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Question-59803

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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 ))
$${let}\:{A}\:=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\frac{\left({lnx}\right)^{\mathrm{2}} }{\:\sqrt{{x}}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:{dx}\:\Rightarrow\pi^{\mathrm{2}} \:{A}\:=_{\sqrt{{x}}={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left({ln}\left({t}^{\mathrm{2}} \right)\right)^{\mathrm{2}} }{{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{8}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:\:\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:=_{{t}=\frac{\mathrm{1}}{{u}}} \:\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\left({lnu}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right) \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\left({lnu}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:{for}\:\mid{x}\mid<\mathrm{1}\:\Rightarrow\sum_{{n}=\mathrm{1}} ^{\infty} \:{nx}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{nt}^{\mathrm{2}{n}−\mathrm{2}} \:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\:{t}^{\mathrm{2}{n}−\mathrm{2}} \right)\left({lnt}\right)^{\mathrm{2}} {dt} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{n}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{t}^{\mathrm{2}{n}−\mathrm{2}} \left({ln}\left({t}\right)\right)^{\mathrm{2}} {dt}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\:{A}_{{n}} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{\mathrm{2}{n}−\mathrm{2}} \left({ln}\left({t}\right)\right)^{\mathrm{2}} \:{dt}\:\:{by}\:{parts}\:\:{u}^{'} ={t}^{\mathrm{2}{n}−\mathrm{2}} \:\:\:{and}\:\:{v}\:=\left({ln}\left({t}\right)^{\mathrm{2}} \:\Rightarrow\right. \\ $$$${A}_{{n}} =\left[\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:{t}^{\mathrm{2}{n}−\mathrm{1}} \:\left({ln}\left({t}\right)\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}{t}^{\mathrm{2}{n}−\mathrm{1}} \:\frac{\mathrm{2}{ln}\left({t}\right)}{{t}}\:{dt} \\ $$$$=−\frac{\mathrm{2}}{\mathrm{2}{n}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:{t}^{\mathrm{2}{n}−\mathrm{2}} \:{ln}\left({t}\right){dt}\:\:{again}\:{by}\:{parts}\:{u}^{'} ={t}^{\mathrm{2}{n}−\mathrm{2}} \:{and}\:{v}={ln}\left({t}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{\mathrm{2}{n}−\mathrm{2}} {ln}\left({t}\right){dt}\:=\left[\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:{t}^{\mathrm{2}{n}−\mathrm{1}} {ln}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:{t}^{\mathrm{2}{n}−\mathrm{1}} \:\frac{\mathrm{1}}{{t}}\:{dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}−\mathrm{2}} \:{dt}\:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\:{A}_{{n}} =\frac{\mathrm{2}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\left({lnt}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}{n}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow\:{A}\:=\mathrm{16}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}{n}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\mathrm{32}\:\sum_{{n}=\mathrm{1}} ^{\infty\:\:\:} \:\:\:\frac{{n}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }\:\:{let}\:{determine}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }\:={S} \\ $$$${S}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}{n}−\mathrm{1}\:+\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$ \\ $$
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) .
$${we}\:{have}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\frac{\mathrm{3}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${and}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }\:=_{{n}={p}+\mathrm{1}} \:\:\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{p}+\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\: \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{3}} }\:=\xi\left(\mathrm{3}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{3}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:=\frac{\mathrm{7}}{\mathrm{8}}\xi\left(\mathrm{3}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }\:\:\Rightarrow\:{S}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:+\frac{\mathrm{7}}{\mathrm{16}}\xi\left(\mathrm{3}\right)\:\Rightarrow \\ $$$$\pi^{\mathrm{2}} {A}\:=\mathrm{32}\left\{\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\:+\frac{\mathrm{7}}{\mathrm{16}}\:\xi\left(\mathrm{3}\right)\right\}\:\:\Rightarrow\pi^{\mathrm{2}} \:{A}\:=\mathrm{2}\pi^{\mathrm{2}} \:+\mathrm{14}\:\xi\left(\mathrm{3}\right)\:\Rightarrow{A}\:=\mathrm{2}\:+\frac{\mathrm{14}}{\pi^{\mathrm{2}} }\:\xi\left(\mathrm{3}\right)\:. \\ $$
Commented by bhanukumarb2@gmail.com last updated on 16/May/19
very osm
$${very}\:{osm} \\ $$

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