advanced-calculus-prove-that-0-1-ln-ln-1-x-dx-ln-1-x-pi-ln-4-

Question Number 137190 by mnjuly1970 last updated on 30/Mar/21

....advanced ....... calculus..... prove that::: 𝛗=∫_0 ^( 1) ln(ln((1/x))).(dx/( (√(ln((1/x)))))) =−(√π) (γ+ln(4))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:…….\:\:{calculus}….. \\ $$$$\:\:\:\:\:\:\:\:{prove}\:{that}::: \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({ln}\left(\frac{\mathrm{1}}{{x}}\right)\right).\frac{{dx}}{\:\sqrt{{ln}\left(\frac{\mathrm{1}}{{x}}\right)}}\:=−\sqrt{\pi}\:\left(\gamma+{ln}\left(\mathrm{4}\right)\right) \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 30/Mar/21

log((1/x))=u⇒x=e^(−u) ⇒1=−e^(−u) (du/dx) =−∫_∞ ^0 t^(−(1/2)) e^(−u) log(t)dt =Γ′((1/2))=−(√π) (γ+log(4))

$${log}\left(\frac{\mathrm{1}}{{x}}\right)={u}\Rightarrow{x}={e}^{−{u}} \Rightarrow\mathrm{1}=−{e}^{−{u}} \frac{{du}}{{dx}} \\ $$$$=−\int_{\infty} ^{\mathrm{0}} {t}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{−{u}} {log}\left({t}\right){dt} \\ $$$$=\Gamma'\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\sqrt{\pi}\:\left(\gamma+{log}\left(\mathrm{4}\right)\right) \\ $$

Commented by mnjuly1970 last updated on 30/Mar/21

$${thank}\:{you}\:{very}\:{much}… \\ $$

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