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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
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\theta}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\theta}{\mathrm{2}}{log}\left(\mathrm{2}\right)+{sin}\theta\frac{{log}\left({sin}\theta\right)}{\mathrm{4}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};{sin}^{\mathrm{2}} \theta\right)+\frac{{sin}\theta}{\mathrm{16}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};{sin}^{\mathrm{2}} \theta\right) \\ $$$${Prove}\:{or}\:{disprove}\:\:\:\:\:\mathrm{0}<\theta<\pi \\ $$
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))
$$\theta=\pi,{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\theta=\mathrm{0} \\ $$$${left}=\mathrm{0} \\ $$$${right}=\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{16}} \\ $$
Commented by Dwaipayan Shikari last updated on 21/Mar/21
Sorry i have edited the question
$${Sorry}\:{i}\:{have}\:{edited}\:{the}\:{question} \\ $$
Commented by Dwaipayan Shikari last updated on 21/Mar/21
Valid for 0<θ<π
$${Valid}\:{for}\:\:\:\mathrm{0}<\theta<\pi \\ $$
Commented by mindispower last updated on 22/Mar/21
ok i will try but tchek it again
$${ok}\:{i}\:{will}\:{try}\:{but}\:\:{tchek}\:{it}\:{again} \\ $$
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 ))
$$\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\theta}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\int={S}=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({n}\theta\right)}{{n}^{\mathrm{2}} }−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left(\mathrm{2}\theta\right)}{\mathrm{4}{n}^{\mathrm{2}} } \\ $$$${recal}\:\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{sin}\left({n}\theta\right)}{{n}^{\mathrm{2}} }={Cl}_{\mathrm{2}} \left(\theta\right),{clausen}\:{function} \\ $$$${we}\:{get}\:{S}={Cl}_{\mathrm{2}} \left(\theta\right)−\frac{{Cl}_{\mathrm{2}} \left(\mathrm{2}\theta\right)}{\mathrm{4}} \\ $$$${clausen}\:{integration}\:\:{representation} \\ $$$${Cl}_{\mathrm{2}} \left(\theta\right)=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\theta}{\mathrm{2}}} {ln}\left(\mathrm{2}{sin}\left({t}\right)\right){dt} \\ $$$$={u}={sin}\left({t}\right)=−\mathrm{2}\int_{\mathrm{0}} ^{{sin}\left(\frac{\theta}{\mathrm{2}}\right)} \frac{{ln}\left(\mathrm{2}{u}\right)}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}{du}={cl}_{\mathrm{2}} \left(\theta\right) \\ $$$$\left(\mathrm{1}+{x}\right)^{{a}} =\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}\leqslant{n}−\mathrm{1}} {\prod}\left({a}−{k}\right)}{{n}!}.{x}^{{n}} \\ $$$${cl}_{\mathrm{2}} \left(\theta\right)=−\mathrm{2}\int_{\mathrm{0}} ^{{sin}\left(\frac{\theta}{\mathrm{2}}\right)} \left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{n}} \frac{\underset{{k}\leqslant{n}−\mathrm{1}} {\prod}\left(−\frac{\mathrm{1}}{\mathrm{2}}−{k}\right){u}^{\mathrm{2}{n}} }{{n}!}\right){ln}\left(\mathrm{2}{u}\right){du} \\ $$$$\int{u}^{\mathrm{2}{n}} {ln}\left(\mathrm{2}{u}\right){du}=\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}{n}+\mathrm{1}}{ln}\left(\mathrm{2}{u}\right)−\frac{{u}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${it}\:{seems}\:{good}\:{starte}\:{we}\:{can}\:{fin}\:_{{p}} {F}_{{q}} \:\: \\ $$$${sinlog}\left({sin}..\right)\:{also}\:{i}\:{think}\:{verry}\:{long}\:{worck} \\ $$$$=−\mathrm{2}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}\right)\left(\frac{{sin}^{\mathrm{2}{n}+\mathrm{1}} \left(\frac{\theta}{\mathrm{2}}\right){ln}\left(\mathrm{2}{sin}\left(\frac{\theta}{\mathrm{2}}\right)\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)}−\frac{{sin}^{\mathrm{2}{n}+\mathrm{1}} \left(\frac{\theta}{\mathrm{2}}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=−\mathrm{2}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{\mathrm{2}.\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{{n}} }\right. \\ $$
Commented by Dwaipayan Shikari last updated on 22/Mar/21
Thanks sir! I will post my work later
$${Thanks}\:{sir}!\:{I}\:{will}\:{post}\:{my}\:{work}\:{later} \\ $$
Commented by mindispower last updated on 22/Mar/21
you are welcom
$${you}\:{are}\:{welcom} \\ $$

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