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Question Number 112119 by mnjuly1970 last updated on 06/Sep/20

....calculus.... prove that::: if Ω =∫_(0 ) ^( 1) ln(ln(1−(√x) ))dx then Re(Ω) := −γ + ln(2).... m.n. july 1970#

....calculus....provethat:::ifΩ=∫01ln(ln(1−x))dxthenRe(Ω):=−γ+ln(2)....You can't use 'macro parameter character #' in math mode

Answered by maths mind last updated on 06/Sep/20

u=ln(1−(√x))⇒x=(1−e^u )^2 ⇒dx=−2e^u (1−e^u )du Ω=∫_0 ^(−∞) ln(u).−2e^u (1−e^u )du we use ln(z)=ln∣z∣+iarg(z) z∉IR_− ^∗ put u=−t⇒=∫_0 ^∞ ln(−t).2e^(−t) (1−e^(−t) )dt =∫_0 ^∞ (ln(t)+iπ)2e^t (1−e^t )dt Re(Ω)=∫_0 ^∞ 2ln(t)e^(−t) dt−2∫_0 ^∞ ln(t)e^(−2t) dt Γ(z)=∫_0 ^∞ t^(z−1) e^(−t) dt⇒Γ′(1)=∫_0 ^(+∞) ln(t)e^(−t) dt=γ Re(Ω)=−2γ−2∫_0 ^(+∞) ln((s/2))e^(−s) .(ds/2) =−2γ−∫_0 ^(+∞) ln(s)e^(−s) ds+ln(2)∫_0 ^(+∞) e^(−s) ds =−2γ+γ+ln(2)[−e^(−s) ]_0 ^(+∞) =−γ+ln(2)

u=ln(1−x)⇒x=(1−eu)2⇒dx=−2eu(1−eu)duΩ=∫0−∞ln(u).−2eu(1−eu)duweuseln(z)=ln∣z∣+iarg(z)z∉IR−∗putu=−t⇒=∫0∞ln(−t).2e−t(1−e−t)dt=∫0∞(ln(t)+iπ)2et(1−et)dtRe(Ω)=∫0∞2ln(t)e−tdt−2∫0∞ln(t)e−2tdtΓ(z)=∫0∞tz−1e−tdt⇒Γ′(1)=∫0+∞ln(t)e−tdt=γRe(Ω)=−2γ−2∫0+∞ln(s2)e−s.ds2=−2γ−∫0+∞ln(s)e−sds+ln(2)∫0+∞e−sds=−2γ+γ+ln(2)[−e−s]0+∞=−γ+ln(2)

Commented by mnjuly1970 last updated on 06/Sep/20

peace be upon you master ..gratful for your attantion and favor...

peacebeuponyoumaster..gratfulforyourattantionandfavor...

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