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Question Number 75984 by Ajao yinka last updated on 21/Dec/19

Answered by mind is power last updated on 23/Dec/19

y=x^2 ⇒Ω=∫_0 ^1 ((log(y))/(4(1+y+y^2 )))dy let z=−log(y)⇒y=e^(−z) ⇒dy=−e^(−z) dz ⇒Ω=∫_0 ^(+∞) ((−ze^(−z) )/(4(e^(−2z) +e^(−z) +1))) 4Ω=∫_0 ^(+∞) ((−z(e^(−z) )(e^(−z) −1))/(e^(−3z) −1))dz 4Ω=∫_0 ^(+∞) ((−ze^(−2z) )/(e^(−3z) −1))dz+∫_0 ^(+∞) ((ze^(−z) )/(e^(−3z) −1))dz Ψ^m (z)=(−1)^(m+1) .∫_0 ^(+∞) ((t^m e^(−zt) )/(1−e^(−t) ))dt Ω=∫_0 ^(+∞) ((ze^(−2z) )/(1−e^(−3z) ))dz−∫_0 ^(+∞) ((ze^(−z) )/(1−e^(−3z) )) 3z=r⇒ 4Ω=(1/9).∫_0 ^(+∞) ((re^(−(2/3)r) )/(1−e^(−r) ))dr−(1/9)∫_0 ^(+∞) ((re^(−(r/3)) )/(1−e^(−r) ))dr 4Ω=(1/9)Ψ^1 ((2/3))−(1/9)Ψ^1 ((1/3))=(1/9)(Ψ^1 ((2/3))−Ψ^1 ((1/3))) Ω=(1/(36)){Ψ^1 ((2/3))−Ψ^1 ((1/3))}

y=x2⇒Ω=∫01log(y)4(1+y+y2)dyletz=−log(y)⇒y=e−z⇒dy=−e−zdz⇒Ω=∫0+∞−ze−z4(e−2z+e−z+1)4Ω=∫0+∞−z(e−z)(e−z−1)e−3z−1dz4Ω=∫0+∞−ze−2ze−3z−1dz+∫0+∞ze−ze−3z−1dzΨm(z)=(−1)m+1.∫0+∞tme−zt1−e−tdtΩ=∫0+∞ze−2z1−e−3zdz−∫0+∞ze−z1−e−3z3z=r⇒4Ω=19.∫0+∞re−23r1−e−rdr−19∫0+∞re−r31−e−rdr4Ω=19Ψ1(23)−19Ψ1(13)=19(Ψ1(23)−Ψ1(13))Ω=136{Ψ1(23)−Ψ1(13)}

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