積分(メモ)

\begin{align} \displaystyle\int\frac{\ln^3x\ln(1-x)}{1-x}dx&=-\int\sum_{k\geq1}H_kx^k\ln^3k\,dx\\ &=-\sum_{k\geq1}H_k\int x^k\ln^3k\,dx\\ &=-\sum_{k\geq1}H_k\frac{\partial^3}{\partial k^3}\left[\int x^k dx\right]\\ &=-\sum_{k\geq1}H_k\frac{\partial^3}{\partial k^3}\left[\frac{x^{k+1}}{k+1}\right]\\ &=-\sum_{k\geq1}H_k\left[\frac{x^{k+1}\ln^3x}{k+1}-\frac{3x^{k+1}\ln^2x}{(k+1)^2}+\frac{6x^{k+1}\ln x}{(k+1)^3}-\frac{6x^{k+1}}{(k+1)^4}\right]\\ &=-\ln^3x\sum_{k\geq1}\frac{H_{k+1}x^{k+1}}{k+1}+\ln^3x\sum_{k\geq1}\frac{x^{k+1}}{(k+1)^2}+3\ln^2x\sum_{k\geq1}\frac{H_{k+1}x^{k+1}}{(k+1)^2} -3\ln^2x\sum_{k\geq1}\frac{x^{k+1}}{(k+1)^3} -6\ln x\sum_{k\geq1}\frac{H_{k+1}x^{k+1}}{(k+1)^3} +6\ln x\sum_{k\geq1}\frac{x^{k+1}}{(k+1)^4} +6\sum_{k\geq1}\frac{H_{k+1}x^{k+1}}{(k+1)^4} -6\ln x\sum_{k\geq1}\frac{x^{k+1}}{(k+1)^4}\\ &=-H_1(x)\ln^3x+\mathrm{Li}_2(x)\ln^3x+3H_2(x)\ln^2x-3\mathrm{Li}_3(x)\ln^2x-6\mathrm{H_3}(x)\ln x+6\mathrm{Li}_4(x)\ln x+6H_4(x)-6\mathrm{Li_5}(x) \end{align}

\begin{align} \displaystyle\int_0^1\frac{\ln x\ln(1+x)}{1-x}dx&=\left[\int_0^x\frac{\ln y}{1-y}dy\ln(1+x)\right]_0^1-\int_0^1\frac{J(x)}{1+x}dx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left(J(x)=\int_0^x\frac{\ln y}{1-y}dy\,\,\,\,x\in[0,1]\right)\\ &=-\zeta(2)\ln2-\int_0^1\int_0^1\frac{x\ln xy}{(1-xy)(1+x)}dydx\\ &=-\zeta(2)\ln2-\int_0^1\left(\int_0^1\frac{x\ln y}{(1-xy)(1+x)}dx\right)dy-\int_0^1\left(\int_0^1\frac{x\ln x}{(1-xy)(1+x)}dy\right)dx\\ &=-\zeta(2)\ln2+\int_0^1\left[\frac{\ln(1-xy)}{y(1+y)}+\frac{\ln(1+x)}{1+y}\right]_{x=0}^{x=1}\ln y\,dy+\int_0^1\left[\frac{\ln(1-xy)}{1+x}\right]_{y=0}^{y=1}\ln x\,dx\\ &=-\zeta(2)\ln2+\int_0^1\frac{\ln(1-y)\ln y}{y(1+y)}dy+\ln2\int_0^1\frac{\ln y}{1+y}dy+\int_0^1\frac{\ln(1-x)\ln x}{1+x}dx\\ &=-\zeta(2)\ln2+\int_0^1\frac{\ln(1-y)\ln y}{y}dy+\ln2\int_0^1\frac{\ln y}{1+y}dy\\ &=-\zeta(2)\ln2+\frac{1}{2}\left(\ln^2x\ln(1-x)+\int_0^1\frac{\ln^2 y}{1-y}dy\right)+\ln2\left(\int_0^1\frac{\ln y}{1-y}dy-\int_0^1\frac{2y\ln y}{1-y^2}dy\right)\\ &=-\zeta(2)\ln2+\frac{1}{2}\int_0^1\frac{\ln^2y}{1-y}dy+\ln2\left(\int_0^1\frac{\ln y}{1-y}dy-\frac{1}{2}\int_0^1\frac{\ln y}{1-y}dy\right)\\ &=-\frac{3}{2}\zeta(2)\ln2+\zeta(3) \end{align}

\begin{align} \displaystyle I=\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^22\ln^2(1-x)}{1-x}\right)dx&=\int_0^1\frac{\ln^2(1-x)}{1-x}(\ln^2(1+x)-\ln^22)dx\\ &\overset{IBP}{=}\frac{2}{3}\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{1+x}dx \end{align}
ここで
\begin{align} a^3b=\frac{(a+b)^4}{8}-\frac{(a-b)^4}{8}-ab^3\xrightarrow[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{align}
であるから
\begin{align} \displaystyle I&=\frac{1}{12}\int_0^1\frac{\ln^4(1-x^2)}{1+x}dx -\frac{1}{12}\int_0^1\frac{\ln^4(1-x)-\ln^4(1+x)}{1+x}dx-\frac{2}{3}\int_0^1\frac{\ln(1-x)\ln^3(1+x)}{1+x}dx\\ &=\left(\frac{16}5\ln^52-16\ln^32\zeta(2)+48\ln^22\zeta(3)-54\ln2\zeta(4)-24\zeta(2)\zeta(3)+72\zeta(5)\right)-\left(\frac{45}2\zeta(5)\right)-\left(-6\operatorname{Li}_5\left(\frac12\right)+6\zeta(5)-6\ln2\zeta(4)+3\ln^22\zeta(3)-\ln^32\zeta(2)+\frac14\ln^52\right)\\ &=\frac{1}{8}\zeta(5)-\frac{1}{2}\ln2\zeta(4)+2\ln^22\zeta(3)-\frac{2}{3}\ln^32\zeta(2)-2\zeta(2)\zeta(3)+\frac{1}{10}\ln^52+4\mathrm{Li}_5\left(\frac{1}{2}\right) \end{align}