$$ \begin{aligned} \int \frac {\ln ^{2}x}{x-a}dx &=\ln \left (1-\frac{x}a\right )\ln ^{2}x-2\int \frac {\ln x}x\ln \left (1-\frac {x}a\right )dx\\ &=\ln \left (1-\frac {x}a\right )\ln ^{2}x+2\operatorname{Li} _{2}\left (\frac {x}a\right )\ln x-2\int \frac {\operatorname {Li} _2 \left (\frac {x}a\right ) }xdx\\ &=\ln \left (1-\frac {x}a\right )\ln ^{2}x+2\operatorname {Li} _2 \left (\frac {x}a\right )\ln x-2\operatorname{Li} _{3}\left (\frac {x}a\right )+Const. \end{aligned} $$

$$ \begin {aligned} \int _{0}^{1}\frac {\ln ^{2}x}{1+x^{2}}dx &=\int _{1}^{\infty }\frac {\ln ^{2}x}{1+x^{2}}dx\\ &=\frac {1}4\left (\int _{0}^{\infty }\frac {\ln ^{2}x}{1+x^{2}}dx+\Re \int _{0}^{\infty }\frac {(\ln x+\pi i)^2+\pi ^{2}}{1+x^{2}}dx\right )\\ &=\frac {1}4\left (\int _{-\infty }^{\infty }\frac {\ln ^{2}x}{1+x^{2}}dx+\frac {\pi ^{3}}2\right )\\ &=\frac {1}4\left (\frac {2\pi i}{2i}\ln ^{2}i+\frac {\pi ^{3}}2\right )\\ &=\frac {1}4\left (-\frac {\pi ^{3}}4+\frac {\pi ^{3}}2\right )\\ &=\textcolor {blue}{\frac {\pi ^{3}}{16}}. \\ \end {aligned} $$

$$ \begin {aligned} \\ \int _{0}^{\infty }\frac {\ln x\ln |1-x|}{1+x^{2}}dx &=\int _{0}^{1}\frac {\ln x\ln (x-1)}{1+x^{2}}dx+\int _{1}^{\infty }\frac {\ln x\ln (x-1)}{1+x^{2}}dx\\ &=\int _{0}^{1}\frac {\ln x\ln (1-x)}{1+x^{2}}dx-\int _{0}^{1}\frac {\ln x(\ln (1-x)-\ln x)}{1+x^{2}}dx\\ &=\int _{0}^{1}\frac {\ln ^{2}x}{1+x^{2}}dx\\ &=\textcolor {blue}{\frac {\pi ^{3}}{16}}. \\ \end {aligned} $$

$$ \begin {aligned} \\ \int _{0}^{\infty }\frac {\ln x\ln (1+x)}{1+x^{2}}dx &=\Re \int _{-\infty }^{0}\frac {\ln (x+0i)\ln (1-x)}{1+x^{2}}dx\\ &=\Re\left (2\pi i \underset{z=i}{\operatorname{Res}\ }\frac {\ln (z)\ln (1-z)}{1+z^{2}}-\int _{0}^{\infty }\frac {\ln x\ln (1-x-0i)}{1+x^2}dx\right )\\ &=\Re \left (\pi \cdot \frac {\pi i}2\left (\frac {\ln 2}2-\frac {\pi i}4\right )\right )-\int _{0}^{\infty }\frac {\ln x\ln |1-x|}{1+x^{2}}dx\\ &=\frac {\pi ^{3}}8-\frac {\pi ^{3}}{16}\\ &=\textcolor {blue}{\frac {\pi ^{3}}{16}}. \end {aligned} $$

$$ \begin {aligned} \int _{0}^{\infty }\frac {\ln ^2(1+x)}{1+x^{2}}dx &=\int _{1}^{\infty }\frac {\ln ^2x}{1+(x-1)^{2}}dx\\ &=\int _{0}^{1}\frac {\ln ^{2}x}{x^{2}+(1-x)^{2}}dx\\ &=\int _{0}^{1}\frac {\ln ^{2}x}{2x^2-2x+1}dx\\ &=2\int _{0}^{1}\frac {\ln ^{2}x}{(2x-1)^{2}+1}dx\\ &=2\Im \int _{0}^{1}\frac {\ln ^{2}x}{2x-1-i}dx\\ &=\Im\int _{0}^{1}\frac {\ln ^{2}x}{x-\frac {1}{1-i}}dx\\ &=\Im \left [\ln (1-(1-i)x)\ln ^{2}x+2\operatorname {Li} _2 ((1-i)x)\ln x-2\operatorname{Li} _{3}((1-i)x) \right ]_0^1\\ &=-2\Im \operatorname{Li} _{3}(1-i) \\ &=\textcolor {blue}{2\Im\operatorname{Li} _{3}(1+i) }. \end {aligned} $$

$$ \begin {aligned} \int _{0}^{1}\frac {\ln (1-x)\ln (1+x^{2})}xdx &=\int _{0}^{1}\sum _{n,m=1}^\infty \frac {x^{n}(-x^{2})^{m}}{nmx}dx\\ &=\sum _{n,m=1}^\infty \frac {(-1)^m}{nm}\int _{0}^{1}x^{n+2m-1}dx\\ &=\sum _{n,m=1}^\infty \frac {(-1)^{m}}{nm(n+2m)}\\ &=\sum _{m=1}^\infty \frac {(-1)^{m}}m\sum _{n=1}^{\infty }\frac {1}{n(n+2m)}\\ &=\sum _{m=1}^\infty \frac {(-1)^{m}}{2m^2}\sum _{n=1}^\infty \left (\frac {1}n-\frac {1}{n+2m}\right )\\ &=2\sum _{m=1}^\infty \frac {(-1)^{m}H_{2m}}{4m^{2}}\\ &=2\Re \sum _{n=1}^\infty \frac {i^{n}H_n}{n^2}\\ &=2\Re \int _{0}^{i}\frac {1}z\sum _{n=1}^\infty \frac {H_n}nz^ndz\\ &=2\Re \int _{0}^{i}\frac {1}z\int _{0}^{z}\frac {1}t\sum _{n=1}^\infty H_nt^ndtdz\\ &=2\Re \int _{0}^{i}\frac {1}z\int _{0}^{z}\frac {1}t\cdot \frac {-\log (1-t)}{1-t}dtdz\\ &=2\Re \int _{0}^{i}\frac {1}z\int _{0}^{z}\left (\frac {-\log (1-t)}t+\frac {-\log (1-t)}{1-t}\right )dtdz\\ &=2\Re \int _{0}^{i}\frac {1}z\left (\operatorname{Li} _{2}(z)+\frac {\log ^{2}(1-z)}2 \right )dz\\ &=2\Re \left [\operatorname{Li} _{3}(i)-\frac {1}2\int _{1}^{1-i}\frac {\log ^{2}z}{1-z} dz\right ]\\ &=-\frac {3}{16}\zeta (3)-\Re \left [-\log (1-z)\log ^{2}z-2\operatorname{Li} _{2}(z)\log z +2\operatorname{Li} _{3}(z) \right ]_1^{1-i}\\ &=-\frac {3}{16}\zeta (3)-\Re \left (-\frac {\pi i}2\log ^{2}(1-i)-2\left (\frac {\pi ^{2}}{16}+i\left (\beta (2)+\frac {\pi \ln 2}4\right )\right )\log (1-i)+2\operatorname{Li} _{3}(1-i)-2\zeta (3) \right )\\ &=-\frac {3}{16}\zeta (3)+\frac {\pi ^2\ln 2}8+\frac {\pi ^{2}\ln 2}{16}-\frac {\pi }2\left (\beta (2)+\frac {\pi \ln 2}4\right )-\frac {35}{32}\zeta (3)-\frac {\pi ^{2}\ln 2}{16}+2\zeta (3)\\ &=\textcolor {blue}{\frac {23}{32}\zeta (3)-\frac {1}2\pi \beta (2)}. \end {aligned} $$

$$ \begin {aligned} \int _{0}^{1}\frac {\ln ^{2}(1-x)}{1+x^{2}}dx &=\Im \int _{0}^{1}\frac {\ln ^{2}(1-x)}{x-i}dx\\ &=\Im \int _{0}^{1}\frac {\ln ^{2}x}{1-i-x}dx\\ &=\Im \left [-\ln \left (1-\frac {x}{1-i}\right )\ln ^{2}x-2\operatorname {Li} _2 \left (\frac {x}{1-i}\right )\ln x+2\operatorname{Li} _{3} \left (\frac {x}{1-i}\right )\right ]_0^1\\ &=\textcolor {blue}{2\Im \operatorname{Li} _{3}\left (\frac {1+i}2\right ) }. \end {aligned} $$

$$ \begin {aligned} \int _{0}^{1}\frac {x\ln ^2(1-x)}{1+x^2}dx &=2\Re \operatorname{Li} _{3}\left (\frac {1+i}2\right ) \\ &=\textcolor {blue}{\frac {35}{32}\zeta (3)-\frac {5\pi ^{2}\ln 2}{96}+\frac {\ln ^{3}2}{24}}. \end {aligned} $$

$$ \begin {aligned} \int _{0}^{1}\frac {\ln x\ln (1-x)}{1+x^{2}} &=\frac {1}\pi \Im\int _{0}^{1}\frac {\ln x\ln (1-x)\ln (x-1)}{1+x^{2}}dx\\ &=\frac {1}\pi \Im\left (\underset{z=i}{\mathrm{Res}}\frac {\ln z\ln (1-z)\ln (z-1)}{1+z^{2}}-\left (\int _{-\infty }^{0}+\int _{1}^{\infty }\right )\frac {\ln x\ln (1-x)\ln (x-1)}{1+x^{2}}dx\right )\\ &=\frac {1}\pi \Im \left (\pi \cdot \frac {\pi i}2\left (\frac {\ln 2}2-\frac {\pi i}4\right )\left (\frac {\ln 2}2+\frac {3\pi i}4\right )-\int _{0}^{\infty }\frac {(\ln x+\pi i)\ln (1+x)(\ln (1+x)+\pi i)}{1+x^{2}}dx-\int _{1}^{\infty }\frac {\ln x\ln (x-1)(\ln (x-1)-\pi i)}{1+x^{2}}dx\right )\\ &=\frac {\pi }2\left (\frac {\ln ^{2}2}4+\frac {3\pi ^{2}}{16}\right )-\int _{0}^{\infty }\frac {\ln ^{2}(1+x)+\ln x\ln (1+x)}{1+x^2}dx+\int _{1}^{\infty }\frac {\ln x\ln (x-1)}{1+x^{2}}dx\\ &=\frac {\pi \ln ^{2}2}{8}+\frac {3\pi ^{3}}{32}-2\Im\operatorname{Li} _{3}(1+i)-\frac {\pi ^{3}}{16}-\int _{0}^{1}\frac {\ln x(\ln (1-x)-\ln x)}{1+x^{2}}dx \\ &=\frac {\pi \ln ^{2}2}8+\frac {3\pi ^{3}}{32}-2\Im \operatorname{Li} _{3}(1+i)-\frac {\pi ^{3}}{16}-\int _{0}^{1}\frac {\ln x\ln (1-x)}{1+x^{2}}dx+\frac {\pi ^{3}}{16} \\ &=\textcolor{blue}{\frac{3\pi^3}{64}+\frac {\pi \ln ^{2}2}{16}-\Im \operatorname{Li} _{3}(1+i) }\\ &=\textcolor{blue}{-\frac {\pi ^{3}}{128}-\frac {\pi \ln ^{2}2}{32}+\Im \operatorname{Li} _{3}\left (\frac {1+i}2\right ) }. \end {aligned} $$

$$ \int_0^1 \frac{\ln x \ln(1+x^2)}{1+x^2}dx $$ を求めます。 $$ \begin {aligned} \int _{0}^{\frac {\pi }4}\frac {x\ln \sin x}{\sin 2x}dx &=\Re \int _{0}^{\frac {\pi }4}\frac {x\ln \frac {1-e^{2ix}}2}{\sin 2x}dx\\ &=\Re \int _{1}^{i}\frac {\frac {\ln z}{2i}\ln \frac {1-z}2}{\frac {z^{2}-1}{2iz}}\frac {dz}{2iz}\\ &=\frac {1}2\Im\int _{1}^{i}\frac {\ln z\ln \frac {1-z}2}{z^{2}-1}dz\\ &=\frac {1}2\Im \int _{0}^{1}\frac {\ln (it)\ln \frac {1-it}2}{-t^{2}-1}idt+\frac {1}2\Im\int _{0}^{1}\frac {\ln x\ln \frac {1-x}2}{1-x^{2}}dx\\ &=-\frac {1}2\Re \int _{0}^{1}\frac {\left (\ln t+\frac {\pi i}2\right )\left (\frac {1}2\ln (1+t^2)-\ln 2-i\arctan t\right )}{1+t^2}dt\\ &=-\frac {1}4\int _{0}^{1}\frac {\ln t\ln (1+t^{2})}{1+t^{2}}dt+\frac {\ln 2}2\int _{0}^{1}\frac {\ln t}{1+t^{2}}dt-\frac {\pi }4\int _{0}^{1}\frac {\arctan t}{1+t^{2}}dt\\ &=-\frac {1}4\int _{0}^{1}\frac {\ln t\ln (1+t^{2})}{1+t^{2}}dt-\frac {G\ln 2}2-\frac {\pi ^{3}}{128}\\ \int _{0}^{1}\frac {\ln x\ln (1+x^{2})}{1+x^{2}}dx&=-\frac {\pi ^{3}}{32}-2G\ln 2-4\int _{0}^{\frac {\pi }4}\frac {x\ln \sin x}{\sin 2x}dx\\ &=-\frac {\pi ^{3}}{32}-2G\ln 2-4\left (-\frac {\pi ^{3}}{32}-\frac {\pi \ln ^{2}2}{32}-\frac {G\ln 2}4+\frac {1}2\Im \operatorname{Li} _{3}(1+i) \right )\\ &=\textcolor {blue}{\frac {3\pi ^{3}}{32}+\frac {\pi \ln ^{2}2}8-G\ln 2-2\Im \operatorname{Li} _{3}(1+i) }. \end {aligned} $$

$$ \begin {aligned} \int _{0}^{1}\frac {\ln (1+x^{2})\arctan x}xdx &=\int _{0}^{1}\Im \frac {\ln ^2(1+ix)}xdx\\ &=\int _{1}^{1+i}\Im \frac {\ln ^2 z}{(z-1)/i}\frac {dz}i\\ &=\left [\Im \left (\ln (1-z)\ln ^{2}z+2\operatorname{Li} _{2}(z)\ln z-2\operatorname{Li} _{3}(z) \right )\right ]_{1+0i}^{1+i}\\ &=\Im \left (-\frac {\pi i}2\left (\frac {\ln 2}2+\frac {\pi i}4\right )^2+2\operatorname {Li} _2 (1+i)\left (\frac {\ln 2}2+\frac {\pi i}4\right )-2\operatorname{Li} _{3}(1+i)-2\zeta (3)\right )\\ &=-\frac {\pi }2\left (\frac {\ln ^{2}2}4-\frac {\pi ^{2}}{16}\right )+2\left (\frac {\pi ^{2}}{16}\cdot \frac {\pi }4+\left (G+\frac {\pi \ln 2}4\right )\cdot \frac {\ln 2}2\right )-2\Im \operatorname{Li} _{3}(1+i)\\ &=\frac {\pi ^{3}}{32} -\frac {\pi \ln ^{2}2}{8}+\frac {\pi ^{3}}{32}+G\ln 2+\frac {\pi \ln ^{2}2}4-2\Im \operatorname{Li} _{3}(1+i) \\ &=\textcolor {blue}{\frac {\pi ^{3}}{16}+\frac {\pi \ln ^{2}2}8+G\ln 2-2\Im \operatorname{Li} _{3}(1+i) }. \end {aligned} $$

$$ \begin {aligned} \int _{0}^{\frac {\pi }4}\frac {x\ln \sin x}{\sin 2x}dx &=\int _{0}^{\frac {\pi }4}\frac {x\ln \cos +x\ln \tan x}{\sin 2x}dx\\ &=-\frac {1}4\int _{0}^{1}\frac {\arctan x\ln (1+x^{2})}{x}dx+\frac {1}2\int _{0}^{1}\frac {\arctan x\ln x}xdx\\ &=-\frac {1}4\left (\frac {\pi ^{3}}{16}+\frac {\pi \ln ^{2}2}8+G\ln 2-2\Im \operatorname{Li} _{3}(1+i) \right )-\frac {1}4\int _{0}^{1}\frac {\ln ^{2}x}{1+x^{2}}dx\\ &=-\frac {\pi ^{3}}{64}-\frac {\pi \ln ^{2}2}{32}-\frac {G\ln 2}4+\frac {1}2\Im\operatorname{Li} _{3}(1+i)-\frac {\pi ^{3}}{64} \\ &=\textcolor {blue}{-\frac {\pi ^{3}}{32}-\frac {\pi \ln ^{2}2}{32}-\frac {G\ln 2}{4}+\frac {1}2\Im \operatorname{Li} _{3}(1+i) }. \end {aligned} $$