Question Number 137190 by mnjuly1970 last updated on 30/Mar/21
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:....{advanced}\:.......\:\:{calculus}..... \\ $$$$\:\:\:\:\:\:\:\:{prove}\:{that}::: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({ln}\left(\frac{\mathrm{1}}{{x}}\right)\right).\frac{{dx}}{\:\sqrt{{ln}\left(\frac{\mathrm{1}}{{x}}\right)}}\:=ā\sqrt{\pi}\:\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 30/Mar/21
$${log}\left(\frac{\mathrm{1}}{{x}}\right)={u}\Rightarrow{x}={e}^{ā{u}} \Rightarrow\mathrm{1}=ā{e}^{ā{u}} \frac{{du}}{{dx}} \\ $$$$=ā\int_{\infty} ^{\mathrm{0}} {t}^{ā\frac{\mathrm{1}}{\mathrm{2}}} {e}^{ā{u}} {log}\left({t}\right){dt} \\ $$$$=\Gamma'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=ā\sqrt{\pi}\:\left(\gamma+{log}\left(\mathrm{4}\right)\right) \\ $$
Commented by mnjuly1970 last updated on 30/Mar/21
$${thank}\:{you}\:{very}\:{much}... \\ $$