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Question Number 117403 by bemath last updated on 11/Oct/20

$$\:\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\mathrm{arc}\:\mathrm{tan}\:\mathrm{x}\right)^{\mathrm{2}} \:\mathrm{dx}\:=? \\$$

Commented by MJS_new last updated on 11/Oct/20

$$\mathrm{use}\:\mathrm{arctan}\:{x}\:=\frac{\mathrm{ln}\:\left(\mathrm{1}+\mathrm{i}{x}\right)\:−\mathrm{ln}\:\left(\mathrm{1}−\mathrm{i}{x}\right)}{\mathrm{2i}}\:\Rightarrow \\$$$$−\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left(\mathrm{ln}^{\mathrm{2}} \:\left(\mathrm{1}+\mathrm{i}{x}\right)\:−\mathrm{2ln}\:\left(\mathrm{1}+\mathrm{i}{x}\right)\:\mathrm{ln}\:\left(\mathrm{1}−\mathrm{i}{x}\right)\:+\mathrm{ln}^{\mathrm{2}} \:\left(\mathrm{1}−\mathrm{i}{x}\right)\right){dx} \\$$$$\mathrm{and}\:\mathrm{these}\:\mathrm{are}\:\mathrm{solveable}\:\left(\mathrm{by}\:\mathrm{parts}\right) \\$$

Answered by mindispower last updated on 11/Oct/20

$$=\left[{xarctan}^{\mathrm{2}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{arctan}\left({x}\right){dx} \\$$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}−{I} \\$$$${let}\:{arctan}\left({x}\right)={t} \\$$$${I}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{2}{tg}\left({t}\right){t}}{\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)}.\left(\mathrm{1}+{tg}^{\mathrm{2}} \left({t}\right)\right){dt} \\$$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {tg}\left({t}\right){tdt}\:{by}\:{part}=−\mathrm{2}\left(\left[−{tln}\left({cost}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \right. \\$$$$+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({t}\right)\right){dt} \\$$$$=−\frac{\pi}{\mathrm{4}}{ln}\left(\mathrm{2}\right)−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({t}\right)\right){dt} \\$$$${one}\:{way}\:{too}\:{find}\:{it} \\$$$${let}\:{a}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cos}\left({t}\right)\right){dt} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}}{\mathrm{2}}{ln}\left({sin}\left({t}\right){cos}\left({t}\right).\frac{\mathrm{1}}{{tg}\left({t}\right)}\right){dt} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\frac{{sin}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)}{\mathrm{2}}{dt}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({t}\right)\right){dt}_{{tg}\left({t}\right)={s}} \\$$$$\mathrm{2}{t}={u}\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({u}\right){du}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\mathrm{2}\right){du}\:{in}\:{first} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({tg}\left({t}\right)\right){dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({s}\right)}{\mathrm{1}+{s}^{\mathrm{2}} }{ds}=\Sigma\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{k}} {s}^{\mathrm{2}{k}+\mathrm{1}} {ln}\left({s}\right){ds} \\$$$$=−\Sigma\left(−\mathrm{1}\right)^{{k}} .\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{1}\right)^{\mathrm{2}} }=−\beta\left(\mathrm{2}\right)\:=−{G}\:,{G}\:{catalan}\:{constante} \\$$$${knowing}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({t}\right)\right){dt}\:{give}\:{us}\:{close}\:{forme} \\$$$$\mathrm{2}{nd}\:{use}\:{fourier}\:{serie}\:{of}\: \\$$$${ln}\left({cos}\left({t}\right)\right)=−{ln}\left(\mathrm{2}\right)−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left(\mathrm{2}\left(\mathrm{2}{n}−\mathrm{1}\right){t}\right)}{\mathrm{2}{n}−\mathrm{1}} \\$$$${give}\:{use}\:{resulte}\:{imediatly} \\$$$${withe}\:\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }\:=\beta\left(\mathrm{2}\right) \\$$