 \\ = \frac{1}{5} \ln^{5} 2 + \ln^{3} 2 ζ (2) + 3 \ln^{2} 2 ζ (3) + 6 \ln 2 ζ (4) + 6 ζ (5) - 6 {Li}_{5} (2) - i \frac{π}{4} \ln^{4} 2 \end{aligned}$
ここで
${Li}_{5} (x) = - \frac{7}{4} ζ (4) \ln (- x) - \frac{1}{6} ζ (2) \ln^{3} (- x) - \frac{1}{120} \ln^{5} (- x) + {Li}_{5} (\frac{1}{x})$
$x = 2$ を代入して
${Li}_{5} (2) = 2 \ln 2 ζ (4) + \frac{1}{3} \ln^{3} 2 ζ (2) - \frac{1}{120} \ln^{5} 2 + {Li}_{5} (\frac{1}{2}) - i \frac{π}{24} \ln^{4} 2$
これを先の結果に代入して
$\int_{0}^{1} \frac{\ln (1 - x) \ln^{3} (1 + x)}{1 + x} d x = - 6 {Li}_{5} (\frac{1}{2}) + 6 ζ (5) - 6 \ln 2 ζ (4) + 3 \ln^{2} 2 ζ (3) - \ln^{3} 2 ζ (2) + \frac{1}{4} \ln^{5} 2$