Legendre chi function

$${\displaystyle \chi _{-1}\left(x\right)={\frac {x\left(1+x^{2}\right)}{\left(1-x^{2}\right)^{2}}}}$$ $${\displaystyle \chi _{0}\left(x\right)={\frac {x}{1-x^{2}}}}$$ $${\displaystyle \chi _{1}(x)=\operatorname {arctanh} (x)={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)}$$ $${\displaystyle \chi _{2}(x)=\int _{0}^{x}{\frac {\operatorname {arctanh} (t)}{t}}\mathrm {d} t}$$ $${\displaystyle \chi _{\infty }(x)=x}$$ $${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\chi _{\nu }(x)={\frac {\chi _{\nu -1}(x)}{x}}}$$ $${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\chi _{2}(x)={\frac {\operatorname {arctanh} (x)}{x}}}$$ $${\displaystyle \chi _{2}(x)+\chi _{2}(1/x)={\frac {\pi ^{2}}{4}}-{\frac {i\pi }{2}}\ln |x|}$$ $${\displaystyle \chi _{2}(x)+\chi _{2}\left({\frac {1-x}{1+x}}\right)={\frac {\pi ^{2}}{8}}+\ln(x)\operatorname {arctanh} (x),\quad x\in (0,1)}$$

It takes the special values:

$${\displaystyle \chi _{2}(1)={\frac {\pi ^{2}}{8}}}$$ $${\displaystyle \chi _{2}(-1)=-{\frac {\pi ^{2}}{8}}}$$ $${\displaystyle \chi _{2}({\sqrt {2}}-1)={\frac {\pi ^{2}}{16}}-{\frac {1}{4}}\ln ^{2}({\sqrt {2}}+1)}$$ $${\displaystyle \chi _{2}\left({\frac {{\sqrt {5}}-1}{2}}\right)={\frac {\pi ^{2}}{12}}-{\frac {3}{4}}\ln ^{2}\left({\frac {{\sqrt {5}}+1}{2}}\right)}$$ $${\displaystyle \chi _{2}({\sqrt {5}}-2)={\frac {\pi ^{2}}{24}}-{\frac {3}{4}}\ln ^{2}\left({\frac {{\sqrt {5}}+1}{2}}\right)}$$ $${\displaystyle \chi _{2}(i)=iK,}$$

where $i$ is the imaginary unit and K is Catalan's constant.^[1] Other special values include:

$${\displaystyle \chi _{n}(1)=\lambda (n)}$$ $${\displaystyle \chi _{n}(i)=i\beta (n),}$$

where $\lambda (n)$ is the Dirichlet lambda function and $\beta (n)$ is the Dirichlet beta function.^[1]

$${\displaystyle \int _{0}^{\pi /2}\arcsin(r\sin \theta )\,\mathrm {d} \theta =\chi _{2}(r),\qquad \int _{0}^{\pi /2}\arccos(r\cos \theta )\,\mathrm {d} \theta =\left({\frac {\pi }{2}}\right)^{2}-\chi _{2}(r)\qquad {\rm {if}}~~|r|\leq 1}$$ $${\displaystyle \int _{0}^{\pi /2}\arctan(r\sin \theta )\,\mathrm {d} \theta =-{\frac {1}{2}}\int _{0}^{\pi }{\frac {r\theta \cos \theta }{1+r^{2}\sin ^{2}\theta }}\,\mathrm {d} \theta =2\chi _{2}\!\!\left({\frac {{\sqrt {1+r^{2}}}-1}{r}}\right)}$$ $${\displaystyle \int _{0}^{\pi /2}\arctan(p\sin \theta )\arctan(q\sin \theta )\,\mathrm {d} \theta =\pi \chi _{2}\!\!\left({\frac {{\sqrt {1+p^{2}}}-1}{p}}\cdot {\frac {{\sqrt {1+q^{2}}}-1}{q}}\right)}$$ $${\displaystyle \int _{0}^{\alpha }\int _{0}^{\beta }{\frac {\mathrm {d} x\,\mathrm {d} y}{1-x^{2}y^{2}}}=\chi _{2}(\alpha \beta )\qquad {\rm {if}}~~|\alpha \beta |\leq 1}$$