Question Number 132324 by mnjuly1970 last updated on 13/Feb/21
![....advanced calculus... evaluation : π=β«_(0^( ) ) ^( β) ((ln(1+x))/(x(1+x^2 ))) dx solution: π=[β«_0 ^( 1) ((ln(1+x))/(x(1+x^2 )))dx=π_1 ]+[β«_1 ^( β) ((ln(1+x))/(x(1+x^2 ))) dx=π_2 ] π_2 =^(x=(1/t)) β«_0 ^( 1) ((ln(1+(1/t)))/((1/t)(1+(1/t^2 ))))(dt/t^2 )=β«_0 ^( 1) ((tln(1+(1/t)))/(1+t^2 ))dt =β«_0 ^( 1) ((tln(1+t))/(1+t^2 ))dtββ«_0 ^( 1) ((tln(t))/(1+t^2 ))dt β΄ π=π_1 +π_2 =β«_0 ^( 1) ((1/x)+x)((ln(1+x))/(1+x^2 ))dxβΞ¦ π=β«_0 ^( 1) ((ln(1+x))/x)dxβΞ¦=βli_2 (β1)βΞ¦ Ξ¦=β«_0 ^( 1) ((xln(x))/(1+x^2 ))dx=Ξ£_(n=0) ^β β«_0 ^( 1) (β1)^n x^(2n+1) ln(x)dx =Ξ£_(n=0) ^β (β1)^n {[(x^(2n+2) /(2n+2)) ln(x)]_0 ^1 β(1/(2n+2))β«_0 ^( 1) x^(2n+1) dx} =Ξ£_(n=0) ^β (β1)^(n+1) (1/(4(n+1)^2 ))=β(1/4) Ξ£_(n=1) ^β (((β1)^(nβ1) )/n^2 ) =((β1)/4) Ξ·(2)=((βΟ^2 )/(48)) .... β΄ π=βli_2 (β1)βΞ¦=(Ο^2 /(12))+(Ο^2 /(48)) π=((5Ο^2 )/(48))](Q132324.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:....{advanced}\:\:\:{calculus}... \\ $$$$\:\:\:{evaluation}\:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}\:^{\:\:} } ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx} \\ $$$$\:\:\:\:{solution}: \\ $$$$\:\:\boldsymbol{\phi}=\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}=\boldsymbol{\phi}_{\mathrm{1}} \right]+\left[\int_{\mathrm{1}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx}=\boldsymbol{\phi}_{\mathrm{2}} \right] \\ $$$$\:\:\:\boldsymbol{\phi}_{\mathrm{2}} \overset{{x}=\frac{\mathrm{1}}{{t}}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)}{\frac{\mathrm{1}}{{t}}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}\frac{{dt}}{{t}^{\mathrm{2}} }=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{tln}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{tln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}β\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{tln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\therefore\:\:\boldsymbol{\phi}=\boldsymbol{\phi}_{\mathrm{1}} +\boldsymbol{\phi}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{\mathrm{1}}{{x}}+{x}\right)\frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}β\Phi \\ $$$$\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}{dx}β\Phi=β{li}_{\mathrm{2}} \left(β\mathrm{1}\right)β\Phi \\ $$$$\:\Phi=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{xln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(β\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({x}\right){dx} \\ $$$$\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(β\mathrm{1}\right)^{{n}} \left\{\left[\frac{{x}^{\mathrm{2}{n}+\mathrm{2}} }{\mathrm{2}{n}+\mathrm{2}}\:{ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} β\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} {dx}\right\} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }=β\frac{\mathrm{1}}{\mathrm{4}}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\mathrm{1}\right)^{{n}β\mathrm{1}} }{{n}^{\mathrm{2}} } \\ $$$$=\frac{β\mathrm{1}}{\mathrm{4}}\:\eta\left(\mathrm{2}\right)=\frac{β\pi^{\mathrm{2}} }{\mathrm{48}}\:\:\:.... \\ $$$$\:\:\:\:\therefore\:\boldsymbol{\phi}=β{li}_{\mathrm{2}} \left(β\mathrm{1}\right)β\Phi=\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$\: \\ $$$$\:\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 13/Feb/21
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{a}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}{n}+{a}} {dt} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+{a}+\mathrm{1}\right)}\:\:\:\:\:\:\:\:\:\Rightarrow{I}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{a}} {log}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+{a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${a}=\mathrm{1}\:\:\:\:\:{I}'\left({a}\right)=β\frac{\mathrm{1}}{\mathrm{4}}.\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=β\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$
Commented by Dwaipayan Shikari last updated on 13/Feb/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{7}} {log}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=β\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(β\mathrm{1}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{8}\right)^{\mathrm{2}} } \\ $$$$=β\left(\frac{\pi^{\mathrm{2}} }{\mathrm{12}}β\left(\mathrm{1}β\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{9}}β\frac{\mathrm{1}}{\mathrm{16}}\right)\right)=\frac{\mathrm{115}}{\mathrm{144}}β\frac{\pi^{\mathrm{2}} }{\mathrm{12}}... \\ $$
Commented by mnjuly1970 last updated on 13/Feb/21
$${grateful}... \\ $$