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Question Number 222411 by Nicholas666 last updated on 25/Jun/25
[ 1 .] ∫_0 ^( 1) ((ln(x) ln(1−x^2 )ln(1+x^2 ))/(1−x^2 )) dx [ 2 .] ∫_0 ^1 ((ln(x) ln(1−x) ln(1+x) ln(1+x^2 ))/(1+x)) dx [ 3 .] ∫_0 ^1 ((ln(x) ln(1−x^2 ) ln(1+x^2 ))/x) dx [ 4 .] ∫_0 ^( 1) ((ln(x) ln(1−x) ln(1+x) ln(1−x^2 ))/x) dx
$$\:\: \\ $$$$\:\:\:\left[\:\mathrm{1}\:.\right]\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left({x}\right)\:\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:{dx}\:\:\:\:\: \\ $$$$\:\:\:\left[\:\mathrm{2}\:.\right]\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{ln}\left({x}\right)\:\mathrm{ln}\left(\mathrm{1}−{x}\right)\:\mathrm{ln}\left(\mathrm{1}+{x}\right)\:\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}\:{dx} \\ $$$$\:\left[\:\mathrm{3}\:.\right]\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left({x}\right)\:\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\:\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}}\:\:{dx}\:\:\:\: \\ $$$$\:\:\:\left[\:\mathrm{4}\:.\right]\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left({x}\right)\:\mathrm{ln}\left(\mathrm{1}−{x}\right)\:\mathrm{ln}\left(\mathrm{1}+{x}\right)\:\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{{x}}\:\:{dx}\:\:\:\:\: \\ $$$$ \\ $$
Answered by MrGaster last updated on 27/Jun/25
(1): I=∫_0 ^1 ln x ln(1−x^2 )ln(1−x^2 )Σ_(k=0) ^∞ x^(2k) dx =Σ_(k=0) ^∞ ∫_0 ^1 x^(2k) ln x ln(1−x^2 )ln(1+x^2 )dx =Σ_(k=0) ^∞ ∫_(0 ) ^1 x^(2k) ln x(−Σ_(m=1) ^∞ (x^(2m) /m))(Σ_(n=1) ^∞ (((−1)^(n+1) x^(2n) )/n))dx =Σ_(k=0) ^∞ Σ_(m=1) ^∞ Σ_(n=1) ^∞ −(1/m) (((−1)^(n+1) )/n)∫_0 ^1 x^(2k+2m+2n) ln xdx =Σ_(k=0) ^∞ Σ_(m=1) ^∞ Σ_(n=1) ^∞ −(1/m) (((−1)^(n+1) )/n)(−(1/((2k+2m+2n+1)^2 ))) =Σ_(k=0) ^∞ Σ_(m=1) ^∞ Σ_(n=1) ^∞ (1/m) (((−1)^(n+1) )/n) (1/((2k+2m+2n+1))) =Σ_(s=2) ^∞ (1/((2s−1)^2 ))Σ_ { ((k=0)),((m=1)),((n=1)) :} ^(k+m+n=s) (1/m) (((−1)^(n+1) )/n) =(π^2 /(54))+(1/(405))
$$\left(\mathrm{1}\right): \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\:{x}\:\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{k}} {dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{k}} \mathrm{ln}\:{x}\:\mathrm{ln}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\int_{\mathrm{0}\:} ^{\mathrm{1}} {x}^{\mathrm{2}{k}} \mathrm{ln}\:{x}\left(−\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{2}{m}} }{{m}}\right)\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {x}^{\mathrm{2}{n}} }{{n}}\right){dx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}−\frac{\mathrm{1}}{{m}}\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}{k}+\mathrm{2}{m}+\mathrm{2}{n}} \mathrm{ln}\:{xdx} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}−\frac{\mathrm{1}}{{m}}\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\left(−\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{2}{m}+\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{m}}\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}}\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}+\mathrm{2}{m}+\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$$=\underset{{s}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{s}−\mathrm{1}\right)^{\mathrm{2}} }\underset{\begin{cases}{{k}=\mathrm{0}}\\{{m}=\mathrm{1}}\\{{n}=\mathrm{1}}\end{cases}} {\overset{{k}+{m}+{n}={s}} {\sum}}\frac{\mathrm{1}}{{m}}\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{54}}+\frac{\mathrm{1}}{\mathrm{405}} \\ $$
Commented by Nicholas666 last updated on 27/Jun/25
thank you, good approach
$$\mathrm{thank}\:\mathrm{you},\:\mathrm{good}\:\mathrm{approach} \\ $$
Commented by Nicholas666 last updated on 03/Jul/25
can you provided?
$$\mathrm{can}\:\mathrm{you}\:\mathrm{provided}?\: \\ $$
Commented by Tawa11 last updated on 26/Jul/25
Funny. I found closed form answer for (3) and (4).
$$\mathrm{Funny}. \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{answer}\:\mathrm{for}\:\left(\mathrm{3}\right)\:\mathrm{and}\:\left(\mathrm{4}\right). \\ $$
Commented by Tawa11 last updated on 08/Nov/25
Sir, please can you provide more step for this sum? I want to learn from it. MrGaster The number 1. Though your result is just an approximate value. I have provided the complete exact value
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{provide}\:\mathrm{more}\:\mathrm{step}\:\mathrm{for}\:\mathrm{this}\:\mathrm{sum}? \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{learn}\:\mathrm{from}\:\mathrm{it}. \\ $$$$\mathrm{MrGaster} \\ $$$$\mathrm{The}\:\mathrm{number}\:\mathrm{1}. \\ $$$$\mathrm{Though}\:\mathrm{your}\:\mathrm{result}\:\mathrm{is}\:\mathrm{just}\:\mathrm{an}\:\mathrm{approximate}\:\mathrm{value}. \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{provided}\:\mathrm{the}\:\mathrm{complete}\:\mathrm{exact}\:\mathrm{value} \\ $$
Answered by MrGaster last updated on 13/Jul/25
(2):ln(1+x)=Σ_(m=1) ^∞ (−1)^(m+1) (x^m /m)Σ_(m=1) ^∞ ,ln(1+x^2 )=Σ_(n=1) ^∞ (−1)^(n+1) (x^(2n) /n) ∫_0 ^1 ((ln x ln(1−x))/(1+x))Σ_(m=1) ^∞ (−1)^(m+1) (x^m /m)Σ_(n=1) ^∞ (−1)^(n+1) (x^(2n) /n)dx =Σ_(m=1) ^∞ Σ_(n=1) ^∞ (−1)^(m+1) (−1)^(n+1) (1/(mn))∫_(0 ) ^1 ((lnxln(1−x))/(1+x))x^(m+2n) dx k=m+2n,m=k−2n,n∈{1,2…,⌊((k+1)/2)⌋},k≥2 (−1)^((k+2n)−1) (−1)^(n+1) =(−1)^(k−n) ∫_(0 ) ^1 ((lnxln(1−x))/(1+x))x^k dx=C(k) C(k)=∫_0 ^1 lnxln(1−x)Σ_(p=0) ^∞ (−1)^p x^p x^k dx=Σ_(p=0) ^∞ (−1)^p ∫_0 ^1 x^(k+p) lnxln(1−x)dx ln(1−x)=−Σ_(i=1) ^∞ (x^i /i) ∫_0 ^1 x^a lnxln(1−x)dx=(H_(a>1) /(a+1)),a>−1 ∫_0 ^1 x^(k+p) lnxln(1−x)dx=(H_(k+p+1) /(k+p+1)) C(k)=Σ_(p=0) ^∞ (−1)^p (−(H_(k+p+1) /(k+p+1)))=Σ_(p=0) ^∞ (−1)^p (H_(k+p+1) /(k+p+1)) I=Σ_(k=2) ^∞ (−1)^k C(k)Σ_(n=1) ^(⌊((k−1)/2)⌋) (((−1)^n )/(n(k−2n))) ζ(3)=Σ_(k=1) ^∞ (1/k^3 ),ζ(4)=(π^4 /(90)) I=(7/8)ζ(3)ln2−(π^4 /(480))
$$\left(\mathrm{2}\right):\mathrm{ln}\left(\mathrm{1}+{x}\right)=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} \frac{{x}^{{m}} }{{m}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}},\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{{x}^{\mathrm{2}{n}} }{{n}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} \frac{{x}^{{m}} }{{m}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{{x}^{\mathrm{2}{n}} }{{n}}{dx} \\ $$$$=\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{m}+\mathrm{1}} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{mn}}\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{\mathrm{ln}{x}\mathrm{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{x}^{{m}+\mathrm{2}{n}} {dx} \\ $$$${k}={m}+\mathrm{2}{n},{m}={k}−\mathrm{2}{n},{n}\in\left\{\mathrm{1},\mathrm{2}\ldots,\lfloor\frac{{k}+\mathrm{1}}{\mathrm{2}}\rfloor\right\},{k}\geq\mathrm{2} \\ $$$$\left(−\mathrm{1}\right)^{\left({k}+\mathrm{2}{n}\right)−\mathrm{1}} \left(−\mathrm{1}\right)^{{n}+\mathrm{1}} =\left(−\mathrm{1}\right)^{{k}−{n}} \\ $$$$\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{\mathrm{ln}{x}\mathrm{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{1}+{x}}{x}^{{k}} {dx}={C}\left({k}\right) \\ $$$${C}\left({k}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}{x}\mathrm{ln}\left(\mathrm{1}−{x}\right)\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{p}} {x}^{{p}} {x}^{{k}} {dx}=\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{p}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}+{p}} \mathrm{ln}{x}\mathrm{ln}\left(\mathrm{1}−{x}\right){dx} \\ $$$$\mathrm{ln}\left(\mathrm{1}−{x}\right)=−\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{i}} }{{i}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} \mathrm{ln}{x}\mathrm{ln}\left(\mathrm{1}−{x}\right){dx}=\frac{{H}_{{a}>\mathrm{1}} }{{a}+\mathrm{1}},{a}>−\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}+{p}} \mathrm{ln}{x}\mathrm{ln}\left(\mathrm{1}−{x}\right){dx}=\frac{{H}_{{k}+{p}+\mathrm{1}} }{{k}+{p}+\mathrm{1}} \\ $$$${C}\left({k}\right)=\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{p}} \left(−\frac{{H}_{{k}+{p}+\mathrm{1}} }{{k}+{p}+\mathrm{1}}\right)=\underset{{p}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{p}} \frac{{H}_{{k}+{p}+\mathrm{1}} }{{k}+{p}+\mathrm{1}} \\ $$$${I}=\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}\left({k}\right)\underset{{n}=\mathrm{1}} {\overset{\lfloor\frac{{k}−\mathrm{1}}{\mathrm{2}}\rfloor} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}\left({k}−\mathrm{2}{n}\right)} \\ $$$$\zeta\left(\mathrm{3}\right)=\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{3}} },\zeta\left(\mathrm{4}\right)=\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$$${I}=\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)\mathrm{ln2}−\frac{\pi^{\mathrm{4}} }{\mathrm{480}} \\ $$
Answered by Tawa11 last updated on 08/Nov/25
Finally: (1) ∫_( 0) ^( 1) ((ln(x) ln(1 − x^2 ) ln(1 + x^2 ))/(1 − x^2 )) dx = ((68)/(99)) ln^4 (2) − ((46)/(11)) ln^2 (2) 𝛇(2) − ((14)/(99)) ln(2) 𝛇(3) + ((10)/(11)) 𝛇(4) + (8/(33)) Li_4 ((1/2)) + (1/3) 𝛇(3) + ((64)/(63)) (3) ∫_( 0) ^( 1) ((ln(x) ln(1 − x^2 ) ln(1 + x^2 ))/x) dx = (1/(48))ln^4 (2) − ((3π^4 )/(640)) − (π^2 /(48)) ln^2 (2) + (7/(16))ln(2)𝛇(3) + (1/2)Li_4 ((1/2)) (4) ∫_( 0) ^( 1) ((ln(x) ln(1 − x) ln(1 + x) ln(1 − x^2 ))/x) dx = (2/(15))ln^5 (2) − (2/3)𝛇(2)ln^3 (2) + (7/4)𝛇(3)ln^2 (2) − (3/8)𝛇(2)𝛇(3) − (7/2)𝛇(5) + 4Li_4 ((1/2))ln(2) + 4Li_5 ((1/2)) + ((7π^2 )/(48))𝛇(3) − ((25)/(16))𝛇(5)
$$\mathrm{Finally}: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:\:\:\:−\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\:\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:\:\:\:+\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}{\mathrm{1}\:\:\:\:\:−\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\:\boldsymbol{\mathrm{dx}} \\ $$$$\:=\:\:\:\:\frac{\mathrm{68}}{\mathrm{99}}\:\boldsymbol{\mathrm{ln}}^{\mathrm{4}} \left(\mathrm{2}\right)\:−\:\frac{\mathrm{46}}{\mathrm{11}}\:\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{2}\right)\:\boldsymbol{\zeta}\left(\mathrm{2}\right)\:\:−\:\:\frac{\mathrm{14}}{\mathrm{99}}\:\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\:\boldsymbol{\zeta}\left(\mathrm{3}\right)\:+\:\:\frac{\mathrm{10}}{\mathrm{11}}\:\boldsymbol{\zeta}\left(\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\frac{\mathrm{8}}{\mathrm{33}}\:\boldsymbol{\mathrm{Li}}_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}\:\boldsymbol{\zeta}\left(\mathrm{3}\right)\:\:+\:\:\frac{\mathrm{64}}{\mathrm{63}} \\ $$$$ \\ $$$$\left(\mathrm{3}\right)\:\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\left(\boldsymbol{\mathrm{x}}\right)\:\mathrm{ln}\left(\mathrm{1}\:−\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)\:\mathrm{ln}\left(\mathrm{1}\:+\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}{\boldsymbol{\mathrm{x}}}\:\mathrm{d}\boldsymbol{\mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{48}}\boldsymbol{\mathrm{ln}}^{\mathrm{4}} \left(\mathrm{2}\right)\:\:−\:\:\frac{\mathrm{3}\pi^{\mathrm{4}} }{\mathrm{640}}\:\:−\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{48}}\:\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)\:\:+\:\:\frac{\mathrm{7}}{\mathrm{16}}\mathrm{ln}\left(\mathrm{2}\right)\boldsymbol{\zeta}\left(\mathrm{3}\right)\:\:+\:\:\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{Li}}_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:\:\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:−\:\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:+\:\boldsymbol{\mathrm{x}}\right)\:\boldsymbol{\mathrm{ln}}\left(\mathrm{1}\:−\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \right)}{\boldsymbol{\mathrm{x}}}\:\boldsymbol{\mathrm{dx}} \\ $$$$\:\:\:\:\:=\:\:\frac{\mathrm{2}}{\mathrm{15}}\boldsymbol{\mathrm{ln}}^{\mathrm{5}} \left(\mathrm{2}\right)\:−\:\frac{\mathrm{2}}{\mathrm{3}}\boldsymbol{\zeta}\left(\mathrm{2}\right)\boldsymbol{\mathrm{ln}}^{\mathrm{3}} \left(\mathrm{2}\right)\:+\:\frac{\mathrm{7}}{\mathrm{4}}\boldsymbol{\zeta}\left(\mathrm{3}\right)\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \left(\mathrm{2}\right)\:−\:\frac{\mathrm{3}}{\mathrm{8}}\boldsymbol{\zeta}\left(\mathrm{2}\right)\boldsymbol{\zeta}\left(\mathrm{3}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\frac{\mathrm{7}}{\mathrm{2}}\boldsymbol{\zeta}\left(\mathrm{5}\right)\:+\:\mathrm{4}\boldsymbol{\mathrm{Li}}_{\mathrm{4}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\boldsymbol{\mathrm{ln}}\left(\mathrm{2}\right)\:+\:\mathrm{4}\boldsymbol{\mathrm{Li}}_{\mathrm{5}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\:\frac{\mathrm{7}\pi^{\mathrm{2}} }{\mathrm{48}}\boldsymbol{\zeta}\left(\mathrm{3}\right)\:−\:\frac{\mathrm{25}}{\mathrm{16}}\boldsymbol{\zeta}\left(\mathrm{5}\right) \\ $$

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