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




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

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