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Evaluate-I-0-1-x-5-x-4-x-3-x-2-x-1-dx-

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Question Number 193377 by Humble last updated on 12/Jun/23
Evaluate I=∫_0 ^( ∞) (1/(x^5 +x^4 +x^3 +x^2 +x+1))dx
EvaluateI=∫0∞1x5+x4+x3+x2+x+1dx
Commented by Frix last updated on 12/Jun/23
x^5 +x^4 +x^3 +x^2 +x+1= =(x+1)(x^2 −x+1)(x^2 +x+1) Just decompose and solve it!
x5+x4+x3+x2+x+1==(x+1)(x2−x+1)(x2+x+1)Justdecomposeandsolveit!
Answered by Mathspace last updated on 16/Jun/23
I=∫_0 ^∞ (dx/(1+x+x^2 +x^3 +x^4 +x^5 )) I=∫_0 ^∞ (dx/((1−x^6 )/(1−x)))=∫_0 ^∞ ((1−x)/(1−x^6 ))dx =∫_0 ^1 ((1−x)/(1−x^6 ))dx +∫_1 ^∞ ((1−x)/(1−x^6 ))dx(→x=(1/t)) =∫_0 ^1 ((1−x)/(1−x^6 ))dx−∫_0 ^1 ((1−(1/t))/(1−(1/t^6 )))×((−dt)/t^2 ) =∫_0 ^1 ((1−x)/(1−x^6 ))dx+∫_0 ^1 (1/t^3 ).((t−1)/(t^6 −1)).t^6 dt =∫_0 ^1 ((1−x)/(1−x^6 ))dx+∫_0 ^1 ((t^4 −t^3 )/(t^6 −1))dt =∫_0 ^1 ((1−x)/(1−x^6 ))dx+∫_0 ^1 ((x^3 −x^4 )/(1−x^6 ))dx =∫_0 ^1 ((1−x+x^3 −x^4 )/(1−x^6 ))dx =∫_0 ^1 (1−x+x^3 −x^4 )Σ_(n=0) ^∞ x^(6n) dx =Σ_(n=0) ^∞ ∫_0 ^1 (x^(6n) −x^(6n+1) +x^(6n+3) −x^(6n+4) )dx =Σ_(n=0) ^∞ ((1/(6n+1))−(1/(6n+2))+(1/(6n+4))−(1/(6n+5))) =Σ_(n=0) ^∞ (1/((6n+1)(6n+2))) +Σ_(n=0) ^∞ (1/((6n+4)(6n+5))) we have Σ_(n=0) ^∞ (1/((6n+1)(6n+2))) =(1/(36))Σ_(n=0) ^∞ (1/((n+(1/6))(n+(1/3)))) =(1/(36)).((Ψ((1/3))−Ψ((1/6)))/((1/3)−(1/6))) =(1/6){Ψ((1/3))−Ψ((1/6))} Σ_(n=0) ^∞ (1/((6n+4)(6n+5))) =(1/(36))Σ_(n=0) ^∞ (1/((n+(2/3))(n+(5/6)))) =(1/(36)).((Ψ((5/6))−Ψ((2/3)))/((5/6)−(2/3))) =(1/6)(Ψ((5/6))−Ψ((2/3))) ⇒I=(1/6){Ψ((1/3))−Ψ((1/6))+Ψ((5/6))−Ψ((2/3))} after use special values of Ψ....
I=∫0∞dx1+x+x2+x3+x4+x5I=∫0∞dx1−x61−x=∫0∞1−x1−x6dx=∫011−x1−x6dx+∫1∞1−x1−x6dx(→x=1t)=∫011−x1−x6dx−∫011−1t1−1t6×−dtt2=∫011−x1−x6dx+∫011t3.t−1t6−1.t6dt=∫011−x1−x6dx+∫01t4−t3t6−1dt=∫011−x1−x6dx+∫01x3−x41−x6dx=∫011−x+x3−x41−x6dx=∫01(1−x+x3−x4)∑n=0∞x6ndx=∑n=0∞∫01(x6n−x6n+1+x6n+3−x6n+4)dx=∑n=0∞(16n+1−16n+2+16n+4−16n+5)=∑n=0∞1(6n+1)(6n+2)+∑n=0∞1(6n+4)(6n+5)wehave∑n=0∞1(6n+1)(6n+2)=136∑n=0∞1(n+16)(n+13)=136.Ψ(13)−Ψ(16)13−16=16{Ψ(13)−Ψ(16)}∑n=0∞1(6n+4)(6n+5)=136∑n=0∞1(n+23)(n+56)=136.Ψ(56)−Ψ(23)56−23=16(Ψ(56)−Ψ(23))⇒I=16{Ψ(13)−Ψ(16)+Ψ(56)−Ψ(23)}afterusespecialvaluesofΨ….

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