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Question Number 136420 by Dwaipayan Shikari last updated on 21/Mar/21

Σ_(n=1) ^∞ ((sin(2n+1)θ)/((2n+1)^2 ))=(θ/2)log(2)+sinθ((log(sinθ))/4) _2 F_1 ((1/2),(1/2);(3/2);sin^2 θ)+((sinθ)/(16)) _2 F_1 ((1/2),(1/2),(1/2);(3/2);(3/2);sin^2 θ)  Prove or disprove     0<θ<π

n=1sin(2n+1)θ(2n+1)2=θ2log(2)+sinθlog(sinθ)42F1(12,12;32;sin2θ)+sinθ162F1(12,12,12;32;32;sin2θ)Proveordisprove0<θ<π

Answered by mindispower last updated on 21/Mar/21

θ=π,sin(2n+1)θ=0  left=0  right=((πln(2))/2)+(1/4)+(1/(16))

θ=π,sin(2n+1)θ=0left=0right=πln(2)2+14+116

Commented by Dwaipayan Shikari last updated on 21/Mar/21

Sorry i have edited the question

Sorryihaveeditedthequestion

Commented by Dwaipayan Shikari last updated on 21/Mar/21

Valid for   0<θ<π

Validfor0<θ<π

Commented by mindispower last updated on 22/Mar/21

ok i will try but  tchek it again

okiwilltrybuttchekitagain

Answered by mindispower last updated on 22/Mar/21

Σ_(n≥0) ((sin(2n+1)θ)/((2n+1)^2 ))∫=S=Σ_(n≥1) ((sin(nθ))/n^2 )−Σ_(n≥1) ((sin(2θ))/(4n^2 ))  recal Σ_(n≥1) ((sin(nθ))/n^2 )=Cl_2 (θ),clausen function  we get S=Cl_2 (θ)−((Cl_2 (2θ))/4)  clausen integration  representation  Cl_2 (θ)=−2∫_0 ^(θ/2) ln(2sin(t))dt  =u=sin(t)=−2∫_0 ^(sin((θ/2))) ((ln(2u))/( (√(1−u^2 ))))du=cl_2 (θ)  (1+x)^a =1+Σ_(n≥1) ((Π_(k≤n−1) (a−k))/(n!)).x^n   cl_2 (θ)=−2∫_0 ^(sin((θ/2))) (1+Σ_(n≥1) (−1)^n ((Π_(k≤n−1) (−(1/2)−k)u^(2n) )/(n!)))ln(2u)du  ∫u^(2n) ln(2u)du=(u^(2n+1) /(2n+1))ln(2u)−(u^(2n+1) /((2n+1)^2 ))  it seems good starte we can fin _p F_q     sinlog(sin..) also i think verry long worck  =−2(1+Σ_(n≥1) ((((1/2))_n )/(n!)))(((sin^(2n+1) ((θ/2))ln(2sin((θ/2))))/((2n+1)))−((sin^(2n+1) ((θ/2)))/((2n+1)^2 )))  =−2(1+Σ_(n≥1) ((((1/2))_n ((1/2))_n )/(2.((3/2))_n ))

n0sin(2n+1)θ(2n+1)2=S=n1sin(nθ)n2n1sin(2θ)4n2recaln1sin(nθ)n2=Cl2(θ),clausenfunctionwegetS=Cl2(θ)Cl2(2θ)4clausenintegrationrepresentationCl2(θ)=20θ2ln(2sin(t))dt=u=sin(t)=20sin(θ2)ln(2u)1u2du=cl2(θ)(1+x)a=1+n1kn1(ak)n!.xncl2(θ)=20sin(θ2)(1+n1(1)nkn1(12k)u2nn!)ln(2u)duu2nln(2u)du=u2n+12n+1ln(2u)u2n+1(2n+1)2itseemsgoodstartewecanfinpFqsinlog(sin..)alsoithinkverrylongworck=2(1+n1(12)nn!)(sin2n+1(θ2)ln(2sin(θ2))(2n+1)sin2n+1(θ2)(2n+1)2)=2(1+n1(12)n(12)n2.(32)n

Commented by Dwaipayan Shikari last updated on 22/Mar/21

Thanks sir! I will post my work later

Thankssir!Iwillpostmyworklater

Commented by mindispower last updated on 22/Mar/21

you are welcom

youarewelcom

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