Question Number 96365 by Ar Brandon last updated on 31/May/20
Answered by M±th+et+s last updated on 01/Jun/20
![x=(1/t) dx=−(dt/t^2 ) I=∫_∞ ^0 (((1/t^2 ).(−ln(t)).(−(dt/t^2 )))/((((1+t^2 ))/t^2 ).(((1+t)^2 )/t^2 ))) I=−∫_0 ^∞ ((ln(x))/((1+x^2 )(1+x)^2 ))dx....2 2I=∫_0 ^∞ ((ln(x)[x^2 −1])/((1+x^2 )(1+x)^2 ))dx ∫_0 ^∞ =∫_0 ^1 +∫_1 ^∞ ((ln(x)[x^2 −1])/((1+x)^2 (1+x^2 )))dx x=(1/u) dx=((−du)/u^2 ) ∫_1 ^0 ((−ln(u).(([1−u^2 ])/u^2 ) .((−du)/u^2 ))/((((1+u)^2 )/u^2 ).(((1+u^2 ))/u^2 ))) 2I=∫_0 ^1 ((ln(u)(u^2 −1))/((1+u)^2 (u^2 +1)))du I=∫_0 ^1 (((u−1)ln(u))/((u+1)(u^2 +1)))du I=∫_0 ^1 ln(u)[(u/(u^2 +1))−(1/(u+1))]du I=∫_0 ^1 ((uln(u))/(u^2 +1))du−∫_0 ^1 ((ln(u))/(u+1))du i will continue later ......](Q96475.png)
$${x}=\frac{\mathrm{1}}{{t}}\:\:\:{dx}=−\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$${I}=\int_{\infty} ^{\mathrm{0}} \frac{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }.\left(−{ln}\left({t}\right)\right).\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)}{\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }.\frac{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} }} \\ $$$${I}=−\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx}....\mathrm{2} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left({x}\right)\left[{x}^{\mathrm{2}} −\mathrm{1}\right]}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx} \\ $$$$\int_{\mathrm{0}} ^{\infty} =\int_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left({x}\right)\left[{x}^{\mathrm{2}} −\mathrm{1}\right]}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$${x}=\frac{\mathrm{1}}{{u}}\:\:\:\:\:\:{dx}=\frac{−{du}}{{u}^{\mathrm{2}} } \\ $$$$\int_{\mathrm{1}} ^{\mathrm{0}} \frac{−{ln}\left({u}\right).\frac{\left[\mathrm{1}−{u}^{\mathrm{2}} \right]}{{u}^{\mathrm{2}} }\:.\frac{−{du}}{{u}^{\mathrm{2}} }}{\frac{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }{{u}^{\mathrm{2}} }.\frac{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{{u}^{\mathrm{2}} }} \\ $$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({u}\right)\left({u}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} \left({u}^{\mathrm{2}} +\mathrm{1}\right)}{du} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({u}−\mathrm{1}\right){ln}\left({u}\right)}{\left({u}+\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{du} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({u}\right)\left[\frac{{u}}{{u}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right]{du} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{uln}\left({u}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({u}\right)}{{u}+\mathrm{1}}{du} \\ $$$${i}\:{will}\:{continue}\:{later}\:...... \\ $$
Answered by M±th+et+s last updated on 01/Jun/20
![I_1 =∫_0 ^1 ((uln(u))/(u^2 +1))du=∫_0 ^1 ln(u) d((1/2)ln∣u^2 +1∣) =[(1/2)ln(u)ln(u^2 +1)]_0 ^1 −(1/2)∫_0 ^1 ((ln(1+u^2 ))/u)du =−(1/2)∫_0 ^1 Σ_(n=1) ^∞ (((−1)^(n+1) )/n).u^(2n−1) du =−(1/2)Σ_(n=1) ^∞ (((−1)^(n+1) )/n)∫_0 ^1 u^(2n−1) du =−(1/4)Σ_(n=1) ^∞ (((−1)^(n+1) )/n^2 )=((−π^2 )/(48)) I_2 =∫_0 ^1 ((ln(u))/(1+u))du=−(1/2)((π^2 /6))=((−π^2 )/(12)) I=I_1 −I_2 I=−(π^2 /(48))+(π^2 /(12))=(π^2 /(16)) hence proved](Q96477.png)
$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{uln}\left({u}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du}=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({u}\right)\:{d}\left(\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{u}^{\mathrm{2}} +\mathrm{1}\mid\right) \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({u}\right){ln}\left({u}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{{u}}{du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}.{u}^{\mathrm{2}{n}−\mathrm{1}} {du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\mathrm{2}{n}−\mathrm{1}} {du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}^{\mathrm{2}} }=\frac{−\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({u}\right)}{\mathrm{1}+{u}}{du}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\right)=\frac{−\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$${I}={I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${I}=−\frac{\pi^{\mathrm{2}} }{\mathrm{48}}+\frac{\pi^{\mathrm{2}} }{\mathrm{12}}=\frac{\pi^{\mathrm{2}} }{\mathrm{16}} \\ $$$$ \\ $$$${hence}\:{proved} \\ $$