Question Number 114135 by mnjuly1970 last updated on 17/Sep/20

....Advanced mathematics ... i:: prove that : Ω=(1/π)∫_0 ^( ∞) (1/((x^2 −x+1)^2 (√x)))dx =1 ii::evaluate :: Φ = ∫_0 ^( 1) x^2 ln(x) ln(1−x)dx=??? ....m.n.july. 1970....

Answered by MJS_new last updated on 17/Sep/20

∫(dx/((x^2 −x+1)^2 (√x)))= [t=(√x) → dx=2(√x)dt] =2∫(dt/((t^4 −t^2 +1)^2 ))= [Ostrogradski′s Method] =((t(t^2 +1))/(3(t^4 −t^2 +1)))+(1/3)∫((t^2 +5)/(t^4 −t^2 +1))dt (1/3)∫((t^2 +5)/(t^4 −t^2 +1))dt= =(1/(18))∫(((4(√3)t+15)/(t^2 +(√3)t+1))−((4(√3)t−15)/(t^2 −(√3)t+1)))dt= [using formula] =((√3)/9)ln ((t^2 +(√3)t+1)/(t^2 −(√3)t+1)) +arctan (2t−(√3)) +arctan (2t+(√3)) ⇒ ∫(dx/((x^2 −x+1)^2 (√x)))= =(((x+1)(√x))/(3(x^2 −x+1)))+ln ((x+1+(√(3x)))/(x+1−(√(3x)))) +arctan (2(√x)−(√3)) +arctan (2(√x)+(√3)) +C ⇒ (1/π)∫_0 ^∞ (dx/((x^2 −x+1)^2 (√x)))=1

Commented bymnjuly1970 last updated on 17/Sep/20

thank you so much mr

Answered by mathmax by abdo last updated on 18/Sep/20

let prove thst ∫_0 ^∞ (dx/((x^2 −x+1)^2 (√x))) =π changement (√x)=t give I =∫_0 ^∞ ((2tdt)/((t^4 −t^2 +1)^2 t)) =2 ∫_0 ^∞ (dt/((t^4 −t^2 +1)^2 )) =∫_(−∞) ^(+∞) (dt/((t^4 −t^2 +1)^2 )) let ϕ(z) =(1/((z^4 −z^2 +1)^2 )) poles of ϕ? z^4 −z^2 +1 =0 ⇒u^2 −u+1 =0 (u=z^2 ) Δ =−3 ⇒u_1 =((1+3i)/2) =e^((iπ)/3) and u_2 =((1−3i)/2) =e^(−((iπ)/3)) ⇒z^4 −z^2 +1 =(z^2 −e^((iπ)/3) )(z^2 −e^(−((iπ)/3)) ) =(z−e^((iπ)/6) )(z+e^((iπ)/6) )(z−e^(−((iπ)/6)) )(z+e^(−i(π/6)) ) ⇒ϕ(z) =(1/((z−e^((iπ)/6) )^2 (z+e^((iπ)/6) )^2 (z−e^(−((iπ)/6)) )^2 (z+e^(−((iπ)/6)) )^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/6) )+Res(ϕ,−e^(−((iπ)/6)) )} Res(ϕ,e^((iπ)/6) ) =lim_(z→e^((iπ)/6) ) (1/((2−1)!)){(z−e^((iπ)/6) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/6) ) {(1/((z+e^((iπ)/6) )^2 (z^2 −e^(−((iπ)/3)) )^2 ))}^((1)) =−lim_(z→e^((iπ)/6) ) ((2(z+e^((iπ)/6) )(z^2 −e^(−((iπ)/3)) )^(2 ) +4z(z^2 −e^(−((iπ)/3)) )(z+e^((iπ)/6) )^2 )/((z+e^((iπ)/6) )^4 (z^2 −e^(−((iπ)/3)) )^4 )) =−2lim_(z→e^((iπ)/6) ) (((z^2 −e^(−((iπ)/3)) )+2z(z+e^((iπ)/6) ))/((z+e^((iπ)/6) )^3 (z^2 −e^(−((iπ)/3)) )^3 )) ....be continued....

Commented bymathmax by abdo last updated on 18/Sep/20

let I =∫_(−∞) ^(+∞) (dx/((x^4 −x^2 +1)^2 )) let try parametric method f(a) =∫_(−∞) ^(+∞) (dx/(x^4 −x^(2 ) +a)) with a>(1/4) we have f^′ (a) =−∫_(−∞) ^(+∞) (dx/((x^4 −x^2 +a)^2 )) I =−f^′ (1) let explicit f(a) ϕ(z) =(1/(z^4 −z^2 +a)) poles of ϕ? u^2 −u+a =0 (u=z^2 )→Δ =1−4a<0 ⇒u_1 =((1+i(√(4a−1)))/2) u_2 =((1−i(√(4a−1)))/2) we have ∣u_1 ∣ =(1/2)(√(1+4a−1))=(√a) ⇒ u_1 =(√a)e^(iar4tan(√(4a−1))) and u_2 =(√a)e^(−iarctan((√(4a−1)))) ⇒ z^4 −z^2 +a =(z^2 −(√a)e^(iarctan((√(4a))−1)) )(z^2 −(√a)e^(−iarftan((√(4a−1)))) ) ⇒ ϕ(z) =(1/((z−(√(√a))e^((i/2)arctan((√(4a−1)))) )(z+(√(√a)) e^((i/2)arctan((√(4a−1)))) )(z−(√(√a))e^(−(i/2)arctan((√(4a−1)))) )(z+(√(√a))e^(−(i/2)arctan((√({a−1)))) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,(√(√a))e^((i/2)arctan((√(4a))−1) )+Res(ϕ,−(√(√a))e^(−(i/2)arctan((√(4a−1)))) } Res(ϕ,(^4 (√a))e^((i/2)arctan((√(4a−1)))) ) =(1/(2(^4 (√a))e^((i/2)ar4tan((√(4a−1)))) ×(√a)(2i sin(arctan((√(4a−1))))) =(e^(−(i/2)arctan((√(4a−1)))) /(4i a^(3/4) sin(arctan((√(4a−1))))) Res(ϕ,−(^4 (√a))e^(−(i/2)arctan((√(4a−1)))) ) =(1/(−2(^4 (√a))e^(−(i/2)arctan((√(4a−1)))) ((√a))(−2i sin(arctan((√(4a−1))))) =(e^((i/2)arctan((√(4a−1)))) /(4i a^(3/4) sin(arctan((√(4a−1))))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(a^(−(3/4)) /(4i))×(e^(−(i/2)arctan((√(4a−1)))) /(sin(arctan((√(4a−1)))))+(a^(−(3/4)) /(4i))×(e^((i/2)arctan((√(4a−1)))) /(sin(arctan((√(4a−1)))))} =(π/(2 a^(3/4) sin(arctan((√(4a−1)))))×2cos(arctan((√(4a−1))) ....be continued....