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Question Number 132324 by mnjuly1970 last updated on 13/Feb/21
            ....advanced   calculus...     evaluation :        𝛗=∫_(0^(  ) ) ^( ∞) ((ln(1+x))/(x(1+x^2 ))) dx      solution:    𝛗=[∫_0 ^( 1) ((ln(1+x))/(x(1+x^2 )))dx=𝛗_1 ]+[∫_1 ^( ∞) ((ln(1+x))/(x(1+x^2 ))) dx=𝛗_2 ]     𝛗_2 =^(x=(1/t)) ∫_0 ^( 1) ((ln(1+(1/t)))/((1/t)(1+(1/t^2 ))))(dt/t^2 )=∫_0 ^( 1) ((tln(1+(1/t)))/(1+t^2 ))dt  =∫_0 ^( 1) ((tln(1+t))/(1+t^2 ))dtβˆ’βˆ«_0 ^( 1) ((tln(t))/(1+t^2 ))dt  ∴  𝛗=𝛗_1 +𝛗_2 =∫_0 ^( 1) ((1/x)+x)((ln(1+x))/(1+x^2 ))dxβˆ’Ξ¦  𝛗=∫_0 ^( 1) ((ln(1+x))/x)dxβˆ’Ξ¦=βˆ’li_2 (βˆ’1)βˆ’Ξ¦   Ξ¦=∫_0 ^( 1) ((xln(x))/(1+x^2 ))dx=Ξ£_(n=0) ^∞ ∫_0 ^( 1) (βˆ’1)^n x^(2n+1) ln(x)dx    =Ξ£_(n=0) ^∞ (βˆ’1)^n {[(x^(2n+2) /(2n+2)) ln(x)]_0 ^1 βˆ’(1/(2n+2))∫_0 ^( 1) x^(2n+1) dx}  =Ξ£_(n=0) ^∞ (βˆ’1)^(n+1) (1/(4(n+1)^2 ))=βˆ’(1/4) Ξ£_(n=1) ^∞ (((βˆ’1)^(nβˆ’1) )/n^2 )  =((βˆ’1)/4) Ξ·(2)=((βˆ’Ο€^2 )/(48))   ....      ∴ 𝛗=βˆ’li_2 (βˆ’1)βˆ’Ξ¦=(Ο€^2 /(12))+(Ο€^2 /(48))                  𝛗=((5Ο€^2 )/(48))
$$\:\:\:\:\:\:\:\:\:\:\:\:….{advanced}\:\:\:{calculus}… \\ $$$$\:\:\:{evaluation}\:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}\:^{\:\:} } ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx} \\ $$$$\:\:\:\:{solution}: \\ $$$$\:\:\boldsymbol{\phi}=\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}=\boldsymbol{\phi}_{\mathrm{1}} \right]+\left[\int_{\mathrm{1}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx}=\boldsymbol{\phi}_{\mathrm{2}} \right] \\ $$$$\:\:\:\boldsymbol{\phi}_{\mathrm{2}} \overset{{x}=\frac{\mathrm{1}}{{t}}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)}{\frac{\mathrm{1}}{{t}}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}\frac{{dt}}{{t}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{tln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{tln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}βˆ’\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{tln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\therefore\:\:\boldsymbol{\phi}=\boldsymbol{\phi}_{\mathrm{1}} +\boldsymbol{\phi}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}+{x}\right)\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}βˆ’\Phi \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}βˆ’\Phi=βˆ’{li}_{\mathrm{2}} \left(βˆ’\mathrm{1}\right)βˆ’\Phi \\ $$$$\:\Phi=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{xln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(βˆ’\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({x}\right){dx} \\ $$$$\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{1}\right)^{{n}} \left\{\left[\frac{{x}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}{n}+\mathrm{2}}\:{ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} βˆ’\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} {dx}\right\} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=βˆ’\frac{\mathrm{1}}{\mathrm{4}}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}βˆ’\mathrm{1}} }{{n}^{\mathrm{2}} } \\ $$$$=\frac{βˆ’\mathrm{1}}{\mathrm{4}}\:\eta\left(\mathrm{2}\right)=\frac{βˆ’\pi^{\mathrm{2}} }{\mathrm{48}}\:\:\:…. \\ $$$$\:\:\:\:\therefore\:\boldsymbol{\phi}=βˆ’{li}_{\mathrm{2}} \left(βˆ’\mathrm{1}\right)βˆ’\Phi=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\: \\ $$$$\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 13/Feb/21
I(a)=∫_0 ^1 (t^a /(1+t^2 ))dt=Ξ£_(n=1) ^∞ (βˆ’1)^(n+1) ∫_0 ^1 t^(2n+a) dt  =Ξ£_(n=1) ^∞ (((βˆ’1)^(n+1) )/((2n+a+1)))         β‡’Iβ€²(a)=∫_0 ^1 ((t^a log(t))/(1+t^2 ))dt=Ξ£_(n=1) ^∞ (((βˆ’1)^n )/((2n+a+1)^2 ))  a=1     Iβ€²(a)=βˆ’(1/4).(Ο€^2 /(12))=βˆ’(Ο€^2 /(48))
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{a}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(βˆ’\mathrm{1}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{n}+{a}} {dt} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+{a}+\mathrm{1}\right)}\:\:\:\:\:\:\:\:\:\Rightarrow{I}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{a}} {log}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+{a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}=\mathrm{1}\:\:\:\:\:{I}'\left({a}\right)=βˆ’\frac{\mathrm{1}}{\mathrm{4}}.\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Feb/21
∫_0 ^1 ((t^7 log(t))/(1+t^2 ))dt=βˆ’Ξ£_(n=1) ^∞ (((βˆ’1)^(n+1) )/((2n+8)^2 ))  =βˆ’((Ο€^2 /(12))βˆ’(1βˆ’(1/4)+(1/9)βˆ’(1/(16))))=((115)/(144))βˆ’(Ο€^2 /(12))...
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{7}} {log}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=βˆ’\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(βˆ’\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{8}\right)^{\mathrm{2}} } \\ $$$$=βˆ’\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}βˆ’\left(\mathrm{1}βˆ’\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}βˆ’\frac{\mathrm{1}}{\mathrm{16}}\right)\right)=\frac{\mathrm{115}}{\mathrm{144}}βˆ’\frac{\pi^{\mathrm{2}} }{\mathrm{12}}… \\ $$
Commented by mnjuly1970 last updated on 13/Feb/21
grateful...
$${grateful}… \\ $$

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