Question Number 148285 by Khalmohmmad last updated on 26/Jul/21
$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \mathrm{ln}{x}}{dx}=? \\ $$
Commented by ahmedNasser last updated on 26/Jul/21
$${let}\:{u}\:=\:{lnx}\:\:\:\:\therefore\:{du}\:=\:\frac{\mathrm{1}}{{x}}\:{dx}\:\:\:\:{x}\:=\:{e}^{{u}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{{e}^{\mathrm{2}{u}} \centerdot{u}}\:\centerdot{e}^{{u}} \:{du}\:\:=\int_{\mathrm{0}} ^{\:\infty} {u}^{−\mathrm{1}} {e}^{−{u}} {du}\:=\Gamma\left(\mathrm{0}\right)\:=\:\infty \\ $$
Answered by puissant last updated on 26/Jul/21
$${diverge}.. \\ $$