复杂的对数积分（五）

[Largedisplaystyle int_0^1frac{ln^3(1+x)\,ln^3x}xmathrm{d}x ]

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(Largemathbf{Solution:})
Using the following known results:

[egin{align*} sum_{n=1}^{infty }frac{H_n}{n2^{n}}&=frac{pi ^{2}}{12}\ sum^infty_{n=1}frac{H_n}{n^22^n}&=zeta(3)-frac{pi^2}{12}ln{2}\ sum^infty_{n=1}frac{H_n}{n^32^n}&={ m Li}_4left(frac{1}{2} ight)+frac{pi^4}{720}-frac{1}{8}zeta(3)ln{2}+frac{1}{24}ln^4{2}\ sum^infty_{n=1}frac{H_n}{n^42^n} &=2{ m Li}_5left(frac{1}{2} ight)+frac{1}{32}zeta(5)+{ m Li}_4left(frac{1}{2} ight)ln{2}-frac{pi^4}{720}ln{2}+frac{1}{2}zeta(3)ln^2{2}\&~~~-frac{pi^2}{12}zeta(3)-frac{pi^2}{36}ln^3{2}+frac{1}{40}ln^5{2}\ sum^infty_{n=1}frac{H_n}{n^52^n} &=3{ m Li}_6left(frac{1}{2} ight)-frac{1}{2}zeta(ar{5},1)-frac{17pi^6}{10080}+{ m Li}_5left(frac{1}{2} ight)ln{2}-frac{1}{32}zeta(5)ln{2}+frac{pi^4}{1440}ln^2{2}\ &~~~-frac{1}{4}zeta^2(3)-frac{1}{6}zeta(3)ln^3{2}+frac{pi^2}{12}zeta(3)ln{2}+frac{pi^2}{144}ln^4{2}-frac{1}{240}ln^6{2} end{align*}]

For the simplicity, let

[color{blue}{mathbf{H}_{m}^{(k)}(x)}=sum_{n=1}^infty frac{H_{n}^{(k)}x^n}{n^m}qquadRightarrowqquadcolor{blue}{mathbf{H}(x)}=sum_{n=1}^infty H_{n}x^n ]

Introduce a generating function for the generalized harmonic numbers for (|x|<1)

[color{blue}{mathbf{H}^{(k)}(x)}=sum_{n=1}^infty H_{n}^{(k)}x^n=frac{operatorname{Li}_k(x)}{1-x}qquadRightarrowqquadcolor{blue}{mathbf{H}(x)}=-frac{ln(1-x)}{1-x} ]

and the following identity

[H_{n+1}^{(k)}-H_{n}^{(k)}=frac1{(n+1)^k}qquadRightarrowqquad H_{n+1}-H_{n}=frac1{n+1} ]

Using (mathcal{I}) to denote the integral in question.Start with integration by parts (IBP),we have

[egin{align*} mathcal{I}&=-frac{3}{4} int_{0}^{1}frac{ln^4xln^2left ( 1+x ight )}{1+x}\, mathrm{d}x\&=-frac{3}{4}int_{1}^{2}frac{ln^2xln^4left ( 1-x ight )}{x}\, mathrm{d}x\ &=-frac{3}{4}{underbrace{int_{frac{1}{2}}^{1}frac{ln^4left ( 1-x ight )ln^2x}{x}\, mathrm{d}x}_{mathcal{I}_1}+3{underbrace{int_{frac{1}{2}}^{1}frac{ln^3left ( 1-x ight )ln^3x}{x}\, mathrm{d}x}_{mathcal{I}_2}-frac{9}{2}{underbrace{int_{frac{1}{2}}^{1}frac{ln^2left ( 1-x ight )ln^4x}{x}\, mathrm{d}x}_{mathcal{I}_3}}}}\ &~~~+3underbrace{int_{frac{1}{2}}^{1}frac{lnleft ( 1-x ight )ln^5x}{x}\, mathrm{d}x}_{mathcal{I}_4}-frac{1}{7}ln^72 end{align*}]

Let us integrating the indefinite form of (mathcal{I}_4),

[egin{align*} int frac{lnleft ( 1-x ight )ln^5x}{x}&=-int sum_{n=1}^{infty }frac{x^{n-1}}{n}ln^5x\, mathrm{d}x\ &=-sum_{n=1}^{infty }frac{1}{n}frac{partial ^{5}}{partial n^{5}}left [ int x^{n-1}\, mathrm{d}x ight ]\ &=-sum_{n=1}^{infty }frac{1}{n}frac{partial ^{5}}{partial n^{5}}left [ frac{x^{n}}{n} ight ]\ &=-sum_{n=1}^{infty }frac{1}{n}Bigg [ frac{x^{n}ln^5x}{n}-frac{5x^{n}ln^4x}{n^{2}}+frac{20x^{n}ln^3x}{n^{3}}-frac{60x^{n}ln^2x}{n^{4}}\ &~~~~~~~~~~~~~~~+frac{120x^{n}ln x}{n^{5}}-frac{120x^{n}}{n^{6}}Bigg ]\ &=sum_{n=1}^{infty }Bigg [ -frac{x^{n}ln^5x}{n^{2}}+frac{5x^{n}ln^4x}{n^{3}}-frac{20x^{n}ln^3x}{n^{4}}+frac{60x^{n}ln^2x}{n^{5}}\ &~~~~~~~~~~~~~~~-frac{120x^{n}ln x}{n^{6}}+frac{120x^{n}}{n^{7}} Bigg ]\ &=-ln^5xmathrm{Li}_2left ( x ight )+5ln^4xmathrm{Li}_3left ( x ight )-20ln^3xmathrm{Li}_4left ( x ight )+60ln^2xmathrm{Li}_5left ( x ight )\ &~~~-120ln xmathrm{Li}_6left ( x ight )+120mathrm{Li}_7left ( x ight ) end{align*}]

Therefore,

[egin{align*} color{Teal}{mathcal{I}_4}&=-ln^52mathrm{Li}_2left ( frac{1}{2} ight )+5ln^42mathrm{Li}_3left ( frac{1}{2} ight )-20ln^32mathrm{Li}_4left ( frac{1}{2} ight )-60ln^22mathrm{Li}_5left ( frac{1}{2} ight )\ &~~~-120ln 2mathrm{Li}_6left ( frac{1}{2} ight )-120mathrm{Li}_7left ( frac{1}{2} ight )\ &=color{Teal}{-frac{1}{3}ln^72+frac{1}{3}pi ^{2}ln^52-frac{35}{8}zeta left ( 3 ight )ln^42-20ln^32mathrm{Li}_{4}left ( frac{1}{2} ight )-60ln^22mathrm{Li}_5left ( frac{1}{2} ight )}\ &~~~color{Teal}{-120ln 2mathrm{Li}_6left ( frac{1}{2} ight )-120mathrm{Li}_7left ( frac{1}{2} ight )} end{align*}]

Applying IBP again to evaluate (mathcal{I}_3),

[mathcal{I}_3=-frac{1}{5}ln^72+frac{2}{5}underbrace{color{Red}{int_{frac{1}{2}}^{1}frac{ln^5xlnleft ( 1-x ight )}{1-x}\, mathrm{d}x}}_mathcal{J} ]

Using the similar approach as calculating the red integral, then

[{smallegin{align*} intfrac{ln^5xlnleft ( 1-x ight )}{1-x}\, mathrm{d}x &= -int sum_{n=1}^{infty }H_nx^{n}ln^5x\, mathrm{d}x\ &=-sum_{n=1}^{infty }H_nint x^{n}ln^5x\, mathrm{d}x\ &=-sum_{n=1}^{infty }H_nfrac{partial ^{5}}{partial n^{5}}left [ int x^{n}\, mathrm{d}x ight ]\ &=-sum_{n=1}^{infty }H_nfrac{partial ^{5}}{partial n^{5}}left [ frac{x^{n+1}}{n+1} ight ]\ &=-sum_{n=1}^{infty }H_nBigg [ frac{x^{n+1}ln^5x}{n+1}-frac{5x^{n+1}ln^4x}{left (n+1 ight )^{2}}+frac{20x^{n+1}ln^3x}{left (n+1 ight )^{3}}-frac{60x^{n+1}ln^2x}{left (n+1 ight )^{4}} \ &~~~~~~~~~~~~~~~+frac{120x^{n+1}ln x}{left (n+1 ight )^{5}}-frac{120x^{n+1}}{left (n+1 ight )^{6}}Bigg ]\ &=-ln^5xsum_{n=1}^{infty }frac{H_{n+1}x^{n+1}}{n+1}+ln^5xsum_{n=1}^{infty }frac{x^{n+1}}{left ( n+1 ight )^{2}}+5ln^4xsum_{n=1}^{infty }frac{H_{n+1}x^{n+1}}{left ( n+1 ight )^{2}}\ &~~~-5ln^4xsum_{n=1}^{infty }frac{x^{n+1}}{left ( n+1 ight )^{3}}-20ln^3xsum_{n=1}^{infty }frac{H_{n+1}x^{n+1}}{left ( n+1 ight )^{3}}+20ln^3xsum_{n=1}^{infty }frac{x^{n+1}}{left ( n+1 ight )^{4}}\ &~~~+60ln^2xsum_{n=1}^{infty }frac{H_{n+1}x^{n+1}}{left ( n+1 ight )^{4}}-60ln^2xsum_{n=1}^{infty }frac{x^{n+1}}{left ( n+1 ight )^{5}}-120ln xsum_{n=1}^{infty }frac{H_{n+1}x^{n+1}}{left ( n+1 ight )^{5}}\ &~~~+120ln xsum_{n=1}^{infty }frac{x^{n+1}}{left ( n+1 ight )^{6}}+120ln^2xsum_{n=1}^{infty }frac{H_{n+1}x^{n+1}}{left ( n+1 ight )^{6}}-120ln xsum_{n=1}^{infty }frac{x^{n+1}}{left ( n+1 ight )^{7}}\ &=-sum_{n=1}^{infty }Bigg [ frac{H_nx^{n}ln^5x}{n}-frac{x^{n}ln^5x}{n^{2}}-frac{5H_nx^{n}ln^4x}{n^{2}}+frac{5x^{n}ln^4x}{n^{3}}\ &~~~~~~~~~+frac{20H_nx^{n}ln^3x}{n^{3}}-frac{20x^{n}ln^3x}{n^{4}}-frac{60H_nx^{n}ln^2x}{n^{4}}+frac{60x^{n}ln^2x}{n^{5}}\ &~~~~~~~~~+frac{120H_nx^{n}ln x}{n^5}-frac{120x^{n}ln x}{n^{6}}-frac{120H_nx^{n}}{n^{6}}-frac{120x^{n}}{n^{7}} Bigg ]\ &=-color{blue}{mathbf{H}_{1}left ( x ight )}ln^5x+mathrm{Li}_{2}left ( x ight )ln^5x+5color{Blue}{mathbf{H}_{2}left ( x ight )}ln^4x-5mathrm{Li}_{3}left ( x ight )ln^4x-20color{Blue}{mathbf{H}_{3}left ( x ight )}ln^3x\ &~~~+20mathrm{Li}_{4}left ( x ight )ln^3x+60color{Blue}{mathbf{H}_4left ( x ight )}ln^2x-60mathrm{Li}_{5}left ( x ight )ln^2x-120color{Blue}{mathbf{H}_5left ( x ight )}ln x\ &~~~+120mathrm{Li}_6left ( x ight )ln x+120color{blue}{mathbf{H}_6left ( x ight )}-120mathrm{Li}_7left ( x ight ) end{align*}}]

Hence

[egin{align*} mathcal{J}&=-color{blue}{mathbf{H}_1left ( frac{1}{2} ight )}ln^52+mathrm{Li}_2left ( frac{1}{2} ight )ln^52-5color{blue}{mathbf{H}_2left ( frac{1}{2} ight )}ln^42+5mathrm{Li}_3left ( frac{1}{2} ight )ln^42\ &~~~-20color{blue}{mathbf{H}_3left ( frac{1}{2} ight )}ln^32+20mathrm{Li}_4left ( frac{1}{2} ight )ln^32-60color{blue}{mathbf{H}_4left ( frac{1}{2} ight )}ln^22+60mathrm{Li}_5left ( frac{1}{2} ight )ln^22\ &~~~-120color{blue}{mathbf{H}_5left ( frac{1}{2} ight )}ln2+120mathrm{Li}_6left ( frac{1}{2} ight )ln2-120color{blue}{mathbf{H}_6left ( frac{1}{2} ight )}+120mathrm{Li}_7left ( frac{1}{2} ight )\ &=-2ln^72-frac{225}{8}zeta left ( 3 ight )ln^42+frac{1}{18}pi ^{4}ln^22-60ln^22mathrm{Li}_5left ( frac{1}{2} ight )\ &~~~-60ln^32mathrm{Li}_4left ( frac{1}{2} ight )-frac{15}{8}ln^22zeta left ( 5 ight )+5pi ^{3}ln^22zeta left ( 3 ight )+frac{5}{3}pi ^{2}ln^52\ &~~~-120ln 2color{blue}{mathbf{H}_5left ( frac{1}{2} ight )}+120ln2mathrm{Li}_6left ( frac{1}{2} ight )--120color{blue}{mathbf{H}_6left ( frac{1}{2} ight )}+120mathrm{Li}_7left ( frac{1}{2} ight ) end{align*}]

Combine these all together,we get

[{smalloxed{displaystyle egin{align*} int_{0}^{1}frac{ln^{3}left ( 1+x ight )ln^{3}x}{x}mathrm{d}x =& color{Teal} {-int_{frac{1}{2}}^{1}frac{ln^{3}xln^{3}left ( 1-x ight )}{1-x}mathrm{d}x+3int_{frac{1}{2}}^{1}frac{ln^{3}left ( 1-x ight )ln^{3}x}{x}mathrm{d}x}\ &color{Teal} {+frac{15}{4}ln^{7}2+frac{75}{2}zeta left ( 3 ight )ln^{4}2-frac{1}{10}pi ^{4}ln^{3}2-2pi ^{2}ln^{5}2+frac{27}{8}zeta left ( 5 ight )ln^22}\ &color{Teal} {-9pi ^{3}zeta left ( 3 ight )ln^22-12ln^22mathrm{Li}_{5}left ( frac{1}{2} ight )+48ln^32mathrm{Li}_{3}left ( frac{1}{2} ight )-576ln2mathrm{Li}_{6}left ( frac{1}{2} ight )}\ &color{Teal} {+216ln2sum_{n=1}^{infty }frac{H_n}{n^{5}2^{n}}+216sum_{n=1}^{infty }frac{H_n}{n^{6}2^{n}}-576mathrm{Li}_{7}left ( frac{1}{2} ight )} end{align*}}}]

In order to use the Beta function,we make the following transformations

[egin{align*} &-int_{frac{1}{2}}^{1}frac{ln^{3}xln^{3}left ( 1-x ight )}{1-x}mathrm{d}x+3int_{frac{1}{2}}^{1}frac{ln^{3}left ( 1-x ight )ln^{3}x}{x}mathrm{d}x\ =&-int_{0}^{1}frac{ln^3xln^{3}left ( 1-x ight )}{1-x}mathrm{d}x+4int_{frac{1}{2}}^{1}frac{ln^{3}left ( 1-x ight )ln^{3}x}{x}mathrm{d}x end{align*}]

And using the defination of Nielsen Generalized Polylogarithm Function

[S_{n,p}left ( z ight )=frac{left ( -1 ight )^{n+p-1}}{left ( n-1 ight )!p!}int_{0}^{1}frac{ln^{n-1}tln^pleft ( 1-zt ight )}{t}mathrm{d}t ]

and

[sum_{m=1}^{infty }frac{z^{m}H_m}{m^{n+1}}=S_{n,2}left ( z ight )+mathrm{Li}_{n+2}left ( z ight ) ]

Hence we get

[{smalloxed{displaystyle egin{align*} int_{0}^{1}frac{ln^{3}left ( 1+x ight )ln^{3}x}{x}mathrm{d}x =& color{Teal} {-frac{partial ^{6}}{partial a^{3}partial b^{3}}mathrm{B}left ( 1,0^+ ight )+4int_{frac{1}{2}}^{1}frac{ln^{2}left ( 1-x ight )ln^{4}x}{1-x}mathrm{d}x}\ &color{Teal} {+frac{77}{20}ln^{7}2+frac{75}{2}zeta left ( 3 ight )ln^{4}2-frac{51}{140}pi ^{6}ln2+frac{1}{20}pi ^{4}ln^{3}2-frac{1}{2}pi ^{2}ln^{5}2}\ &color{Teal} {-frac{27}{8}zeta left ( 5 ight )ln^22-9pi ^{3}zeta left ( 3 ight )ln^22+18pi ^{2}zeta left ( 3 ight )ln^22-36zeta left ( 3 ight )ln^42}\ &color{Teal} {-54zeta ^{2}left ( 3 ight )ln2-108ln2zeta left ( ar{5},1 ight )+204ln^22mathrm{Li}_{5}left ( frac{1}{2} ight )+48ln^32mathrm{Li}_{3}left ( frac{1}{2} ight )}\ &color{Teal} {-72ln2mathrm{Li}_{6}left ( frac{1}{2} ight )+216S_{5,2}left ( frac{1}{2} ight )-360mathrm{Li}_{7}left ( frac{1}{2} ight )} end{align*}}}]

Let

[mathbf{O}=frac{partial ^{6}}{partial a^{3}partial b^{3}}mathrm{B}left ( 1,0^+ ight )~~,~~mathbf{P}=frac{partial ^{6}}{partial a^{4}partial b^{2}}mathrm{B}left ( 1,0^+ ight )~~,~~mathbf{Q}=frac{partial ^{6}}{partial a^{2}partial b^{4}}mathrm{B}left ( 1,0^+ ight ) ]

For (mathcal{I}_1)

[mathcal{I}_1=-frac{1}{4}ln^72+frac{4}{3}underbrace{color{Red} {int_{frac{1}{2}}^{1}frac{ln^3xln^3left ( 1-x ight )}{1-x}mathrm{d}x}}_mathcal{M} ]

For (mathcal{I}_2)

[mathcal{I}_2=frac{1}{4}ln^72+frac{3}{4}underbrace{color{Red} {int_{frac{1}{2}}^{1}frac{ln^4xln^2left ( 1-x ight )}{1-x}mathrm{d}x}}_mathcal{N} ]

Therefore,

[egin{align*} &4int_{frac{1}{2}}^{1}frac{ln^4xln^2left ( 1-x ight )}{1-x}mathrm{d}x+4mathcal{I}_1-4mathcal{I}_1\ &=4mathbf{P}-4mathcal{I}_1\ &=4mathbf{P}+frac{4}{3}ln^72-frac{16}{3}mathcal{M}-frac{16}{3}mathcal{I}_2\ &=4mathbf{P}+frac{4}{3}ln^72-frac{16}{3}mathbf{O}+frac{16}{3}mathcal{I}_2\ &=4mathbf{P}+frac{8}{3}ln^72-frac{16}{3}mathbf{O}-frac{3}{4}mathbf{Q}+frac{3}{4}mathcal{I}_3\ &=4mathbf{P}+frac{151}{60}ln^72-frac{16}{3}mathbf{O}-frac{3}{4}mathbf{Q}+frac{3}{10}mathcal{J} end{align*}]

With the help of Mathematica,we can easily get the value of (mathbf{O},mathbf{P},mathbf{Q}),and combine with the previous results,we get the Final Result:

[{smalloxed{displaystyle color{Blue}{egin{align*} color{black}{int_{0}^{1}frac{ln^{3}left ( 1+x ight )ln^{3}x}{x}mathrm{d}x =}&~frac{1}{5}pi ^{4}zeta left ( 3 ight )+frac{1}{4}pi ^{2}zeta left ( 5 ight )-frac{3}{20}zeta left ( 7 ight )+42ln^32zeta left ( 3 ight )+8ln^62\ &-4pi ^{2}ln^32+frac{71}{12}ln^{7}2+frac{87}{2}zeta left ( 3 ight )ln^{4}2-frac{17}{56}pi ^{6}ln2+frac{1}{24}pi ^{4}ln^{3}2\ &-frac{5}{4}pi ^{2}ln^{5}2-frac{45}{16}zeta left ( 5 ight )ln^22-frac{21}{2}pi ^{3}zeta left ( 3 ight )ln^22+15pi ^{2}zeta left ( 3 ight )ln^22\ &-frac{711}{16}zeta left ( 3 ight )ln^42-45zeta ^{2}left ( 3 ight )ln2-90ln2zeta left ( ar{5},1 ight )+150ln^22mathrm{Li}_{5}left ( frac{1}{2} ight )\ &-18ln^32mathrm{Li}_{4}left ( frac{1}{2} ight )-216ln2mathrm{Li}_{6}left ( frac{1}{2} ight )+180S_{5,2}left ( frac{1}{2} ight )-360mathrm{Li}_{7}left ( frac{1}{2} ight ) end{align*}}}}]