Question Number 136922 by malwan last updated on 27/Mar/21
$$\int\:{ln}\mid{tan}^{−\mathrm{1}} {x}\mid\:{dx}\:=\:? \\ $$
Answered by Olaf last updated on 27/Mar/21
$$\mathrm{F}\left({x}\right)\:=\:\int\mathrm{ln}\mid\mathrm{atan}{x}\mid\:{dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{atan}{x} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\int\mathrm{ln}\mid{u}\mid\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {u}\right)\:{du} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{ln}\mid{u}\mid\mathrm{tan}{u}−\int\frac{\mathrm{tan}{u}}{{u}}\:{du} \\ $$$$\mathrm{tan}{u}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right){B}_{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}{u}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{ln}\mid{u}\mid\mathrm{tan}{u}−\int\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right){B}_{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}{u}^{\mathrm{2}{n}−\mathrm{2}} {du} \\ $$$$\mathrm{F}\left({u}\right)\:=\:\mathrm{ln}\mid{u}\mid\mathrm{tan}{u}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right){B}_{\mathrm{2}{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}{u}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\mathrm{F}\left({x}\right)\:=\:{x}\mathrm{ln}\mid\mathrm{atan}{x}\mid−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right){B}_{\mathrm{2}{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}\right)!}\mathrm{atan}^{\mathrm{2}{n}−\mathrm{1}} {x} \\ $$
Commented by malwan last updated on 28/Mar/21
$${thank}\:{you}\:{so}\:{much}\:{sir} \\ $$