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Integration Questions

Question Number 209393 by Shrodinger last updated on 08/Jul/24

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}\left({tanx}\right)}{\mathrm{1}+{tanx}}{dx} \\$$

Commented by Frix last updated on 09/Jul/24

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\$$

Commented by Shrodinger last updated on 09/Jul/24

$${Yes}\:{sir}. \\$$

Commented by Frix last updated on 09/Jul/24

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{ln}\:\mathrm{tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:{x}}{dx}\:\overset{{t}=\mathrm{tan}\:{x}} {=}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}= \\$$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}+\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}= \\$$$$\:\:\:\:\:\left[\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:\overset{{u}=\frac{\mathrm{1}}{{t}}} {=}\:−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{u}\mathrm{ln}\:{u}}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{du}\right] \\$$$$=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\left(\mathrm{1}−{t}\right)\mathrm{ln}\:{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:\underset{\mathrm{parts}} {\overset{\mathrm{by}} {=}} \\$$$${u}'=\frac{\mathrm{1}−{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\:\rightarrow\:{u}=\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:−\frac{\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}} \\$$$${v}=\mathrm{ln}\:{t}\:\rightarrow\:{v}'=\frac{\mathrm{1}}{{t}} \\$$$$=\left[\left(\mathrm{ln}\:\left({t}+\mathrm{1}\right)\:−\frac{\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}}\right)\mathrm{ln}\:{t}\right]_{\mathrm{0}} ^{\mathrm{1}} −\:\:\:\:\:\left\{=\mathrm{0}\right\} \\$$$$\:\:\:\:\:−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left({t}+\mathrm{1}\right)}{{t}}{dt}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\:\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{{t}}{dt}= \\$$$$=\left[\mathrm{Li}_{\mathrm{2}} \:\left(−{t}\right)−\frac{\mathrm{Li}_{\mathrm{2}} \:\left(−{t}^{\mathrm{2}} \right)}{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{3Li}_{\mathrm{2}} \:\left(−\mathrm{1}\right)}{\mathrm{4}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\$$

Answered by mathmax last updated on 09/Jul/24

$${I}=_{{tanx}={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({t}\right)}{\mathrm{1}+{t}}\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}\:+\int_{\mathrm{1}} ^{\infty} \frac{{lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}\left(\rightarrow{t}=\frac{\mathrm{1}}{{u}}\right) \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{lnu}}{\left(\frac{\mathrm{1}}{{u}}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+\mathrm{1}\right)}\left(\frac{−{du}}{{u}^{\mathrm{2}} }\right) \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tlnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{t}\right){lnt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\$$$${f}\left({t}\right)=\frac{\mathrm{1}−{t}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{a}}{{t}+\mathrm{1}}+\frac{{bt}+{c}}{{t}^{\mathrm{2}} +\mathrm{1}} \\$$$${a}=\mathrm{1}\:\:\:{lim}_{\infty} {tf}\left({t}\right)=\mathrm{0}={a}+{b}\:\Rightarrow{b}=−\mathrm{1} \\$$$${f}\left(\mathrm{0}\right)=\mathrm{1}={a}+{c}\:\Rightarrow{c}=\mathrm{0}\:\Rightarrow \\$$$${f}\left({t}\right)=\frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow \\$$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{t}+\mathrm{1}}−\frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){lnt}\:{dt} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tlnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:{we}\:{have} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {lnt}\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{{n}} {dt} \\$$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{n}} {lnt}\:{dt}=\Sigma\left(−\mathrm{1}\right)^{{n}} {u}_{{n}} \\$$$${u}_{{n}} =\left[\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}{lnt}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\frac{{dt}}{{t}} \\$$$$=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\mathrm{1}+{t}}{dt}=−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\$$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\eta\left(\mathrm{2}\right)=\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right) \\$$$$=−\frac{\mathrm{1}}{\mathrm{2}}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tlnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} {tlnt}\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} {t}^{\mathrm{2}{n}} {dt} \\$$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}+\mathrm{1}} {lnt}\:{dt}=\Sigma\left(−\mathrm{1}\right)^{{n}} {v}_{{n}} \\$$$${v}_{{n}} =\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}{t}^{\mathrm{2}{n}+\mathrm{2}} {lnt}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}{n}+\mathrm{2}}\frac{{dt}}{{t}} \\$$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{2}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tlnt}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=−\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} } \\$$$$=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\right)\:\Rightarrow \\$$$${I}=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{4}}.\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=\left(\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}\right)\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\$$$$=−\frac{\mathrm{3}}{\mathrm{4}}×\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\$$$$\bigstar{I}=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\bigstar \\$$$$\\$$