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Question Number 59803 by bhanukumarb2@gmail.com last updated on 15/May/19

Commented by maxmathsup by imad last updated on 16/May/19

let A =(1/π^2 ) ∫_0 ^∞ (((lnx)^2 )/((√x)(1−x)^2 )) dx ⇒π^2 A =_((√x)=t) ∫_0 ^∞ (((ln(t^2 ))^2 )/(t(1−t^2 )^2 )) (2t)dt =8 ∫_0 ^∞ (((lnt)^2 )/((1−t^2 )^2 )) dt but ∫_0 ^∞ (((lnt)^2 )/((1−t^2 )^2 ))dt =∫_0 ^1 (((lnt)^2 )/((1−t^2 )^2 )) dt +∫_1 ^(+∞) (((lnt)^2 )/((1−t^2 )^2 ))dt ∫_1 ^(+∞) (((lnt)^2 )/((1−t^2 )^2 )) dt =_(t=(1/u)) −∫_0 ^1 (((lnu)^2 )/((1−(1/u^2 ))^2 )) (−(du/u^2 )) = ∫_0 ^1 (((lnu)^2 )/((1−u^2 )^2 )) ⇒ ∫_0 ^∞ (((lnt)^2 )/((1−t^2 )^2 )) dt =2 ∫_0 ^1 (((lnt)^2 )/((1−t^2 )^2 )) dt Σ_(n=0) ^∞ x^n =(1/(1−x)) for ∣x∣<1 ⇒Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒ (1/((1−t^2 )^2 )) =Σ_(n=1) ^∞ nt^(2n−2) ⇒∫_0 ^1 (((lnt)^2 )/((1−t^2 )^2 ))dt = ∫_0 ^1 (Σ_(n=1) ^∞ n t^(2n−2) )(lnt)^2 dt =Σ_(n=1) ^∞ n ∫_0 ^1 t^(2n−2) (ln(t))^2 dt =Σ_(n=1) ^∞ n A_n A_n = ∫_0 ^1 t^(2n−2) (ln(t))^2 dt by parts u^′ =t^(2n−2) and v =(ln(t)^2 ⇒ A_n =[(1/(2n−1)) t^(2n−1) (ln(t))^2 ]_0 ^1 −∫_0 ^1 (1/(2n−1))t^(2n−1) ((2ln(t))/t) dt =−(2/(2n−1)) ∫_0 ^1 t^(2n−2) ln(t)dt again by parts u^′ =t^(2n−2) and v=ln(t) ∫_0 ^1 t^(2n−2) ln(t)dt =[(1/(2n−1)) t^(2n−1) ln(t)]_0 ^1 −∫_0 ^1 (1/(2n−1)) t^(2n−1) (1/t) dt =−(1/(2n−1)) ∫_0 ^1 t^(2n−2) dt =−(1/((2n−1)^2 )) ⇒ A_n =(2/((2n−1)^3 )) ⇒ ∫_0 ^1 (((lnt)^2 )/((1−t^2 )^2 )) dt =Σ_(n=1) ^∞ ((2n)/((2n−1)^3 )) ⇒ A =16 Σ_(n=1) ^∞ ((2n)/((2n−1)^3 )) =32 Σ_(n=1) ^(∞ ) (n/((2n−1)^3 )) let determine Σ_(n=1) ^∞ (n/((2n−1)^3 )) =S S =(1/2) Σ_(n=1) ^∞ ((2n−1 +1)/((2n−1)^3 )) =(1/2) Σ_(n=1) ^∞ (1/((2n−1)^2 )) +(1/2) Σ_(n=1) ^∞ (1/((2n−1)^3 ))

letA=1π2∫0∞(lnx)2x(1−x)2dx⇒π2A=x=t∫0∞(ln(t2))2t(1−t2)2(2t)dt=8∫0∞(lnt)2(1−t2)2dtbut∫0∞(lnt)2(1−t2)2dt=∫01(lnt)2(1−t2)2dt+∫1+∞(lnt)2(1−t2)2dt∫1+∞(lnt)2(1−t2)2dt=t=1u−∫01(lnu)2(1−1u2)2(−duu2)=∫01(lnu)2(1−u2)2⇒∫0∞(lnt)2(1−t2)2dt=2∫01(lnt)2(1−t2)2dt∑n=0∞xn=11−xfor∣x∣<1⇒∑n=1∞nxn−1=1(1−x)2⇒1(1−t2)2=∑n=1∞nt2n−2⇒∫01(lnt)2(1−t2)2dt=∫01(∑n=1∞nt2n−2)(lnt)2dt=∑n=1∞n∫01t2n−2(ln(t))2dt=∑n=1∞nAnAn=∫01t2n−2(ln(t))2dtbypartsu′=t2n−2andv=(ln(t)2⇒An=[12n−1t2n−1(ln(t))2]01−∫0112n−1t2n−12ln(t)tdt=−22n−1∫01t2n−2ln(t)dtagainbypartsu′=t2n−2andv=ln(t)∫01t2n−2ln(t)dt=[12n−1t2n−1ln(t)]01−∫0112n−1t2n−11tdt=−12n−1∫01t2n−2dt=−1(2n−1)2⇒An=2(2n−1)3⇒∫01(lnt)2(1−t2)2dt=∑n=1∞2n(2n−1)3⇒A=16∑n=1∞2n(2n−1)3=32∑n=1∞n(2n−1)3letdetermine∑n=1∞n(2n−1)3=SS=12∑n=1∞2n−1+1(2n−1)3=12∑n=1∞1(2n−1)2+12∑n=1∞1(2n−1)3

Commented by maxmathsup by imad last updated on 16/May/19

we have Σ_(n=1) ^∞ (1/n^2 ) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒(3/4) Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒(3/4) (π^2 /6) =(π^2 /8) =Σ_(n=0) ^∞ (1/((2n+1)^2 )) and Σ_(n=1) ^∞ (1/((2n−1)^2 )) =_(n=p+1) Σ_(p=0) ^∞ (1/((2p+1)^2 )) =(π^2 /8) Σ_(n=1) ^∞ (1/n^3 ) =ξ(3) =Σ_(n=1) ^∞ (1/(8n^3 )) +Σ_(n=0) ^∞ (1/((2n+1)^3 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^3 )) =(7/8)ξ(3) =Σ_(n=1) ^∞ (1/((2n−1)^3 )) ⇒ S =(π^2 /(16)) +(7/(16))ξ(3) ⇒ π^2 A =32{(π^2 /(16)) +(7/(16)) ξ(3)} ⇒π^2 A =2π^2 +14 ξ(3) ⇒A =2 +((14)/π^2 ) ξ(3) .

wehave∑n=1∞1n2=14∑n=1∞1n2+∑n=0∞1(2n+1)2⇒34∑n=1∞1n2=∑n=0∞1(2n+1)2⇒34π26=π28=∑n=0∞1(2n+1)2and∑n=1∞1(2n−1)2=n=p+1∑p=0∞1(2p+1)2=π28∑n=1∞1n3=ξ(3)=∑n=1∞18n3+∑n=0∞1(2n+1)3⇒∑n=0∞1(2n+1)3=78ξ(3)=∑n=1∞1(2n−1)3⇒S=π216+716ξ(3)⇒π2A=32{π216+716ξ(3)}⇒π2A=2π2+14ξ(3)⇒A=2+14π2ξ(3).

Commented by bhanukumarb2@gmail.com last updated on 16/May/19

very osm

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