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0-1-ln-ln-1-x-dx-

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Question Number 158697 by amin96 last updated on 07/Nov/21
∫_0 ^1 ln(ln(1−x))dx=?
∫01ln(ln(1−x))dx=?
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=uΩ=Re{∫01ln(ln(u))du}=−ln(u)=t=Re(∫0∞e−tln(−t)dt)=Re(∫0∞e−t(ln(eiπ)+ln(t))dt)=Re(∫0∞e−t(iπ+ln(t))dt)=∫0∞e−t(ln(t))dt=−γγ:euler−mascheruniconstant

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