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Question Number 133050 by mnjuly1970 last updated on 18/Feb/21
           ...nice ......calculus...       𝛗= ∫_(0 ) ^( 1) xli_3 (x)dx=???
$$\:\:\:\:\:\:\:\:\:\:\:…{nice}\:……{calculus}… \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\int_{\mathrm{0}\:} ^{\:\mathrm{1}} {xli}_{\mathrm{3}} \left({x}\right){dx}=??? \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 18/Feb/21
∫_0 ^1 xli_3 (x)dx=∫_0 ^1 xΞ£_(n=1) ^∞ (x^n /n^3 )=Ξ£_(n=1) ^∞ (1/(n^3 (n+2)))  =(1/2)Ξ£_(n=1) ^∞ (1/n^3 )βˆ’(1/2)Ξ£_(n=1) ^∞ (1/(n^2 (n+2)))=((ΞΆ(3))/2)βˆ’(1/4)Ξ£_(n=1) ^∞ (1/n^2 )+(1/4)Ξ£_(n=1) ^∞ (1/(n(n+2)))  =((ΞΆ(3))/2)βˆ’(Ο€^2 /(24))+(1/8)(1+(1/2))=((ΞΆ(3))/2)βˆ’(Ο€^2 /(24))+(3/(16))=(1/(48))(24ΞΆ(3)βˆ’2Ο€^2 +9)
$$\int_{\mathrm{0}} ^{\mathrm{1}} {xli}_{\mathrm{3}} \left({x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{3}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} \left({n}+\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }βˆ’\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} \left({n}+\mathrm{2}\right)}=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}βˆ’\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)} \\ $$$$=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\mathrm{3}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{48}}\left(\mathrm{24}\zeta\left(\mathrm{3}\right)βˆ’\mathrm{2}\pi^{\mathrm{2}} +\mathrm{9}\right) \\ $$
Commented by mnjuly1970 last updated on 18/Feb/21
thank you for your effort..  grateful...
$${thank}\:{you}\:{for}\:{your}\:{effort}.. \\ $$$${grateful}… \\ $$
Answered by mnjuly1970 last updated on 18/Feb/21
    𝛗=[(x^2 /2)li_3 (x)]_0 ^1 βˆ’(1/2)∫_0 ^( 1) xli_2 (x)dx          =(1/2)li_3 (1)βˆ’(Ξ¦/2)      where  Ξ¦=∫_0 ^( 1) xli_2 (x)dx                    =[(x^2 /2)li_2 (x)]_0 ^1 βˆ’(1/2)∫_0 ^( 1) xli_1 (x)dx        =(Ο€^2 /(12))βˆ’(Ξ¨/2)      Ξ¨=∫_0 ^( 1) xli_1 (x)dx=Ξ£_(n=1) ^∞ ∫_0 ^( 1) (x^(n+1) /n)dx        =Ξ£_(n=1) ^∞ (1/(n(n+2)))=(1/2)(Ξ£_(n=1) ^∞ (1/n)βˆ’(1/(n+2)))       =(1/2)((1/1)βˆ’(1/3)+(1/2)βˆ’(1/4)+(1/3)βˆ’(1/5)+(1/4)βˆ’(1/6)+....)       =(3/4)     ∴   𝛗 =((ΞΆ(3))/2) βˆ’(1/2)((Ο€^2 /(12))βˆ’(1/2)((3/4)))           =((ΞΆ(3))/2)βˆ’(Ο€^2 /(24))+(3/(16)) ....
$$\:\:\:\:\boldsymbol{\phi}=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{li}_{\mathrm{3}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {xli}_{\mathrm{2}} \left({x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}{li}_{\mathrm{3}} \left(\mathrm{1}\right)βˆ’\frac{\Phi}{\mathrm{2}} \\ $$$$\:\:\:\:{where}\:\:\Phi=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xli}_{\mathrm{2}} \left({x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\frac{{x}^{\mathrm{2}} }{\mathrm{2}}{li}_{\mathrm{2}} \left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {xli}_{\mathrm{1}} \left({x}\right){dx} \\ $$$$\:\:\:\:\:\:=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}βˆ’\frac{\Psi}{\mathrm{2}} \\ $$$$\:\:\:\:\Psi=\int_{\mathrm{0}} ^{\:\mathrm{1}} {xli}_{\mathrm{1}} \left({x}\right){dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} }{{n}}{dx} \\ $$$$\:\:\:\:\:\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}\left({n}+\mathrm{2}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}βˆ’\frac{\mathrm{1}}{{n}+\mathrm{2}}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}}βˆ’\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}βˆ’\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{3}}βˆ’\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{4}}βˆ’\frac{\mathrm{1}}{\mathrm{6}}+….\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\:\:\therefore\:\:\:\boldsymbol{\phi}\:=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}\:βˆ’\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}βˆ’\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{2}}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{24}}+\frac{\mathrm{3}}{\mathrm{16}}\:…. \\ $$
Commented by Dwaipayan Shikari last updated on 18/Feb/21
There was a mistake on my answer . I have edited
$${There}\:{was}\:{a}\:{mistake}\:{on}\:{my}\:{answer}\:.\:{I}\:{have}\:{edited} \\ $$
Commented by mnjuly1970 last updated on 18/Feb/21
  grateful sir payan...
$$\:\:{grateful}\:{sir}\:{payan}… \\ $$$$ \\ $$

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