Question Number 62232 by aliesam last updated on 18/Jun/19 | ||
Commented by maxmathsup by imad last updated on 18/Jun/19 | ||
$${we}\:{have}\:\mid\alpha+\beta\:{e}^{{i}\theta} \mid\:=\mid\alpha\:+\beta{cos}\theta\:+{i}\beta{sin}\theta\mid\:=\sqrt{\left(\alpha+\beta{cos}\theta\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\:\Rightarrow \\ $$$${ln}\left(\mid\alpha+{ie}^{{i}\theta} \mid\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\:\left(\alpha+\beta\:{cos}\theta\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\right\}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left\{\alpha^{\mathrm{2}} +\mathrm{2}\alpha\beta\:{cos}\theta\:+\beta^{\mathrm{2}} \right\}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {ln}\left(\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \:+\mathrm{2}\alpha\beta\:{cos}\theta\right){d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left\{{ln}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)\:+{ln}\left(\mathrm{1}+\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:{cos}\theta\right)\right\}{d}\theta \\ $$$$=\pi{ln}\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:{ln}\left(\mathrm{1}+\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:{cos}\theta\right){d}\theta\:\:{let}\:{find} \\ $$$${f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {ln}\left(\mathrm{1}+{tcos}\theta\right){d}\theta\:\:\:\:\Rightarrow{f}^{'} \left({t}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cos}\theta}{\mathrm{1}+{tcos}\theta}\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{{t}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{1}+{tcos}\theta−\mathrm{1}}{\mathrm{1}+{tcos}\theta}\:{d}\theta\:=\frac{\mathrm{2}\pi}{{t}}\:−\frac{\mathrm{1}}{{t}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{d}\theta}{\mathrm{1}+{tcos}\theta} \\ $$$${changement}\:{e}^{{i}\theta} \:={z}\:{give}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{d}\theta}{\mathrm{1}+{t}\:{cos}\theta}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\mathrm{2}\:+{tz}\:+{tz}^{−\mathrm{1}} \right)}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{2}{idz}}{\mathrm{2}{z}\:+{tz}^{\mathrm{2}} \:+{t}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{2}{idz}}{{tz}^{\mathrm{2}} \:+\mathrm{2}{z}\:+{t}} \\ $$$${let}\:{w}\left({z}\right)\:=\frac{−\mathrm{2}{i}}{{tz}^{\mathrm{2}} \:+\mathrm{2}{z}\:+{t}}\:\:{poles}\:{of}\:{w}? \\ $$$$\Delta^{'} \:=\mathrm{1}−{t}^{\mathrm{2}} \:\:\:{case}\:\mathrm{1}\:\:\:\:\mid{t}\mid<\mathrm{1}\:\Rightarrow\:{z}_{\mathrm{1}} =−\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:{and}\:{z}_{\mathrm{2}} =−\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\mid−\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\mid−\mathrm{1}\:=\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }−\mathrm{1}\:<\mathrm{0}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\mid−\mathrm{1}−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\mid−\mathrm{1}\:=\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }−\mathrm{1}\:>\mathrm{0}\:\left(\:{to}\:{eliminate}\:{from}\:{residus}\right) \\ $$$$\left.\int_{\mid{z}\mid=\mathrm{1}} \:\:\:{w}−{z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({w},{z}_{\mathrm{1}} \right)\:\:\:{but}\:{w}\left({z}\right)\:=\frac{−\mathrm{2}{i}}{{t}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${Res}\left({w},{z}_{\mathrm{1}} \right)\:=\frac{−\mathrm{2}{i}}{{t}\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}\:=\frac{−\mathrm{2}{i}}{{t}\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:=\frac{−{i}}{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\:{w}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left(\frac{−{i}}{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\right)\:=\frac{\mathrm{2}\pi}{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\Rightarrow \\ $$$${f}^{'} \left({t}\right)\:=\:\frac{\mathrm{2}\pi}{{t}}\:−\frac{\mathrm{2}\pi}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\Rightarrow{f}\left({t}\right)\:=\mathrm{2}\pi{ln}\mid{t}\mid\:−\mathrm{2}\pi\:\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:+{c}\: \\ $$$$\int\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:=\int\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{−\mathrm{1}}{\mathrm{2}}} \:{dt}\:=_{{by}\:{parts}} \:\:\:\:\:−\frac{\mathrm{1}}{{t}}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:−\int\:−\frac{\mathrm{1}}{{t}}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\mathrm{2}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} {dt} \\ $$$$=−\frac{\mathrm{1}}{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:+\:\int\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{dt}\:\:\:{cyangement}\:{t}\:={sin}\alpha\:{give} \\ $$$$\int\:\:\:\frac{{dt}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:=\int\:\:\frac{\mathrm{1}}{{cos}^{\mathrm{3}} \alpha}\:{cos}\alpha\:{d}\alpha\:=\int\:\:\frac{{d}\alpha}{{cos}^{\mathrm{2}} \alpha}\:=\int\:\:\:\frac{\mathrm{2}{d}\alpha}{\mathrm{1}+{cos}\left(\mathrm{2}\alpha\right)} \\ $$$$=_{{tan}\left(\alpha\right)={u}} \:\:\:\:\:\:\:\int\:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)}\:=\int\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+\mathrm{1}−{u}^{\mathrm{2}} }\:=\int\:{du}\:={u}\:+{c} \\ $$$$={tan}\left(\alpha\right)+{c}\:={tan}\left({arcsint}\right)\:+{c}\:\Rightarrow \\ $$$${f}\left({t}\right)\:=\mathrm{2}\pi\:{ln}\mid{t}\mid−\mathrm{2}\pi\:\left\{\frac{−\mathrm{1}}{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:+{tan}\left({arcsint}\right)\right\}\:+{c}\:\Rightarrow \\ $$$${I}\:=\pi{ln}\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:+\frac{\mathrm{1}}{\mathrm{2}}{f}\left(\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:\right)\:+{c} \\ $$$${I}=\pi{ln}\left(\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:+\pi{ln}\mid\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\mid−\pi\left\{−\frac{\mathrm{1}}{\left.\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\sqrt{\mathrm{1}−\left(\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\left(\right.} }\right.}\right)^{\mathrm{2}} }\:+{tan}\left({arcsin}\left(\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\right)\right\}\right. \\ $$$$+{c}\:\:\:{rest}\:{to}\:{find}\:\:{the}\:{value}\:{of}\:{c}\:\:\:....{be}\:{continued}... \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by aliesam last updated on 18/Jun/19 | ||
$${thanks}\:{sir}\:{brilliant}\:{sol} \\ $$ | ||
Commented by maxmathsup by imad last updated on 18/Jun/19 | ||
$$\mid\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\mid<\mathrm{1}\:\:{for}\:{that}\:\:{i}\:{take}\:{t}\:=\frac{\mathrm{2}\alpha\beta}{\alpha^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:\:{without}\:{studying}\:{the}\:{case}\:\mid{t}\mid>\mathrm{1}.. \\ $$ | ||
Commented by maxmathsup by imad last updated on 18/Jun/19 | ||
$${you}\:{are}\:{welcome}\:{sir}\:. \\ $$ | ||