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Question Number 141916 by mnjuly1970 last updated on 24/May/21
provethat::I:=∫0π2arccosh(sin(x)+cos(x))dx=π2ln(2)..
Answered by mindispower last updated on 25/May/21
bypartwithed(arccoh(x))=dx1−x2I=xarccosh(sinx+cos(x))]0π2−∫0π2x(cos(x)−sin(x))sin(2x)dx−∫0π2xcos(x)−xsin(x)sin(2x)dx=−12∫0π2xsin−12(x)cos12(x)+12∫0π2xsin12(x)cos−12(x)12∫0∞arctg(x)(x−1x)1+x2dx=A∫0∞xsarctan(tx)1+x2dx=f(t),s∈]−1,1[f′(t)=∫0∞xs+1(1+x2)(1+t2x2)dx∫−∞∞xs+1(1+x2)(1+t2x2)dx=a=2iπ(.is+12i(1−t2)+is+1t2ts+1.(t2−1).12it)=πis+11−t2(1−t−s)a=f(t)+∫0∞(−x)s+1(1+x2)(1+t2x2)dx=(1+(−1)s+1)f(t)=(1+eiπ(s+1))f(t)f(t)=π1−t2(1−t−s).eiπ2(s+1)1+eiπ(s+1)=π(1−t−s)2(1−t2)cos(π2(s+1))=f′(t)∫01f′(t)=xsarctan(x)1+x2=π2cos(π2(s+1))∫011−t−s1−t2dtt2=x=−π4sin(πs2)∫011−x−s21−xx−12=−π4sin(πs2)∫01x−12−x−s+121−xdx=g(s)Ψ(s+1)=−γ+∫011−xs1−xdxg(s)=−π4sin(πs2).(Ψ(1−s2)−Ψ(12))A=−π4sin(π4)2(Ψ(14)−2Ψ(12)+Ψ(34))Ψ(12)=−2ln(2)−γΨ(14)=(−π2−3ln(2)−γ)Ψ(34)=Ψ(14)+π=−π4(−π−6ln(2)−2γ+π+4ln(2)+2γ)=−π4.−2ln(2)=π2ln(2)
Commented by mnjuly1970 last updated on 25/May/21
verynice.thanksalot...
Commented by mindispower last updated on 25/May/21
withepleasursir
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