复杂的对数积分（一）

[Largeint_0^1frac{ln^3(1+x)ln x}xmathrm dx ]

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(Largemathbf{Solution:})
Start with integration by parts (IBP) by setting (u=ln^3(1+x)) and (mathrm{d}v=dfrac{ln x}{x} mathrm{d}x) yields

[egin{align*} I&=-frac32int_0^1frac{ln^2(1+x)ln^2 x}{1+x} mathrm{d}x\ &=-frac32int_1^2frac{ln^2xln^2 (x-1)}{x} mathrm{d}xquadRightarrowquadcolor{red}{xmapsto1+x}\ &=-frac32int_{largefrac12}^1left[frac{ln^2xln^2 (1-x)}{x}-frac{2ln^3xln(1-x)}{x}+frac{ln^4x}{x} ight] mathrm{d}xquadRightarrowquadcolor{red}{xmapstofrac1x}\ &=-frac32int_{largefrac12}^1frac{ln^2xln^2 (1-x)}{x} mathrm{d}x+3int_{largefrac12}^1frac{ln^3xln(1-x)}{x} mathrm{d}x-left.frac3{10}ln^5x ight|_{largefrac12}^1\ &=-frac32color{red}{int_{largefrac12}^1frac{ln^2xln^2 (1-x)}{x} mathrm{d}x}+3int_{largefrac12}^1frac{ln^3xln(1-x)}{x} mathrm{d}x-frac3{10}ln^52 end{align*}]

Applying IBP again to evaluate the red integral by setting (u=ln^2(1-x)) and (mathrm{d}v=dfrac{ln^2 x}{x} mathrm{d}x) yields

[color{red}{int_{largefrac12}^1frac{ln^2xln^2 (1-x)}{x} mathrm{d}x}=frac13ln^52+frac23color{blue}{int_{largefrac12}^1frac{ln^3xln (1-x)}{1-x} mathrm{d}x} ]

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} ]

Let us integrating the indefinite form of the blue integral.

[egin{align*} color{blue}{intfrac{ln^3xln (1-x)}{1-x} mathrm{d}x}=&-intsum_{n=1}^infty H_nx^nln^3x mathrm{d}x\ =&-sum_{n=1}^infty H_nint x^nln^3x mathrm{d}x\ =&-sum_{n=1}^infty H_nfrac{partial^3}{partial n^3}left[int x^n mathrm{d}x ight]\ =&-sum_{n=1}^infty H_nfrac{partial^3}{partial n^3}left[frac{x^{n+1}}{n+1} ight]\ =&-sum_{n=1}^infty H_nleft[frac{x^{n+1}ln^3x}{n+1}-frac{3x^{n+1}ln^2x}{(n+1)^2}+frac{6x^{n+1}ln x}{(n+1)^3}-frac{6x^{n+1}}{(n+1)^4} ight]\ =&-ln^3xsum_{n=1}^infty frac{H_{n+1}x^{n+1}}{n+1}+ln^3xsum_{n=1}^infty frac{x^{n+1}}{(n+1)^2}+3ln^2xsum_{n=1}^infty frac{H_{n+1}x^{n+1}}{(n+1)^2}\&-3ln^2xsum_{n=1}^infty frac{x^{n+1}}{(n+1)^3}-6ln xsum_{n=1}^infty frac{H_{n+1}x^{n+1}}{(n+1)^3}+6ln xsum_{n=1}^infty frac{x^{n+1}}{(n+1)^4}\&+6sum_{n=1}^infty frac{H_{n+1}x^{n+1}}{(n+1)^4}-6sum_{n=1}^infty frac{x^{n+1}}{(n+1)^5}\ =& -sum_{n=1}^inftyleft[frac{H_nx^{n}ln^3x}{n}-frac{x^{n}ln^3x}{n^2}-frac{3H_nx^{n}ln^2x}{n^2}+frac{3x^{n}ln^2x}{n^3} ight.\& left. +frac{6H_nx^{n}ln x}{n^3}-frac{6x^{n}ln x}{n^4}-frac{6H_nx^{n}}{n^4}+frac{6x^{n}}{n^5} ight]\ =& -color{blue}{mathbf{H}_{1}(x)}ln^3x+operatorname{Li}_2(x)ln^3x+3color{blue}{mathbf{H}_{2}(x)}ln^2x-3operatorname{Li}_3(x)ln^2x\& -6color{blue}{mathbf{H}_{3}(x)}ln x+6operatorname{Li}_4(x)ln x+6color{blue}{mathbf{H}_{4}(x)}-6operatorname{Li}_5(x) end{align*}]

Therefore

[egin{align*} color{blue}{int_{Largefrac12}^1frac{ln^3xln (1-x)}{1-x} mathrm{d}x} =& 6color{blue}{mathbf{H}_{4}(1)}-6operatorname{Li}_5(1)-left[color{blue}{mathbf{H}_{1}left(frac12 ight)}ln^32-operatorname{Li}_2left(frac12 ight)ln^32 ight.\&left. +3color{blue}{mathbf{H}_{2}left(frac12 ight)}ln^22-3operatorname{Li}_3left(frac12 ight)ln^22+6color{blue}{mathbf{H}_{3}left(frac12 ight)}ln 2 ight.\& -6operatorname{Li}_4(x)ln 2+6color{blue}{mathbf{H}_{4}(x)}-6operatorname{Li}_5(x)igg]\ =& 12zeta(5)-pi^2zeta(3)+frac{3}8zeta(3)ln^22-frac{pi^4}{120}ln2-frac{1} {4}ln^52\& -6color{blue}{mathbf{H}_{4}left(frac12 ight)}+6operatorname{Li}_4left(frac12 ight)ln 2+6operatorname{Li}_5left(frac12 ight) end{align*}]

Using the similar approach as calculating the blue integral, then

[egin{align*} intfrac{ln^3xln (1-x)}{x} mathrm{d}x&=-intsum_{n=1}^infty frac{x^{n-1}}{n}ln^3x mathrm{d}x\ &=-sum_{n=1}^infty frac{1}{n}int x^{n-1}ln^3x mathrm{d}x\ &=-sum_{n=1}^infty frac{1}{n}frac{partial^3}{partial n^3}left[int x^{n-1} mathrm{d}x ight]\ &=-sum_{n=1}^infty frac{1}{n}frac{partial^3}{partial n^3}left[frac{x^{n}}{n} ight]\ &=-sum_{n=1}^infty frac{1}{n}left[frac{x^{n}ln^3x}{n}-frac{3x^{n}ln^2x}{n^2}+frac{6x^{n}ln x}{n^3}-frac{6x^{n}}{n^4} ight]\ &=sum_{n=1}^infty left[-frac{x^{n}ln^3x}{n^2}+frac{3x^{n}ln^2x}{n^3}-frac{6x^{n}ln x}{n^4}+frac{6x^{n}}{n^5} ight]\ &=6operatorname{Li}_5(x)-6operatorname{Li}_4(x)ln x+3operatorname{Li}_3(x)ln^2x-operatorname{Li}_2(x)ln^3x end{align*}]

Hence

[int_{largefrac{1}{2}}^1frac{ln^3xln (1-x)}{x} mathrm{d}x=frac{pi^2}{6}ln^32-frac{21}{8}zeta(3)ln^22-6operatorname{Li}_4left(frac{1}{2} ight)ln2-6operatorname{Li}_5left(frac{1}{2} ight)+6zeta(5) ]

Combining altogether, we have

[egin{align*} I=& frac{pi^4}{120}ln2-frac{33}4zeta(3)ln^22+frac{pi^2}2ln^32-frac{11}{20}ln^52+6zeta(5)+pi^2zeta(3)\ & +6color{blue}{mathbf{H}_{4}left(frac12 ight)}-18operatorname{Li}_4left(frac12 ight)ln2-24operatorname{Li}_5left(frac12 ight) end{align*}]

Continuing the answer in: A sum containing harmonic numbers (displaystylesum_{n=1}^inftyfrac{H_n}{n^3\,2^n}),we have

[egin{align*} color{blue}{mathbf{H}_{3}left(x ight)}=&frac12zeta(3)ln x-frac18ln^2xln^2(1-x)+frac12ln xleft[color{blue}{mathbf{H}_{2}left(x ight)}-operatorname{Li}_3(x) ight]\&+operatorname{Li}_4(x)-frac{pi^2}{12}operatorname{Li}_2(x)-frac12operatorname{Li}_3(1-x)ln x+frac{pi^4}{60} ag1 end{align*}]

Dividing ((1)) by (x) and then integrating yields

[egin{align*} color{blue}{mathbf{H}_{4}left(x ight)}=&frac14zeta(3)ln^2 x-frac18intfrac{ln^2xln^2(1-x)}xmathrm dx+frac12intfrac{ln x}xigg[color{blue}{mathbf{H}_{2}left(x ight)}-operatorname{Li}_3(x)igg] mathrm dx\&+operatorname{Li}_5(x)-frac{pi^2}{12}operatorname{Li}_3(x)-frac12intfrac{operatorname{Li}_3(1-x)ln x}x mathrm dx+frac{pi^4}{60}ln x\ =&frac14zeta(3)ln^2 x+frac{pi^4}{60}ln x+operatorname{Li}_5(x)-frac{pi^2}{12}operatorname{Li}_3(x)-frac18color{red}{intfrac{ln^2xln^2(1-x)}x mathrm dx}\&+frac12left[color{purple}{sum_{n=1}^inftyfrac{H_{n}}{n^2}int x^{n-1}ln x mathrm dx}-color{green}{intfrac{operatorname{Li}_3(x)ln x}x mathrm dx}-color{orange}{intfrac{operatorname{Li}_3(1-x)ln x}x mathrm dx} ight] ag2 end{align*}]

Evaluating the red integral using the same technique as the previous one yields

[color{red}{intfrac{ln^2xln^2(1-x)}x mathrm dx}=frac13ln^3xln^2(1-x)-frac23color{blue}{intfrac{ln(1-x)ln^3 x}{1-x} mathrm dx} ]

Evaluating the purple integral yields

[egin{align*} color{purple}{sum_{n=1}^inftyfrac{H_{n}}{n^2}int x^{n-1}ln x mathrm dx}&=sum_{n=1}^inftyfrac{H_{n}}{n^2}frac{partial}{partial n}left[int x^{n-1} mathrm dx ight]\ &=sum_{n=1}^inftyfrac{H_{n}}{n^2}left[frac{x^nln x}{n}-frac{x^n}{n^2} ight]\ &=color{blue}{mathbf{H}_{3}(x)}ln x-color{blue}{mathbf{H}_{4}(x)} end{align*}]

Evaluating the green integral using IBP by setting (u=ln x) and (mathrm dv=dfrac{operatorname{Li}_3(x)}{x}mathrm dx) yields

[egin{align*} color{green}{intfrac{operatorname{Li}_3(x)ln x}x mathrm dx}&=operatorname{Li}_4(x)ln x-intfrac{operatorname{Li}_4(x)}x mathrm dx\ &=operatorname{Li}_4(x)ln x-operatorname{Li}_5(x) end{align*}]

Evaluating the orange integral using IBP by setting (u=operatorname{Li}_3(1-x)) and (mathrm dv=dfrac{ln x}{x} mathrm dx) yields

[color{orange}{intfrac{operatorname{Li}_3(1-x)ln x}x mathrm dx}=frac12operatorname{Li}_3(1-x)ln^2 x+frac12color{maroon}{intfrac{operatorname{Li}_2(1-x)ln^2 x}{1-x} mathrm dx} ]

Applying IBP again to evaluate the maroon integral by setting (u=operatorname{Li}_2(1-x)) and

[mathrm dv=dfrac{ln^2 x}{1-x}mathrm dxquadRightarrowquad v=2operatorname{Li}_3(x)-2operatorname{Li}_2(x)ln x-ln(1-x)ln^2x]

we have

[{egin{align*} color{maroon}{intfrac{operatorname{Li}_2(1-x)ln^2 x}{1-x} mathrm dx}=&left[2operatorname{Li}_3(x)-2operatorname{Li}_2(x)ln x-ln(1-x)ln^2x ight]operatorname{Li}_2(1-x)\ &-2intfrac{operatorname{Li}_3(x)ln x}{1-x} mathrm dx+2intfrac{operatorname{Li}_2(x)ln x}{1-x} mathrm dx+color{blue}{intfrac{ln(1-x)ln^3 x}{1-x} mathrm dx} end{align*}}]

We use the generating function for the generalized harmonic numbers evaluate the above integrals involving polylogarithm.

[egin{align*} intfrac{operatorname{Li}_k(x)ln x}{1-x} mathrm dx&=sum_{n=1}^infty H_{n}^{(k)}int x^nln x mathrm dx\ &=sum_{n=1}^infty H_{n}^{(k)}frac{partial}{partial n}left[int x^n mathrm dx ight]\ &=sum_{n=1}^infty H_{n}^{(k)}left[frac{x^{n+1}ln x}{n+1}-frac{x^{n+1}}{(n+1)^2} ight]\ &=sum_{n=1}^inftyleft[frac{H_{n+1}^{(k)}x^{n+1}ln x}{n+1}-frac{x^{n+1}ln x}{(n+1)^{k+1}}-frac{H_{n+1}^{(k)}x^{n+1}}{(n+1)^2}+frac{x^{n+1}}{(n+1)^{k+2}} ight]\ &=sum_{n=1}^inftyleft[frac{H_{n}^{(k)}x^{n}ln x}{n}-frac{x^{n}ln x}{n^{k+1}}-frac{H_{n}^{(k)}x^{n}}{n^2}+frac{x^{n}}{n^{k+2}} ight]\ &=color{blue}{mathbf{H}_{1}^{(k)}(x)}ln x-operatorname{Li}_{k+1}(x)ln x-color{blue}{mathbf{H}_{2}^{(k)}(x)}+operatorname{Li}_{k+2}(x) end{align*}]

Dividing generating function of (color{blue}{mathbf{H}^{(k)}(x)}) by (x) and then integrating yields

[egin{align*} sum_{n=1}^infty frac{H_{n}^{(k)}x^n}{n}&=intfrac{operatorname{Li}_k(x)}{x(1-x)} mathrm dx\ color{blue}{mathbf{H}_{1}^{(k)}(x)}&=intfrac{operatorname{Li}_k(x)}{x} mathrm dx+intfrac{operatorname{Li}_k(x)}{1-x} mathrm dx\ &=operatorname{Li}_{k+1}(x)+intfrac{operatorname{Li}_k(x)}{1-x} mathrm dx end{align*}]

Repeating the process above yields

[egin{align*} sum_{n=1}^infty frac{H_{n}^{(k)}x^n}{n^2} &=intfrac{operatorname{Li}_{k+1}(x)}{x} mathrm dx+intfrac{operatorname{Li}_k(x)}{x(1-x)} mathrm dx\ color{blue}{mathbf{H}_{2}^{(k)}(x)}&=operatorname{Li}_{k+2}(x)+operatorname{Li}_{k+1}(x)+intfrac{operatorname{Li}_k(x)}{1-x} mathrm dx end{align*}]

where it is easy to show by using IBP that

[egin{align*} intfrac{operatorname{Li}_2(x)}{1-x} mathrm dx&=-intfrac{operatorname{Li}_2(1-x)}{x} mathrm dx\ &=2operatorname{Li}_3(x)-2operatorname{Li}_2(x)ln(x)-operatorname{Li}_2(1-x)ln x-ln (1-x)ln^2x end{align*}]

and

[intfrac{operatorname{Li}_3(x)}{1-x} mathrm dx=-intfrac{operatorname{Li}_3(1-x)}{x} mathrm dx=-frac12operatorname{Li}_2^2(1-x)-operatorname{Li}_3(1-x)ln x ]

Now, all unknown terms have been obtained. Putting altogether to ((2)), we have

[{egin{align*} color{blue}{mathbf{H}_{4}(x)} =& frac1{10}zeta(3)ln^2 x+frac{pi^4}{150}ln x-frac{pi^2}{30}operatorname{Li}_3(x)-frac1{60}ln^3xln^2(1-x)+frac65operatorname{Li}_5(x)\&-frac15left[operatorname{Li}_3(x)-operatorname{Li}_2(x)ln x-frac12ln(1-x)ln^2x ight]operatorname{Li}_2(1-x)-frac15operatorname{Li}_4(x)\&-frac35operatorname{Li}_4(x)ln x+frac15operatorname{Li}_3(x)ln x+frac15operatorname{Li}_3(x)ln^2x-frac1{10}operatorname{Li}_3(1-x)ln^2 x\&-frac1{15}operatorname{Li}_2(x)ln^3x-frac15color{blue}{mathbf{H}_{2}^{(3)}(x)}+frac15color{blue}{mathbf{H}_{2}^{(2)}(x)} +frac15color{blue}{mathbf{H}_{1}^{(3)}(x)}ln x\&-frac15color{blue}{mathbf{H}_{1}^{(2)}(x)}ln x+frac25color{blue}{mathbf{H}_{3}(x)}ln x-frac15color{blue}{mathbf{H}_{2}(x)}ln^2x+frac1{15}color{blue}{mathbf{H}_{1}(x)}ln^3x+C ag3 end{align*}}]

The next step is finding the constant of integration. Setting (x=1) to ((3)) yields

[{egin{align*} color{blue}{mathbf{H}_{4}(1)} &=-frac{pi^2}{30}operatorname{Li}_3(1)+frac65operatorname{Li}_5(1)-frac15operatorname{Li}_4(1)-frac15color{blue}{mathbf{H}_{2}^{(3)}(1)}+frac15color{blue}{mathbf{H}_{2}^{(2)}(1)}+C\ 3zeta(5)+zeta(2)zeta(3)&=-frac{pi^2}{30}operatorname{Li}_3(1)+frac{19}{30}operatorname{Li}_5(1)+frac{3}{5}operatorname{Li}_3(1)+C\ C&=frac{pi^4}{450}+frac{pi^2}{5}zeta(3)-frac35zeta(3)+3zeta(5) end{align*}}]

Thus

[{egin{align*} color{blue}{mathbf{H}_{4}(x)} =& frac1{10}zeta(3)ln^2 x+frac{pi^4}{150}ln x-frac{pi^2}{30}operatorname{Li}_3(x)-frac1{60}ln^3xln^2(1-x)+frac65operatorname{Li}_5(x)\&-frac15left[operatorname{Li}_3(x)-operatorname{Li}_2(x)ln x-frac12ln(1-x)ln^2x ight]operatorname{Li}_2(1-x)-frac15operatorname{Li}_4(x)\&-frac35operatorname{Li}_4(x)ln x+frac15operatorname{Li}_3(x)ln x+frac15operatorname{Li}_3(x)ln^2x-frac1{10}operatorname{Li}_3(1-x)ln^2 x\&-frac1{15}operatorname{Li}_2(x)ln^3x-frac15color{blue}{mathbf{H}_{2}^{(3)}(x)}+frac15color{blue}{mathbf{H}_{2}^{(2)}(x)} +frac15color{blue}{mathbf{H}_{1}^{(3)}(x)}ln x\&-frac15color{blue}{mathbf{H}_{1}^{(2)}(x)}ln x+frac25color{blue}{mathbf{H}_{3}(x)}ln x-frac15color{blue}{mathbf{H}_{2}(x)}ln^2x+frac1{15}color{blue}{mathbf{H}_{1}(x)}ln^3x\&+frac{pi^4}{450}+frac{pi^2}{5}zeta(3)-frac35zeta(3)+3zeta(5) ag4 end{align*}}]

and setting (x=dfrac12) to ((4)) yields

[egin{align*} color{blue}{mathbf{H}_{4}left(frac12 ight)}=& frac{ln^52}{40}-frac{pi^2}{36}ln^32+frac{zeta(3)}{2}ln^22-frac{pi^2}{12}zeta(3)\&+frac{zeta(5)}{32}-frac{pi^4}{720}ln2+operatorname{Li}_4left(frac12 ight)ln2+2operatorname{Li}_5left(frac12 ight) ag5 end{align*}]

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Finally, we obtain

[oxed{egin{align*} int_0^1frac{ln^3(1+x)ln x}xmathrm dx=& color{blue}{frac{pi^2}2zeta(3)+frac{99}{16}zeta(5)-frac25ln^52+frac{pi^2}3ln^32-frac{21}4zeta(3)ln^22}\&color{blue}{-12operatorname{Li}_4left(frac12 ight)ln2-12operatorname{Li}_5left(frac12 ight)} end{align*}}]

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(mathbf{References :})

([1]) : Harmonic number
([2]) : Polylogarithm

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(Largemathbf{Here \,\,\,is \,\,\,another \,\,\,way \,\,\,solving \,\,\,this \,\,\,problem\,\,:})

Step 1: Expressing the integral as a sum
It is easy to derive the formula

[left(sum^{infty}_{n=1}a_nx^n ight)left(sum^{infty}_{n=1}b_nx^n ight)=sum^infty_{n=1}sum^{n}_{k=1}a_kb_{n-k+1}x^{n+1} ]

We apply this formula to derive the Taylor series of (ln^2(1+x)).

[egin{align*} ln^2(1+x) &=left(sum^{infty}_{n=1}frac{(-1)^{n-1}}{n}x^n ight)left(sum^{infty}_{n=1}frac{(-1)^{n-1}}{n}x^n ight)\ &=sum^infty_{n=1}sum^n_{k=1}frac{(-1)^{k-1}(-1)^{n-k}}{k(n-k+1)}x^{n+1}\ &=sum^infty_{n=1}frac{(-1)^{n+1}}{n+1}sum^n_{k=1}left(frac{1}{k}+frac{1}{n-k+1} ight)x^{n+1}\ &=sum^infty_{n=1}frac{(-1)^{n+1}2H_n}{n+1}x^{n+1} end{align*}]

Apply this formula again to obtain the Taylor series of (displaystylefrac{ln^2(1+x)}{1+x}).

[egin{align*} frac{ln^2(1+x)}{1+x} &=left(sum^infty_{n=1}frac{(-1)^{n+1}2H_n}{n+1}x^{n+1} ight)left(sum^{infty}_{n=1}(-1)^{n-1}x^{n-1} ight)\ &=sum^infty_{n=1}sum^n_{k=1}frac{(-1)^{k+1}(-1)^{n-k}2H_k}{k+1}x^{n+1}\ &=sum^infty_{n=1}2(-1)^{n+1}sum^n_{k=1}frac{H_k}{k+1}x^{n+1}\ end{align*}]

The inner sum is

[egin{align*} sum^n_{k=1}frac{H_k}{k+1} &=sum^n_{k=1}frac{H_{k+1}}{k+1}-sum^n_{k=1}frac{1}{(k+1)^2}\ &=sum^{n+1}_{k=1}frac{H_k}{k}-H_{n+1}^{(2)}\ &=sum^{n+1}_{k=1}frac{1}{k}sum^k_{j=1}frac{1}{j}-H_{n+1}^{(2)}\ &=sum^{n+1}_{j=1}frac{1}{j}left(sum^{n+1}_{k=1}frac{1}{k}-sum^{j-1}_{k=1}frac{1}{k} ight)-H_{n+1}^{(2)}\ &=H_{n+1}^2-sum^{n+1}_{j=1}frac{H_j}{j}\ &=frac{H_{n+1}^2-H_{n+1}^{(2)}}{2} end{align*}]

Hence

[frac{ln^2(1+x)}{1+x}=sum^infty_{n=1}(-1)^{n+1}left(H_{n+1}^2-H_{n+1}^{(2)} ight)x^{n+1} ]

Pluck this into the integral.

[egin{align*} int^1_0frac{ln^3(1+x)ln{x}}{x}{ m d}x &=-frac{3}{2}int^1_0frac{ln^2(1+x)ln^2{x}}{1+x}{ m d}x\ &=-frac{3}{2}sum^infty_{n=1}(-1)^{n+1}left(H_{n+1}^2-H_{n+1}^{(2)} ight)int^1_0x^{n+1}ln^2{x} { m d}x\ &=-3sum^infty_{n=1}frac{(-1)^{n+1}left(H_{n+1}^2-H_{n+1}^{(2)} ight)}{(n+2)^3}\ &=3sum^infty_{n=1}frac{(-1)^{n}left(H_{n}^{(2)}-H_{n}^2 ight)}{(n+1)^3}\ end{align*}]

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Step 2: Evaluation of (displaystylesum^infty_{n=1}frac{(-1)^nH_n^{(2)}}{(n+1)^3})
We begin with some simple manipulations of the sum.

[egin{align*} sum^infty_{n=1}frac{(-1)^nH_n^{(2)}}{(n+1)^3} &=sum^infty_{n=1}frac{(-1)^nH_{n+1}^{(2)}}{(n+1)^3}-sum^infty_{n=1}frac{(-1)^n}{(n+1)^5}\ &=-frac{15}{16}zeta(5)-underbrace{sum^infty_{n=1}frac{(-1)^nH_n^{(2)}}{n^3}}_{S} end{align*}]

Consider the function (displaystyle f(z)=frac{picsc(pi z)psi_1(-z)}{z^3}). At the positive integers,

[egin{align*} { m Res}(f,n) &=operatorname*{Res}_{z=n}left[frac{(-1)^n}{z^3(z-n)^3}+frac{(-1)^n(H_n^{(2)}+2zeta(2))}{z^3(z-n)} ight]\ &=frac{6(-1)^n}{n^5}+frac{(-1)^nH_n^{(2)}}{n^3}+frac{2(-1)^nzeta(2)}{n^3} end{align*}]

Summing them up gives

[sum^infty_{n=1} { m Res}(f,n)=-frac{45}{8}zeta(5)+S-frac{3}{2}zeta(2)zeta(3) ]

At the negative integers,

[egin{align*} { m Res}(f,-n) &=-frac{(-1)^npsi_1(n)}{n^3}\ &=frac{(-1)^nH_n^{(2)}}{n^3}-frac{(-1)^nzeta(2)}{n^3}-frac{(-1)^n}{n^5} end{align*}]

Summing them up gives

[sum^infty_{n=1} { m Res}(f,-n)=S+frac{3}{4}zeta(2)zeta(3)+frac{15}{16}zeta(5) ]

At (z=0),

[egin{align*} { m Res}(f,0) &=[z^2]left(frac{1}{z}+zeta(2)z ight)left(frac{1}{z^2}+zeta(2)+2zeta(3)z+3zeta(4)z^2+4zeta(5)z^3 ight)\ &=4zeta(5)+2zeta(2)zeta(3) end{align*}]

Since the sum of the (mathrm{reisudes} =0),

[sum^infty_{n=1}frac{(-1)^nH_n^{(2)}}{(n+1)^3}=-frac{41}{32}zeta(5)+frac{5}{8}zeta(2)zeta(3) ]

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Step 3: Evaluation of (displaystylesum^infty_{n=1}frac{(-1)^nH_n^{2}}{(n+1)^3})
Formula ((45)) in this page states that this sum is equal to

[4{ m Li}_5left(frac{1}{2} ight)+4{ m Li}_4left(frac{1}{2} ight)ln{2}+frac{2}{15}ln^5{2}-frac{107}{32}zeta(5)+frac{7}{4}zeta(3)ln^2{2}-frac{2}{3}zeta(2)ln^2{2}-frac{3}{8}zeta(2)zeta(3) ]

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Step 4: Obtaining the final result
Combining our previous results, we get

[egin{align*} &{int^1_0frac{ln^3(1+x)ln{x}}{x}{ m d}x}\ &={3sum^infty_{n=1}frac{(-1)^nleft(H_{n}^{(2)}-H_n^2 ight)}{(n+1)^3}}\ &=3Bigg(frac{33}{16}zeta(5)+zeta(2)zeta(3)-4{ m Li}_5left(frac{1}{2} ight)-4{ m Li}_4left(frac{1}{2} ight)ln{2}-frac{2}{15}ln^5{2}\ &~~~-frac{7}{4}zeta(3)ln^2{2}+frac{2}{3}zeta(2)ln^3{2}Bigg)\ &=oxed{color{blue}{dfrac{99}{16}zeta(5)+dfrac{pi^2}{2}zeta(3)-12{ m Li}_5left(dfrac{1}{2} ight)-12{ m Li}_4left(dfrac{1}{2} ight)ln{2}-dfrac{2}{5}ln^5{2}-dfrac{21}{4}zeta(3)ln^2{2}+dfrac{pi^2}{3}ln^3{2}}} end{align*}]