Menu Close

Show-that-0-x-2-ln-x-x-4-x-2-1-dx-pi-2-12-

Tinku Tara 0 Comments
Pin




Question Number 130053 by Lordose last updated on 22/Jan/21
Show that ∫_0 ^( ∞) ((x^2 ln(x))/(x^4 +x^2 +1))dx = (π^2 /(12))
Showthat∫0∞x2ln(x)x4+x2+1dx=π212
Answered by Ar Brandon last updated on 22/Jan/21
f(a)=∫_0 ^∞ ((x^2 ln(ax))/(x^4 +x^2 +1))dx f ′(a)=∫_0 ^∞ (x/(x^4 +x^2 +1))dx , x^2 =u =(1/2)∫_0 ^∞ (du/(u^2 +u+1))=(1/2)∫_0 ^∞ (du/((u+(1/2))^2 +(3/4))) =(1/( 2(√3)))[tan^(−1) (((2u+1)/( (√3))))]_0 ^∞ =(1/( 2(√3)))[(π/2)−(π/6)]=(π/( 6(√3))) f(a)=(π/(6(√3)))a+C...
f(a)=∫0∞x2ln(ax)x4+x2+1dxf′(a)=∫0∞xx4+x2+1dx,x2=u=12∫0∞duu2+u+1=12∫0∞du(u+12)2+34=123[tan−1(2u+13)]0∞=123[π2−π6]=π63f(a)=π63a+C…
Answered by mathmax by abdo last updated on 22/Jan/21
let try this way Φ=∫_0 ^∞ ((x^2 lnx)/(x^4 +x^2 +1))dx ⇒ Φ=∫_0 ^1 ((x^2 lnx)/(x^4 +x^2 +1))dx +∫_1 ^∞ ((x^2 lnx)/(x^4 +x^2 +1))dx(→x=(1/t)) =∫_0 ^1 ((x^2 lnx)/(x^4 +x^2 +1))dx−∫_0 ^1 ((−lnt)/(t^2 ((1/t^4 ) +(1/t^2 ) +1)))(−(dt/t^2 )) =∫_0 ^1 ((x^2 lnx)/(x^4 +x^2 +1))dx−∫_0 ^1 ((lnt)/(t^4 +t^2 +1))dt =∫_0 ^1 ((x^2 lnx−lnx)/(1+x^2 +(x^2 )^2 )) =∫_0 ^1 (((1−x^2 )(x^2 −1)lnx)/(1−x^6 ))dx=−∫_0 ^1 (((x^2 −1)^2 lnx)/(1−x^6 ))dx =−∫_0 ^1 ((x^4 −2x^2 +1)/(1−x^6 ))dx =−∫_0 ^1 (x^4 −2x^2 +1)Σ_(n=0) ^∞ x^(6n) =−Σ_(n=0) ^∞ ∫_0 ^1 x^(6n+4) dx+2Σ_(n=0) ^∞ ∫_0 ^1 x^(6n+2) dx−Σ_(n=0) ^∞ ∫_0 ^1 x^(6n) dx =−Σ_(n=0) ^∞ (1/(6n+5))+2Σ_(n=0) ^∞ (1/(6n+3))−Σ_(n=0) ^∞ (1/(6n+1)) rest calculus of those series ...be continued...
lettrythiswayΦ=∫0∞x2lnxx4+x2+1dx⇒Φ=∫01x2lnxx4+x2+1dx+∫1∞x2lnxx4+x2+1dx(→x=1t)=∫01x2lnxx4+x2+1dx−∫01−lntt2(1t4+1t2+1)(−dtt2)=∫01x2lnxx4+x2+1dx−∫01lntt4+t2+1dt=∫01x2lnx−lnx1+x2+(x2)2=∫01(1−x2)(x2−1)lnx1−x6dx=−∫01(x2−1)2lnx1−x6dx=−∫01x4−2x2+11−x6dx=−∫01(x4−2x2+1)∑n=0∞x6n=−∑n=0∞∫01x6n+4dx+2∑n=0∞∫01x6n+2dx−∑n=0∞∫01x6ndx=−∑n=0∞16n+5+2∑n=0∞16n+3−∑n=0∞16n+1restcalculusofthoseseries…becontinued…
Answered by mnjuly1970 last updated on 22/Jan/21
φ=∫_(0 ) ^( 1) ((x^2 ln(x))/(1+x^2 +x^4 ))dx+[∫_1 ^( ∞) ((x^2 ln(x))/(1+x^2 +x^4 ))dx =Φ] Φ=∫_0 ^( 1) ((−ln(x) )/(1+x^2 +x^4 ))dx ∅=∫_0 ^( 1) (((x^2 −1)ln(x))/(1+x^2 +x^4 ))dx=∫_0 ^( 1) ((−(x^2 −1)^2 ln(x))/(1−x^6 ))dx =−∫_0 ^( 1) (((x^4 −2x^2 +1)ln(x))/(1−x^6 )) =^(x^6 =t) −(1/6)∫_0 ^( 1) (((t^(2/3) −2t^(1/3) +1)t^((−5)/6) ln(t))/(1−t))dt =((−1)/(36))∫_0 ^( 1() ((t^((−1)/6) −2t^((−1)/2) +t^((−5)/6) )ln(t))/(1−t))dt f(a)=∫_0 ^( 1) ((t^(a−(1/6)) −2t^(a−(1/2)) +t^(a−(5/6)) )/(1−t))dt φ=((−1)/(36))f ′(0) f(a)=∫_0 ^( 1) ((t^(a−(1/6)) −1+1−t^(a−(1/2)) −1+t^(a−(5/6)) −t^(a−(1/2)) +1)/(1−t))dt =−ψ(a+(5/6))+2ψ(a+(1/2))−ψ(a+(1/6)) =−ψ(a+(5/6))−ψ(a+(1/6))+2ψ(a+(1/2)) f ′(a)=−ψ′(a+(5/6))−ψ′(a+(1/6))+2ψ′(a+(1/2)) f ′(0)=−ψ′((5/6))−ψ′((1/6))+2ψ′((1/2)) =−(π^2 /(sin^2 ((π/6))))+2 (π^2 /2)=−3π^2 φ=(π^2 /(12)) .....✓✓ m.n.july.1970
ϕ=∫01x2ln(x)1+x2+x4dx+[∫1∞x2ln(x)1+x2+x4dx=Φ]Φ=∫01−ln(x)1+x2+x4dx∅=∫01(x2−1)ln(x)1+x2+x4dx=∫01−(x2−1)2ln(x)1−x6dx=−∫01(x4−2x2+1)ln(x)1−x6=x6=t−16∫01(t23−2t13+1)t−56ln(t)1−tdt=−136∫01(t−16−2t−12+t−56)ln(t)1−tdtf(a)=∫01ta−16−2ta−12+ta−561−tdtϕ=−136f′(0)f(a)=∫01ta−16−1+1−ta−12−1+ta−56−ta−12+11−tdt=−ψ(a+56)+2ψ(a+12)−ψ(a+16)=−ψ(a+56)−ψ(a+16)+2ψ(a+12)f′(a)=−ψ′(a+56)−ψ′(a+16)+2ψ′(a+12)f′(0)=−ψ′(56)−ψ′(16)+2ψ′(12)=−π2sin2(π6)+2π22=−3π2ϕ=π212…..✓✓m.n.july.1970
Answered by Lordose last updated on 22/Jan/21
𝛀 = lim_(𝛆→0) ∫_𝛆 ^( ∞) ((x^2 ln(x))/(x^4 +x^2 +1))dx = (∫_1 ^( ∞) ((x^2 ln(x))/(x^4 +x^2 +1))dx + ∫_0 ^( 1) ((x^2 ln(x))/(x^4 +x^2 +1))dx) Ω = Φ + Ψ Ψ = ∫_0 ^( ∞) ((x^2 ln(x))/(x^4 +x^2 +1))dx =^(u=(1/x)) ∫_1 ^( 0) −((−((1/u^2 ))ln(u))/(u^2 ((1/u^4 )+(1/u^2 )+1)))du = −∫_0 ^( 1) ((ln(u))/(u^4 +u^2 +1))du Ω = ∫_0 ^( 1) ((x^2 ln(x)−ln(x))/(x^4 +x^2 +1))dx = ∫_0 ^( 1) (((1−x^2 )(x^2 −1)ln(x))/(1−x^6 ))dx = −∫_0 ^( 1) (((x^2 −1)^2 ln(x))/(1−x^6 ))dx Ω =^(t=x^6 ) −∫_0 ^( 1) (((1/6)(t^(1/3) −1)^2 ln(t))/(1−t)) ((t^(−(5/6)) dt)/6) = −(1/(36))∫_0 ^( 1) (((t^((4/6)−(5/6)) −2t^((2/6)−(5/6)) +t^(−(5/6)) )ln(t))/(1−t))dt Ω(n) = ∫_0 ^( 1) (((t^(−(1/6)+n) −2t^(−(1/2)+n) +t^(−(5/6)+n) ))/(1−t))dt Ω = −(1/(36))Ω^′ (0) Ω(n) = ∫_0 ^( 1) ((t^(n−(1/6)) −2t^(n−(1/2)) +t^(n−(5/6)) )/(1−t))dt Ω(n) = −𝛙^((0)) (n+(1/6)) − 𝛙^((0)) (n+(5/6)) + 2𝛙^((0)) (n+(1/2)) Ω′(n) = −𝛙^((0)^′ ) (n+(1/6)) − 𝛙^((0)^′ ) (n+(5/6)) + 2𝛙^((0)′) (n+(1/2)) Ω′(0) = −(𝛙^((0)′) ((1/6)) + 𝛙^((0)′) ((5/6))) + 2𝛙^((0)′) ((1/2)) Ω′(0) = −(π^2 csc^2 ((π/6)))+2 (π^2 /2)=−3π^2 Ω = −(1/(36))Ω′(0) = (1/(36))∙3π^2 = (π^2 /(12)) ★L𝛗rD ∅sE
Ω=limϵ→0∫ϵ∞x2ln(x)x4+x2+1dx=(∫1∞x2ln(x)x4+x2+1dx+∫01x2ln(x)x4+x2+1dx)Ω=Φ+ΨΨ=∫0∞x2ln(x)x4+x2+1dx=u=1x∫10−−(1u2)ln(u)u2(1u4+1u2+1)du=−∫01ln(u)u4+u2+1duΩ=∫01x2ln(x)−ln(x)x4+x2+1dx=∫01(1−x2)(x2−1)ln(x)1−x6dx=−∫01(x2−1)2ln(x)1−x6dxΩ=t=x6−∫0116(t13−1)2ln(t)1−tt−56dt6=−136∫01(t46−56−2t26−56+t−56)ln(t)1−tdtΩ(n)=∫01(t−16+n−2t−12+n+t−56+n)1−tdtΩ=−136Ω′(0)Ω(n)=∫01tn−16−2tn−12+tn−561−tdtΩ(n)=−ψ(0)(n+16)−ψ(0)(n+56)+2ψ(0)(n+12)Ω′(n)=−ψ(0)′(n+16)−ψ(0)′(n+56)+2ψ(0)′(n+12)Ω′(0)=−(ψ(0)′(16)+ψ(0)′(56))+2ψ(0)′(12)Ω′(0)=−(π2csc2(π6))+2π22=−3π2Ω=−136Ω′(0)=136⋅3π2=π212★LϕrD∅sE

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *