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....