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Question Number 114721 by mnjuly1970 last updated on 20/Sep/20

....nice calculus.... prove that:: Φ=∫_(0 ) ^( 1) xln[ln(x).ln(1−x)]dx=−γ γ ::= euler mascheroni constant. ...m.n.july.1970...

....nicecalculus....provethat::Φ=∫01xln[ln(x).ln(1−x)]dx=−γγ::=eulermascheroniconstant....m.n.july.1970...

Answered by maths mind last updated on 20/Sep/20

by part =∫_0 ^1 (1−x)ln[ln(1−x)ln(x)]dx ⇔2∫_0 ^1 xln(ln(x)ln(1−x))dx=∫_0 ^1 ln[ln(x)ln(1−x)]dx =∫_0 ^1 ln(−ln(x).−ln(1−x))dx =∫_0 ^1 ln(−ln(x))dx+∫_0 ^1 ln(−ln(1−x))dx in 2nd 1−x=t⇒ =2∫_0 ^1 ln(−ln(x))dx t=−ln(x) =2∫_0 ^∞ ln(t)e^(−t) dt Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt Γ′(1)=∫_0 ^∞ ln(t)e^(−t) dt=Ψ(1)=−γ ⇔2∫_0 ^1 xln[ln(x)ln(1−x)]dx=2∫_0 ^1 ln(−ln(x))dx=−2γ ⇒∫_0 ^1 xln[ln(x)ln(1−x)]dx=−γ

bypart=∫01(1−x)ln[ln(1−x)ln(x)]dx⇔2∫01xln(ln(x)ln(1−x))dx=∫01ln[ln(x)ln(1−x)]dx=∫01ln(−ln(x).−ln(1−x))dx=∫01ln(−ln(x))dx+∫01ln(−ln(1−x))dxin2nd1−x=t⇒=2∫01ln(−ln(x))dxt=−ln(x)=2∫0∞ln(t)e−tdtΓ(x)=∫0∞tx−1e−tdtΓ′(1)=∫0∞ln(t)e−tdt=Ψ(1)=−γ⇔2∫01xln[ln(x)ln(1−x)]dx=2∫01ln(−ln(x))dx=−2γ⇒∫01xln[ln(x)ln(1−x)]dx=−γ

Commented by mnjuly1970 last updated on 20/Sep/20

excellent sir..thank you ...

excellentsir..thankyou...

Commented by maths mind last updated on 20/Sep/20

withe pleasur

withepleasur

Answered by mnjuly1970 last updated on 20/Sep/20

my solution.. Φ =∫_0 ^( 1) xln[(−ln(x))(−ln (1−x))]dx =∫_0 ^( 1) xln(−ln(x))dx +∫_0 ^( 1) xln(−ln(1−x))dx =∫_0 ^( 1) xln(−ln(x))dx +∫_0 ^( 1) (1−x)ln(−ln(x))dx =∫_0 ^( 1) ln(−ln(x))dx=^([−ln(x)=t]) −∫_∞ ^( 0) e^(−t) ln(t)dt =∫_0 ^( ∞) e^(−t) ln(t)dt = −γ ✓ m.n.july.1970

mysolution..Φ=∫01xln[(−ln(x))(−ln(1−x))]dx=∫01xln(−ln(x))dx+∫01xln(−ln(1−x))dx=∫01xln(−ln(x))dx+∫01(1−x)ln(−ln(x))dx=∫01ln(−ln(x))dx=[−ln(x)=t]−∫∞0e−tln(t)dt=∫0∞e−tln(t)dt=−γ✓m.n.july.1970

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