\(G\)はカタラン定数です。
\(\ln z\)の偏角は,\(-\pi <\arg z\leq \pi \)とします。
基本的な不定積分
$$
\begin {aligned}
\int \frac {\ln x}{x-a}dx
&=\ln \left (1-\frac {x}a\right )\ln x+\operatorname {Li} _2 \left (\frac {x}a\right )+Const.
\end {aligned}
$$
$$
\begin {aligned}
\int \frac {\ln x}{x-a}dx
&=\ln \left (1-\frac {x}a\right )\ln x-\int \frac {\ln \left (1-\frac {x}a\right )}{x}dx\\
&=\ln \left (1-\frac {x}a\right )\ln x+\operatorname {Li} _2 \left (\frac {x}a\right )+Const.
\end {aligned}
$$
一番単純なやつ
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln x}{1-x}dx&=-\frac {\pi ^{2}}6\\
\int _{0}^{1}\frac {\ln x}{1+x}dx&=-\frac {\pi ^{2}}{12}\\
\int _{0}^{1}\frac {\ln x}{1-x^{2}}dx&=-\frac {\pi ^{2}}8\\
\int _{1}^{\infty }\frac {\ln x}{1-x^{2}}dx&=-\frac {\pi ^{2}}8\\
\int _{0}^{1}\frac {\ln x}{1+x^{2}}dx&=-G\\
\int _{1}^{\infty }\frac {\ln x}{1+x^{2}}dx&=G\\
\int _{0}^{1}\frac {x\ln x}{1-x^{2}}dx&=-\frac {\pi ^{2}}{24}\\
\int _{0}^{1}\frac {x\ln x}{1+x^{2}}dx&=-\frac {\pi ^{2}}{48}
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln x}{1-x}dx
&=-\left [\ln x\ln (1-x)+\operatorname {Li} _2 (x)\right ]_0^1\\
&=-\operatorname {Li} _2 (1)\\
&=\textcolor {blue}{-\frac {\pi ^{2}}6}.
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln x}{1+x}dx
&=\left [\ln x\ln (1+x)+\operatorname {Li} _2 (-x)\right ]_0^1\\
&=\operatorname {Li} _2 (-1)\\
&=\textcolor {blue}{-\frac {\pi ^{2}}{12}}.
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln x}{1-x^{2}}dx&=\frac {1}2\left (\int _{0}^{1}\frac {\ln x}{1-x}dx+\int _{0}^{1}\frac {\ln x}{1+x}dx\right )\\
&=\frac {1}2\left (-\frac {\pi ^{2}}6-\frac {\pi ^{2}}{12}\right )\\
&=\textcolor {blue}{-\frac {\pi ^{2}}8}.\\
\int _{1}^{\infty }\frac {\ln x}{1-x^{2}}dx&=\int _{0}^{1}\frac {\ln x}{1-x^{2}}dx\\
&=\textcolor {blue}{-\frac{\pi ^{2}}{8}}.
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln x}{1+x^{2}}dx
&=\Im \int _{0}^{1}\frac {\ln x}{x-i}dx\\
&=\Im \left [\ln x\ln (1+ix)+\operatorname {Li} _2 \left (-ix\right )\right ]_0^1\\
&=\Im \operatorname {Li} _2 (-i)\\
&=\textcolor {blue}{-G}.\\
\int _{1}^{\infty }\frac {\ln x}{1+x^{2}}dx
&=-\int _{0}^{1}\frac {\ln x}{1+x^{2}}dx\\
&=\textcolor {blue}{G}.
\end {aligned}
$$
複雑なやつ
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln (1+x)}{1+x^{2}}dx
&=\frac {\pi \ln 2}8
\\
\int _{0}^{\infty }\frac {\ln (1+x)}{1+x^{2}}dx
&=\frac {\pi \ln 2}4+G
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{1}\frac {\ln (1+x)}{1+x^{2}}dx
&=\int _{0}^{\frac {\pi }4}\ln (1+\tan x)dx\\
&=\int _{0}^{\frac {\pi }4}\ln \left (1+\tan \left (\frac {\pi }4-x\right )\right )dx\\
&=\int _{0}^{\frac {\pi }4}\ln \left (1+\frac {1-\tan x}{1+\tan x}\right )dx\\
&=\int _{0}^{\frac {\pi }4}\ln 2dx-\int _{0}^{\frac {\pi }4}\ln (1+\tan x)dx\\
&=\frac {1}2\int _{0}^{\frac {\pi }4}\ln 2dx\\
&=\textcolor {blue}{\frac {\pi \ln 2}8}.
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{\infty }\frac {\ln (1+x)}{1+x^{2}}dx
&=\frac {1}\pi \Im \int _{0}^{\infty }\frac {(\ln x+\pi i)\ln (1+x)}{1+x^{2}}dx\\
&=\frac {1}\pi \Im \int _{-\infty }^{0}\frac {\ln (x+0i)\ln (1-x)}{1+x^{2}}dx\\
&=\frac {1}\pi \Im \left (2\pi i\underset{z=i}{\operatorname{Res}}\frac {\ln z\ln (1-z)}{1+z^{2}}-\int _{0}^{\infty }\frac {\ln x\ln (1-x-0i)}{1+x^{2}}dx\right )\\
&=\frac {1}\pi \Im \left (\pi \cdot \frac {\pi i}2\left (\frac {\ln 2}2-\frac {\pi i}4\right )\right )+\frac {1}\pi \int _{1}^{\infty }\frac {\ln x\cdot \pi }{1+x^{2}}dx\\
&=\textcolor {blue}{\frac {\pi \ln 2}4+G}.
\end {aligned}
$$
これ、積分区間を\(\int_0^1\)と\(\int_0^\infty \)に分けて一つ目の積分を使った方が自然なのかもしれないけど、まあいいや
三角関数絡み
$$
\begin {aligned}
\int _{0}^{\frac {\pi }2}\ln (1+\sin x)dx
&=2\beta (2)-\frac {\pi \ln 2}2
\end {aligned}
$$
$$
\begin {aligned}
\int _{0}^{\frac {\pi }2}\ln (1+\sin x)dx
&=\int _{0}^{\frac {\pi }2}\ln \sin xdx+\int _{0}^{\frac {\pi }2}\ln \left (\frac {1}{\sin x}+1\right )dx\\
&=-\frac {\pi \ln 2}2+2\int _{0}^{1}\ln \left (\frac {1+x^{2}}{2x}+1\right )\frac {dx}{1+x^{2}}\\
&=-\frac {\pi \ln 2}2+4\int _{0}^{1}\frac {\ln (1+x)}{1+x^{2}}dx-2\int _{0}^{1}\frac {\ln 2+\ln x}{1+x^{2}}dx\\
& =-\frac {\pi \ln 2}2+\frac {\pi \ln 2}2-\frac {\pi \ln 2}2+2\beta (2)\\
&=\textcolor {blue}{2\beta (2)-\frac {\pi \ln 2}2}.
\end {aligned}
$$