∫_0 ^1 ln(ln(1−x))dx=?


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Question Number 158697 by amin96 last updated on 07/Nov/21

$\int_{0}^{1} l n (l n (1 - x)) d x = ?$
	Answered by mnjuly1970 last updated on 08/Nov/21
	$1−x= u Ω=Re{∫_0 ^( 1) ln(ln(u))du}=^(−ln(u)=t) = Re(∫_( 0) ^( ∞) e^( −t) ln (−t )dt) =Re(∫_0 ^( ∞) e^( −t) (ln(e^( iπ) )+ln(t))dt) = Re (∫_0 ^( ∞) e^( −t) ( iπ +ln(t))dt) = ∫_0 ^( ∞) e^( −t) (ln(t))dt = −γ γ : euler−mascheruni constant$
	$1 - x = u$ $Ω = R e {\int_{0}^{1} l n (l n (u)) d u} \overset{- l n (u) = t}{=}$ $= R e (\int_{0}^{\infty} e^{- t} l n (- t) d t)$ $= R e (\int_{0}^{\infty} e^{- t} (l n (e^{i π}) + l n (t)) d t)$ $= R e (\int_{0}^{\infty} e^{- t} (i π + l n (t)) d t)$ $= \int_{0}^{\infty} e^{- t} (l n (t)) d t = - γ$ $γ : e u l e r - m a s c h e r u n i c o n s t a n t$
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