Integration Questions

Question Number 89052 by M±th+et£s last updated on 15/Apr/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{{tan}\left({x}\right)}{dx} \\$$

Commented by abdomathmax last updated on 15/Apr/20

$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{x}}{{tanx}}{dx}\:\Rightarrow{I}\:=_{{tanx}={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({t}\right)}{{t}}\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\$$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({t}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:\:{let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \frac{{arctan}\left({at}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\$$$${f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left(\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\$$$$=_{{at}={u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{{a}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{u}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)}\frac{{du}}{{a}} \\$$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} \:+{u}^{\mathrm{2}} \right)}\:{decomposition}\:{of} \\$$$${F}\left({u}\right)\:=\frac{\mathrm{1}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)}\:=\frac{\alpha{u}\:+\beta}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\lambda{u}\:+\rho}{{u}^{\mathrm{2}} \:+{a}^{\mathrm{2}} } \\$$$${F}\left(−{u}\right)={F}\left({u}\right)\:\Rightarrow \\$$$$\frac{−\alpha{u}\:+\beta}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{−\lambda{u}\:+\rho}{{u}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:={F}\left({u}\right)\:\Rightarrow\alpha=\lambda=\mathrm{0}\:{and}\:\rho=\beta\:\Rightarrow \\$$$${F}\left({u}\right)=\frac{\beta}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\rho}{{u}^{\mathrm{2}} \:+{a}^{\mathrm{2}} } \\$$$${lim}_{{u}\rightarrow+\infty} \:{u}^{\mathrm{2}} \:{F}\left({u}\right)\:=\mathrm{0}\:=\beta+\rho\:\Rightarrow\rho=−\beta\:\Rightarrow \\$$$${F}\left({u}\right)=\frac{\beta}{{u}^{\mathrm{2}} +\mathrm{1}}−\frac{\beta}{{u}^{\mathrm{2}} \:+{a}^{\mathrm{2}} } \\$$$${F}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:=\beta−\frac{\beta}{{a}^{\mathrm{2}} }\:\Rightarrow\mathrm{1}\:={a}^{\mathrm{2}} \beta−\beta\:=\left({a}^{\mathrm{2}} −\mathrm{1}\right)\beta\:\Rightarrow \\$$$$\beta\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow{F}\left({u}\right)=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\right)\:\Rightarrow \\$$$${f}^{'} \left({a}\right)=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\right){du} \\$$$$\int\:\frac{{du}}{{u}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:=_{{u}={az}} \:\:\int\:\frac{{adz}}{{a}^{\mathrm{2}} \left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{a}}\:{arctan}\left(\frac{{u}}{{a}}\right)\:\Rightarrow \\$$$${f}^{'} \left({a}\right)\:=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\left[{arctanu}\:−\frac{\mathrm{1}}{{a}}{arctan}\left(\frac{{u}}{{a}}\right)\right]_{\mathrm{0}} ^{+\infty} \\$$$$=\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\left\{\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}{a}}\right\}\:=\frac{\pi}{\mathrm{2}}×\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{1}}\left(\mathrm{1}−\frac{\mathrm{1}}{{a}}\right) \\$$$$=\frac{\pi}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\left(\frac{{a}−\mathrm{1}}{{a}}\right)\:=\frac{\pi}{\mathrm{2}{a}\left({a}+\mathrm{1}\right)}\:\Rightarrow \\$$$${f}\left({a}\right)\:=\frac{\pi}{\mathrm{2}}\:\int\:\:\frac{{da}}{{a}\left({a}+\mathrm{1}\right)}\:+{c} \\$$$$=\frac{\pi}{\mathrm{2}}\int\left(\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{a}+\mathrm{1}}\right){da}\:+{c}\:=\frac{\pi}{\mathrm{2}}{ln}\mid\frac{{a}}{{a}+\mathrm{1}}\mid\:+{c} \\$$$${c}\:\:={lim}_{{a}\rightarrow+\infty} \:\:{f}\left({a}\right) \\$$$${I}\:={f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:+{c}\:={c}−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\$$$${rest}\:{to}\:{find}\:{c}... \\$$$$\\$$$$\\$$

Commented by abdomathmax last updated on 15/Apr/20

$${let}\:{try}\:\:{residus}\:{method}\:\:{we}?{hsve} \\$$$${I}?=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({t}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dx}\Rightarrow\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({t}\right)}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\$$$$\\$$$${let}\:\varphi\left({z}\right)=\frac{{arctan}\left({z}\right)}{{z}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{arctanz}}{{z}\left({z}−{i}\right)\left({z}+{i}\right)} \\$$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{i}\right)\:+{Res}\left(\varphi,{o}\right)\right\} \\$$$$=\mathrm{2}{i}\pi×\frac{{arctan}\left({i}\right)}{{i}\left(\mathrm{2}{i}\right)}\:=−{i}\pi\:{arctan}\left({i}\right) \\$$$${but}\:{the}\:{frmula}\:\:{arctan}\left({z}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{iz}}{\mathrm{1}−{iz}}\right)\:{is}\:{not} \\$$$${valide}\:{for}\:{z}={i}...{be}\:{continued}... \\$$$$\\$$

Commented by M±th+et£s last updated on 15/Apr/20

$${thanx}\:{sir} \\$$

Commented by mathmax by abdo last updated on 15/Apr/20

$${you}\:{are}\:{welcome} \\$$