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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(1x))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(1x)x=(1eu)2dx=2eu(1eu)duΩ=0ln(u).2eu(1eu)duweuseln(z)=lnz+iarg(z)zIRputu=t⇒=0ln(t).2et(1et)dt=0(ln(t)+iπ)2et(1et)dtRe(Ω)=02ln(t)etdt20ln(t)e2tdtΓ(z)=0tz1etdtΓ(1)=0+ln(t)etdt=γRe(Ω)=2γ20+ln(s2)es.ds2=2γ0+ln(s)esds+ln(2)0+esds=2γ+γ+ln(2)[es]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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