Question Number 144109 by henderson last updated on 21/Jun/21
$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\mathrm{1}.\:\boldsymbol{\mathrm{calculate}}\::\:\boldsymbol{\mathrm{I}}\:=\int_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{3}}} \boldsymbol{{ln}}\left(\boldsymbol{{tan}}\:\boldsymbol{{x}}\right)\boldsymbol{{dx}}. \\ $$$$\mathrm{2}.\:\boldsymbol{\mathrm{calculate}}\:\::\:\underset{\boldsymbol{{x}}\:\rightarrow\:\boldsymbol{{e}}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{x}}\sqrt{\mathrm{1}β\boldsymbol{{ln}}\:\boldsymbol{{x}}}}{\boldsymbol{{x}}β\boldsymbol{{e}}}\:. \\ $$
Commented by bobhans last updated on 22/Jun/21
$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow\mathrm{e}} {\mathrm{lim}}\:\frac{\mathrm{e}\:\sqrt{\mathrm{1}β\mathrm{ln}\:\mathrm{x}}}{\mathrm{x}β\mathrm{e}}\:= \\ $$$$\:\:\:\:\:\:\:\:\underset{{x}\rightarrow\mathrm{e}} {\mathrm{lim}}\:\frac{\mathrm{e}\left(\frac{β\frac{\mathrm{1}}{\mathrm{x}}}{\mathrm{2}\sqrt{\mathrm{1}β\mathrm{ln}\:\mathrm{x}}}\right)}{\mathrm{1}}\:=β\:\infty \\ $$$$\:\:\:\:\:\:\:\: \\ $$
Answered by Olaf_Thorendsen last updated on 21/Jun/21
$$\mathrm{I}\:=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \left(\mathrm{ln}\left(\mathrm{tan}{x}\right)\:{dx}\right. \\ $$$$\mathrm{Let}\:{u}\:=\:\frac{\pi}{\mathrm{2}}β{x} \\ $$$$\mathrm{I}\:=\:\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \mathrm{ln}\left(\mathrm{cot}{u}\right)\:{du}\:=\:β\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \mathrm{ln}\left(\mathrm{tan}{x}\right)\:{dx} \\ $$$$\mathrm{I}\:=\:β\mathrm{I}\:\:\Rightarrow\:\mathrm{I}\:=\:\mathrm{0} \\ $$
Answered by mathmax by abdo last updated on 21/Jun/21
$$\left.\mathrm{1}\right)\:\mathrm{I}=\int_{\frac{\pi}{\mathrm{6}}} ^{\frac{\pi}{\mathrm{3}}} \mathrm{log}\left(\mathrm{tanx}\right)\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{tanx}\:=\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{I}=\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \:\frac{\mathrm{logt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\:=_{\mathrm{t}=\frac{\mathrm{1}}{\mathrm{u}}} \:\:β\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \:\frac{β\mathrm{logu}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }}\left(β\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} }\right) \\ $$$$=β\int_{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}} ^{\sqrt{\mathrm{3}}} \:\frac{\mathrm{logu}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }=β\mathrm{I}\:\Rightarrow\mathrm{2I}=\mathrm{0}\:\Rightarrow\mathrm{I}=\mathrm{0} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 21/Jun/21
$$\left.\mathrm{2}\right)\mathrm{1}β\mathrm{logx}>\mathrm{0}\:\Rightarrow\mathrm{logx}<\mathrm{1}\:\Rightarrow\mathrm{x}<\mathrm{e} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=β\frac{\mathrm{x}\sqrt{\mathrm{1}β\mathrm{logx}}}{\mathrm{e}β\mathrm{x}}\:\:\mathrm{changement}\:\mathrm{e}β\mathrm{x}=\mathrm{y}\:\mathrm{give} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=β\frac{\left(\mathrm{e}β\mathrm{y}\right)\sqrt{\mathrm{1}β\mathrm{log}\left(\mathrm{e}β\mathrm{y}\right)}}{\mathrm{y}}=\frac{\left(\mathrm{y}β\mathrm{e}\right)\sqrt{\mathrm{1}β\mathrm{log}\left(\mathrm{e}β\mathrm{y}\right)}}{\mathrm{y}}=\mathrm{g}\left(\mathrm{y}\right)\:\:\left(\mathrm{y}\rightarrow\mathrm{o}^{+} \right) \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{log}\left(\mathrm{e}β\mathrm{y}\right)=\mathrm{1}+\mathrm{log}\left(\mathrm{1}β\frac{\mathrm{y}}{\mathrm{e}}\right)\sim\mathrm{1}β\frac{\mathrm{y}}{\mathrm{e}}\:\Rightarrow\mathrm{1}β\mathrm{log}\left(\mathrm{e}β\mathrm{y}\right)\sim\frac{\mathrm{y}}{\mathrm{e}} \\ $$$$\Rightarrow\sqrt{\mathrm{1}β\mathrm{log}\left(\mathrm{e}β\mathrm{y}\right)}\sim\frac{\sqrt{\mathrm{y}}}{\:\sqrt{\mathrm{e}}}\:\Rightarrow \\ $$$$\mathrm{g}\left(\mathrm{y}\right)\sim\frac{\sqrt{\mathrm{y}}}{\:\sqrt{\mathrm{e}}\mathrm{y}}\left(\mathrm{y}β\mathrm{e}\right)\:\Rightarrow\mathrm{g}\left(\mathrm{y}\right)\sim\frac{\mathrm{y}β\mathrm{e}}{\:\sqrt{\mathrm{e}}\sqrt{\mathrm{y}}}\rightarrowβ\infty\:\left(\mathrm{y}\rightarrow\mathrm{0}^{+} \right)\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{e}^{β} } \:\:\:\mathrm{f}\left(\mathrm{x}\right)=β\infty \\ $$