Question Number 152350 by mnjuly1970 last updated on 27/Aug/21
$$ \\ $$$$\:\:\:\mathrm{prove}\:\mathrm{that}\:.... \\ $$$$\:\: \\ $$$$\boldsymbol{\phi}:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}.\:{ln}^{\:\mathrm{2}} \left(\:\mathrm{1}+\:{x}\:\right)}{\mathrm{1}\:+\:{x}^{\:\mathrm{2}} }\:\:\:{dx}\:=\frac{\mathrm{1}}{\mathrm{96}}\:{ln}\left(\mathrm{2}\:\right)\:\left(\:\pi^{\:\mathrm{2}} \:+\:\mathrm{4}\:{ln}^{\:\mathrm{2}} \:\left(\mathrm{2}\:\right)\right)....\blacksquare\:\:\:\:\: \\ $$$$\:\:\:\mathrm{M}.\mathrm{N}.. \\ $$$$ \\ $$$$ \\ $$
Answered by mindispower last updated on 27/Aug/21
![x→((1−t)/(1+t))⇒dx=−((2dt)/((1+t)^2 )) ⇔φ=∫_0 ^1 ((((1−t)/(1+t))ln^2 ((2/(1+t))))/(1+t^2 )).dt ∫_0 ^1 ((1−t)/((1+t)(1+t^2 )))(ln^2 (2)+ln^2 (1+t)−2ln(2)ln(1+t))dt ((1−t)/((1+t)(1+t^2 )))=(1/(1+t))−(t/(1+t^2 )) =∫_0 ^1 ((1/(1+t))−(t/(1+t^2 )))(ln^2 (2)+ln^2 (1+t)−2ln(2)ln(1+t))dt =ln^2 (2)∫_0 ^1 ((1/(1+t))−(t/(1+t^2 )))dt+∫_0 ^1 ((ln^2 (1+t))/(1+t))−φ −2ln(2)∫_0 ^1 ((ln(1+t))/(1+t))dt+2ln(2)∫_0 ^1 ((tln(1+t))/(1+t^2 ))dt 2φ=ln^2 (2)[((ln(2))/2)]+(1/3)ln^3 (2)−2ln(2).((ln^2 (2))/2) +2ln(2)∫_0 ^1 ((tln(1+t))/(1+t^2 ))dt ∫_0 ^1 ((tln(1+t))/(1+t^2 ))=S f(a)=∫_0 ^1 ((tln(1+at))/(1+t^2 ))dt S=∫_0 ^1 ∫_0 ^1 ((t^2 dt)/((1+t^2 )(1+at)))da =∫_0 ^1 ((a(aln(4)−π)+4ln(a+1))/(4(a^3 +a))) =∫_0 ^1 ((aln(4)−π)/(4(1+a^2 )))da+∫_0 ^1 ((ln(1+a))/(a+a^3 ))da =((ln(4))/8)ln(2)−(π^2 /(16))+∫_0 ^1 ((ln(1+a))/a)−((aln(1+a))/(1+a^2 ))da =((ln^2 (2))/4)−(π^2 /(16))+∫_0 ^1 ((ln(1−(−a)))/(−a))d(−a)−S S=((ln^2 (2))/8)−(π^2 /(32))−(1/2)li_2 (−1) =((ln^2 (2))/8)+(π^2 /(96)) 2φ=−((ln^3 (2)/6)+2ln(2)(((ln^2 (2))/8)+(π^2 /(96))) =−((ln^3 (2))/6)+((ln^3 (2))/4)+((π^2 ln(2))/(48))=((ln^3 (2))/(12))+((ln(2))/(48))π^2 =((ln(2))/(48))(π^2 +4ln^2 (2)) φ=((ln(2))/(96))(π^2 +4ln^2 (2))](Q152378.png)
$${x}\rightarrow\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\Rightarrow{dx}=−\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\Leftrightarrow\phi=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}{ln}^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{1}+{t}}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }.{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\left({ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{ln}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{1}+{t}\right)\right){dt} \\ $$$$\frac{\mathrm{1}−{t}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{1}+{t}}−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)\left({ln}^{\mathrm{2}} \left(\mathrm{2}\right)+{ln}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right){ln}\left(\mathrm{1}+{t}\right)\right){dt} \\ $$$$={ln}^{\mathrm{2}} \left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}}−\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}−\phi \\ $$$$−\mathrm{2}{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}}{dt}+\mathrm{2}{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{2}\phi={ln}^{\mathrm{2}} \left(\mathrm{2}\right)\left[\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\right]+\frac{\mathrm{1}}{\mathrm{3}}{ln}^{\mathrm{3}} \left(\mathrm{2}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right).\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$+\mathrm{2}{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }={S} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{tln}\left(\mathrm{1}+{at}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${S}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} {dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{at}\right)}{da} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{a}\left({aln}\left(\mathrm{4}\right)−\pi\right)+\mathrm{4}{ln}\left({a}+\mathrm{1}\right)}{\mathrm{4}\left({a}^{\mathrm{3}} +{a}\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{aln}\left(\mathrm{4}\right)−\pi}{\mathrm{4}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)}{da}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{a}\right)}{{a}+{a}^{\mathrm{3}} }{da} \\ $$$$=\frac{{ln}\left(\mathrm{4}\right)}{\mathrm{8}}{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{a}\right)}{{a}}−\frac{{aln}\left(\mathrm{1}+{a}\right)}{\mathrm{1}+{a}^{\mathrm{2}} }{da} \\ $$$$=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left(−{a}\right)\right)}{−{a}}{d}\left(−{a}\right)−{S} \\ $$$${S}=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{8}}−\frac{\pi^{\mathrm{2}} }{\mathrm{32}}−\frac{\mathrm{1}}{\mathrm{2}}{li}_{\mathrm{2}} \left(−\mathrm{1}\right) \\ $$$$=\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{8}}+\frac{\pi^{\mathrm{2}} }{\mathrm{96}} \\ $$$$\mathrm{2}\phi=−\frac{{ln}^{\mathrm{3}} \left(\mathrm{2}\right.}{\mathrm{6}}+\mathrm{2}{ln}\left(\mathrm{2}\right)\left(\frac{{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{8}}+\frac{\pi^{\mathrm{2}} }{\mathrm{96}}\right) \\ $$$$=−\frac{{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{6}}+\frac{{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\pi^{\mathrm{2}} {ln}\left(\mathrm{2}\right)}{\mathrm{48}}=\frac{{ln}^{\mathrm{3}} \left(\mathrm{2}\right)}{\mathrm{12}}+\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{48}}\pi^{\mathrm{2}} \\ $$$$=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{48}}\left(\pi^{\mathrm{2}} +\mathrm{4}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$$$\phi=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{96}}\left(\pi^{\mathrm{2}} +\mathrm{4}{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$
Commented by mindispower last updated on 29/Aug/21
$${thanx}\:{withe}\:{pleasur} \\ $$
Commented by mnjuly1970 last updated on 28/Aug/21
$$\:\:\:\:{very}\:{nice}\:{sir}\:{power}...{grateful} \\ $$