Question Number 224920 by Tawa11 last updated on 12/Oct/25
$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}\:\mathrm{ln}\left(\mathrm{1}\:\:\:+\:\:\:\mathrm{x}\right)\:\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}\:\:\:\:+\:\:\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$
Answered by MrAjder last updated on 02/Jan/26
![Li_2 (x)=Σ_(n=1) ^∞ (x^n /n^2 ),∣x∣≤1 ln(1+x)=Σ_(k=1) ^∞ (((−1)^(k−1) x^k )/k),∣x∣≤1 I=Σ_(n=1) ^∞ Σ_(k=1) ^∞ Σ_(m=0) ^∞ (((−1)^(k−1+m) )/(kn^2 ))∫_0 ^1 x^(n−k+2m−1) dx ∫_0 ^1 x^(n+k+2m+1) dx=(1/(n+k+2m+2)) Σ_(m=0) ^∞ (((−1)^m )/(n+k+2m+2))=(1/4)[ψ(((n+k+4)/4))−ψ(((n+k+2)/4))] I=(1/4)Σ_(n=1) ^∞ Σ_(k=1) ^∞ (((−1)^(k−1) )/(kn^2 ))[ψ(((n+k+4)/4))−ψ(((n+k+2)/4))] ψ(z)=−γ+Σ_(p=0) ^∞ ((1/(p+1))−(1/(p+z))) ψ(((n+k+4)/4))−ψ(((n+k+2)/4))=Σ_(p=0) ^∞ ((1/(p+((n+k+2)/4)))−(1/(p+((n+k+4)/4)))) I=(1/4)Σ_(n=1) ^∞ Σ_(k=1) ^∞ (((−1)^(k−1) )/(kn^2 ))Σ_(p=0) ^∞ ∫_0 ^1 (x^(p+((n+k−2)/4)) −x^(p+((n+k)/2)) )dx Σ_(p=0) ^∞ x^p =(1/(1−x)),∣x∣<1 ψ(((n+k+4)/4))−ψ(((n+k+2)/4))=Σ_(p=0) ^∞ ((1/(p+((n+k+2)/4)))−(1/(p+((n+k+4)/4)))) I=(1/4)Σ_(n=1) ^∞ Σ_(k=1) ^∞ (((−1)^(k−1) )/(kn^2 ))Σ_(p=0) ^∞ ∫_0 ^1 (x^(p+((n+k−2)/4)) −x^(p+((n+k)/4)) )dx =(1/4)∫_(0 ) ^1 (Σ_(p=0) ^∞ x^p )(Σ_(n=1) ^∞ (x^(n/4) /n^2 ))(Σ_(n=1) ^∞ (((−1)^(k−1) x^(k/4) )/k))(x^(−(1/2)) −1)dx =(1/4)∫_0 ^1 ((1−x^(1/2) )/(1−x))x^(−(1/2)) Li_2 (x^(1/4) )ln(1+x^(1/4) )dx =(1/4) ∫_0 ^1 ((1−x)/((1+x)(1+x^2 )))Li_2 (x^(1/4) )ln(1+x^(1/4) )dx x=t^4 ⇒dx=4t^3 dt I=∫_0 ^1 ((t^3 (1−t^4 ))/(1−t^8 ))Li_2 (t)ln(1+t)dt=∫_0 ^1 (t^3 /(1+t^4 ))Li_2 (t)ln(1+t)dt (t^3 /(1+t^4 ))=(1/4)Σ_(ω^4 =−1) (ω/(ω−t)) I=(1/4)Σ_(ω^4 =−1) ω∫_0 ^1 ((Li_2 (t)ln(1+t))/(ω−t))dt ∫_0 ^1 ((Li_2 (t)ln(1+t))/(ω−t))dt=Σ_(n=1) ^∞ Σ_(k=1) ^∞ (((−1)^(k+1) )/(kn^2 ))∫_0 ^1 (t^(n−k) /(ω−t))dt ∫_(0 ) ^1 (t^m /(ω−t))dt=ω^m [ψ(((m+1)/2))−ψ(((m+2)/2))+π cot(πω)] Σ_(ω^4 =−1) ω^(m+1) cot(πw)= { ((0,),(m≡0),((mod 4))),((π,),(m≡1),((mod 4))),((0,),(m≡2),((mod 4))),((−π,),(m≡3),((mod 4))) :} I=(π/4)Σ_(n=1) ^∞ Σ_(k=1) ^∞ (((−1)^(k+1) )/(kn^2 ))[((1+(−1)^(n+k) )/2)]sgn(n+k−1)−(1/4)Σ_(n=1) ^∞ Σ_(k=1) ^∞ (((−1)^(k−1) )/(kn^2 ))[ψ(((n+k+1)/2))−ψ(((n+k+2)/2))] Σ_(n=1) ^∞ Σ_(k=1) ^∞ (((−1)^(k−1) )/(kn^2 ))cos(((π(n+k))/2))=Re[Li_1 (i)Li_(2 ) (i)]=(π/4)G−(π^2 /8)ln^2 Σ_(n=1) ^∞ Σ_(k=1) ^∞ (((−1)^(k−1) )/(kn^2 ))[ψ(((n+k+1)/2))−ψ(((n+k+2)/2))]=(π^2 /(14))ln 2−(7/(48))ζ(3) I=(π/4)((π/4)G−(π^2 /8)ln 2)−(1/(4 ))((π^2 /(14))ln 2−(7/(48))ζ(3)) I=((πG)/(16))−((π^2 ln 2)/(32))−((π^2 ln 2)/(96))+((7ζ(3))/(192)) =((πG)/(16))+((7ζ(3))/(192))−((π^2 ln2)/(24))](https://www.tinkutara.com/question/Q227133.png)
$$\mathrm{Li}_{\mathrm{2}} \left({x}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{{n}^{\mathrm{2}} },\mid{x}\mid\leq\mathrm{1} \\ $$$$\mathrm{ln}\left(\mathrm{1}+{x}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {x}^{{k}} }{{k}},\mid{x}\mid\leq\mathrm{1} \\ $$$${I}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}+{m}} }{{kn}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−{k}+\mathrm{2}{m}−\mathrm{1}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}+{k}+\mathrm{2}{m}+\mathrm{1}} {dx}=\frac{\mathrm{1}}{{n}+{k}+\mathrm{2}{m}+\mathrm{2}} \\ $$$$\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{n}+{k}+\mathrm{2}{m}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\left[\psi\left(\frac{{n}+{k}+\mathrm{4}}{\mathrm{4}}\right)−\psi\left(\frac{{n}+{k}+\mathrm{2}}{\mathrm{4}}\right)\right] \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{kn}^{\mathrm{2}} }\left[\psi\left(\frac{{n}+{k}+\mathrm{4}}{\mathrm{4}}\right)−\psi\left(\frac{{n}+{k}+\mathrm{2}}{\mathrm{4}}\right)\right] \\ $$$$\psi\left({z}\right)=−\gamma+\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{p}+\mathrm{1}}−\frac{\mathrm{1}}{{p}+{z}}\right) \\ $$$$\psi\left(\frac{{n}+{k}+\mathrm{4}}{\mathrm{4}}\right)−\psi\left(\frac{{n}+{k}+\mathrm{2}}{\mathrm{4}}\right)=\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{p}+\frac{{n}+{k}+\mathrm{2}}{\mathrm{4}}}−\frac{\mathrm{1}}{{p}+\frac{{n}+{k}+\mathrm{4}}{\mathrm{4}}}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{kn}^{\mathrm{2}} }\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{{p}+\frac{{n}+{k}−\mathrm{2}}{\mathrm{4}}} −{x}^{{p}+\frac{{n}+{k}}{\mathrm{2}}} \right){dx} \\ $$$$\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{p}} =\frac{\mathrm{1}}{\mathrm{1}−{x}},\mid{x}\mid<\mathrm{1} \\ $$$$\psi\left(\frac{{n}+{k}+\mathrm{4}}{\mathrm{4}}\right)−\psi\left(\frac{{n}+{k}+\mathrm{2}}{\mathrm{4}}\right)=\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{p}+\frac{{n}+{k}+\mathrm{2}}{\mathrm{4}}}−\frac{\mathrm{1}}{{p}+\frac{{n}+{k}+\mathrm{4}}{\mathrm{4}}}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{kn}^{\mathrm{2}} }\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}^{{p}+\frac{{n}+{k}−\mathrm{2}}{\mathrm{4}}} −{x}^{{p}+\frac{{n}+{k}}{\mathrm{4}}} \right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}\:} ^{\mathrm{1}} \left(\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{p}} \right)\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\frac{{n}}{\mathrm{4}}} }{{n}^{\mathrm{2}} }\right)\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} {x}^{\frac{{k}}{\mathrm{4}}} }{{k}}\right)\left({x}^{−\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{1}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−{x}}{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{Li}_{\mathrm{2}} \left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)\mathrm{ln}\left(\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{4}}} \right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\mathrm{Li}_{\mathrm{2}} \left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)\mathrm{ln}\left(\mathrm{1}+{x}^{\frac{\mathrm{1}}{\mathrm{4}}} \right){dx} \\ $$$${x}={t}^{\mathrm{4}} \Rightarrow{dx}=\mathrm{4}{t}^{\mathrm{3}} {dt} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{3}} \left(\mathrm{1}−{t}^{\mathrm{4}} \right)}{\mathrm{1}−{t}^{\mathrm{8}} }\mathrm{Li}_{\mathrm{2}} \left({t}\right)\mathrm{ln}\left(\mathrm{1}+{t}\right){dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{3}} }{\mathrm{1}+{t}^{\mathrm{4}} }\mathrm{Li}_{\mathrm{2}} \left({t}\right)\mathrm{ln}\left(\mathrm{1}+{t}\right){dt} \\ $$$$\frac{{t}^{\mathrm{3}} }{\mathrm{1}+{t}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{4}}\underset{\omega^{\mathrm{4}} =−\mathrm{1}} {\sum}\frac{\omega}{\omega−{t}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\underset{\omega^{\mathrm{4}} =−\mathrm{1}} {\sum}\omega\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{Li}_{\mathrm{2}} \left({t}\right)\mathrm{ln}\left(\mathrm{1}+{t}\right)}{\omega−{t}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{Li}_{\mathrm{2}} \left({t}\right)\mathrm{ln}\left(\mathrm{1}+{t}\right)}{\omega−{t}}{dt}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{kn}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}−{k}} }{\omega−{t}}{dt} \\ $$$$\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{{t}^{{m}} }{\omega−{t}}{dt}=\omega^{{m}} \left[\psi\left(\frac{{m}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{m}+\mathrm{2}}{\mathrm{2}}\right)+\pi\:\mathrm{cot}\left(\pi\omega\right)\right] \\ $$$$\underset{\omega^{\mathrm{4}} =−\mathrm{1}} {\sum}\omega^{{m}+\mathrm{1}} \mathrm{cot}\left(\pi{w}\right)=\begin{cases}{\mathrm{0},}&{{m}\equiv\mathrm{0}}&{\left(\mathrm{mod}\:\mathrm{4}\right)}\\{\pi,}&{{m}\equiv\mathrm{1}}&{\left(\mathrm{mod}\:\mathrm{4}\right)}\\{\mathrm{0},}&{{m}\equiv\mathrm{2}}&{\left(\mathrm{mod}\:\mathrm{4}\right)}\\{−\pi,}&{{m}\equiv\mathrm{3}}&{\left(\mathrm{mod}\:\mathrm{4}\right)}\end{cases} \\ $$$${I}=\frac{\pi}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{{kn}^{\mathrm{2}} }\left[\frac{\mathrm{1}+\left(−\mathrm{1}\right)^{{n}+{k}} }{\mathrm{2}}\right]\mathrm{sgn}\left({n}+{k}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{kn}^{\mathrm{2}} }\left[\psi\left(\frac{{n}+{k}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{n}+{k}+\mathrm{2}}{\mathrm{2}}\right)\right] \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{kn}^{\mathrm{2}} }\mathrm{cos}\left(\frac{\pi\left({n}+{k}\right)}{\mathrm{2}}\right)=\mathrm{R}{e}\left[\mathrm{Li}_{\mathrm{1}} \left({i}\right)\mathrm{Li}_{\mathrm{2}\:} \left({i}\right)\right]=\frac{\pi}{\mathrm{4}}{G}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{ln}^{\mathrm{2}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{kn}^{\mathrm{2}} }\left[\psi\left(\frac{{n}+{k}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{n}+{k}+\mathrm{2}}{\mathrm{2}}\right)\right]=\frac{\pi^{\mathrm{2}} }{\mathrm{14}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{7}}{\mathrm{48}}\zeta\left(\mathrm{3}\right) \\ $$$${I}=\frac{\pi}{\mathrm{4}}\left(\frac{\pi}{\mathrm{4}}{G}−\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\mathrm{ln}\:\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{4}\:}\left(\frac{\pi^{\mathrm{2}} }{\mathrm{14}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{7}}{\mathrm{48}}\zeta\left(\mathrm{3}\right)\right) \\ $$$${I}=\frac{\pi{G}}{\mathrm{16}}−\frac{\pi^{\mathrm{2}} \mathrm{ln}\:\mathrm{2}}{\mathrm{32}}−\frac{\pi^{\mathrm{2}} \mathrm{ln}\:\mathrm{2}}{\mathrm{96}}+\frac{\mathrm{7}\zeta\left(\mathrm{3}\right)}{\mathrm{192}} \\ $$$$=\frac{\pi{G}}{\mathrm{16}}+\frac{\mathrm{7}\zeta\left(\mathrm{3}\right)}{\mathrm{192}}−\frac{\pi^{\mathrm{2}} \mathrm{ln2}}{\mathrm{24}} \\ $$
Commented by Tawa11 last updated on 06/Jan/26
Commented by Tawa11 last updated on 06/Jan/26
Commented by Tawa11 last updated on 06/Jan/26
$$\mathrm{C}\:\:\:\:=\:\:\:\:\mathrm{catalans}\:\mathrm{constant}. \\ $$
Commented by Tawa11 last updated on 06/Jan/26
$$\mathrm{Nice}\:\mathrm{work}\:\mathrm{sir},\: \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct} \\ $$