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Question Number 158405 by Eric002 last updated on 03/Nov/21
prove: ∫_0 ^∞ (1/(x^5 +x^4 +x^3 +x^2 +x+1))dx=(π/(3(√3)))
prove:∫0∞1x5+x4+x3+x2+x+1dx=π33
Answered by MJS_new last updated on 03/Nov/21
x^5 +x^4 +x^3 +x^2 +x+1=((x^6 −1)/(x−1)) ⇒ ∫(dx/(x^5 +x^4 +x^3 +x^2 +x+1))= =∫(dx/((x+1)(x^2 +x+1)(x^2 +x+1)))= =(1/3)∫(dx/(x+1))+(1/2)∫(dx/(x^2 +x+1))−(1/6)∫((2x−1)/(x^2 −x+1))= =(1/3)ln ∣x+1∣ +((√3)/3)arctan ((2x+1)/( (√3))) −(1/6)ln (x^2 −x+1) +C ⇒ ∫_0 ^∞ (dx/(x^5 +x^4 +x^3 +x^2 +x+1))=(π/(3(√3)))
x5+x4+x3+x2+x+1=x6−1x−1⇒∫dxx5+x4+x3+x2+x+1==∫dx(x+1)(x2+x+1)(x2+x+1)==13∫dxx+1+12∫dxx2+x+1−16∫2x−1x2−x+1==13ln∣x+1∣+33arctan2x+13−16ln(x2−x+1)+C⇒∫∞0dxx5+x4+x3+x2+x+1=π33
Commented by Eric002 last updated on 03/Nov/21
well done
welldone
Commented by Tawa11 last updated on 04/Nov/21
Great sir
Greatsir
Answered by Ar Brandon last updated on 03/Nov/21
I=∫_0 ^∞ (dx/(x^5 +x^4 +x^3 +x^2 +x+1))=∫_0 ^∞ ((1−x)/(1−x^6 ))dx =∫_0 ^1 ((1−x)/(1−x^6 ))dx+∫_1 ^∞ ((1−x)/(1−x^6 ))dx=∫_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=(1/6)∫_0 ^1 ((v^(−(5/6)) −v^(−(4/6)) +v^(−(2/6)) −v^(−(1/6)) )/(1−v))dv =(1/6)[−ψ((1/6))+ψ((2/6))−ψ((4/6))+ψ((5/6))] =(1/6)[(ψ((5/6))−ψ((1/6)))+(ψ((1/3))−ψ((2/3)))] =(1/6)[−πcot((5/6)π)−πcot((π/3))]=(1/6)[π(√3)−(π/( (√3)))]=((2(√3)π)/(18))=((√3)/9)π
I=∫0∞dxx5+x4+x3+x2+x+1=∫0∞1−x1−x6dx=∫011−x1−x6dx+∫1∞1−x1−x6dx=∫011−x1−x6dx+∫01x3−x41−x6dx=∫011−x+x3−x41−x6dx=16∫01v−56−v−46+v−26−v−161−vdv=16[−ψ(16)+ψ(26)−ψ(46)+ψ(56)]=16[(ψ(56)−ψ(16))+(ψ(13)−ψ(23))]=16[−πcot(56π)−πcot(π3)]=16[π3−π3]=23π18=39π
Commented by Ar Brandon last updated on 03/Nov/21
Φ=∫_1 ^∞ ((1−x)/(1−x^6 ))dx, u=(1/x)⇒x=(1/u)⇒dx=−(1/u^2 )du ⇒Φ=∫_0 ^1 ((1−(1/u))/(1−(1/u^6 )))∙(du/u^6 )=∫_0 ^1 ((u^4 −u^3 )/(u^6 −1))du=∫_0 ^1 ((x^3 −x^4 )/(1−x^6 ))dx
Φ=∫1∞1−x1−x6dx,u=1x⇒x=1u⇒dx=−1u2du⇒Φ=∫011−1u1−1u6⋅duu6=∫01u4−u3u6−1du=∫01x3−x41−x6dx

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