How to find ∫10ln3(1+x)lnxxdx{\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx

Please help me to find a closed form for this integral:
I=10ln3(1+x)lnxxdx
I suspect it might exist because there are similar integrals having closed forms:
10ln3(1x)lnxxdx=12ζ(5)π2ζ(3)10ln2(1+x)lnxxdx=π42416ln42+π26ln2272ζ(3)ln24Li4(12)10ln3(1+x)lnxx2dx=34ζ(3)634ζ(3)ln2+23π412034ln422ln32+3π24ln2218Li4(12).
Thanks!

Answer

Start with integration by parts (IBP) by setting u=ln3(1+x) and dv=lnxx dx yields
I=3210ln2(1+x)ln2x1+x dx=3221ln2xln2(x1)x dxx1+x=32112[ln2xln2(1x)x2ln3xln(1x)x+ln4xx] dxx1x=32112ln2xln2(1x)x dx+3112ln3xln(1x)x dx310ln5x|112=32112ln2xln2(1x)x dx+3112ln3xln(1x)x dx310ln52.
Applying IBP again to evaluate the red integral by setting u=ln2(1x) and dv=ln2xx dx yields
112ln2xln2(1x)x dx=13ln52+23112ln3xln(1x)1x dx.

For the simplicity, let
H(k)m(x)=n=1H(k)nxnnmH(x)=n=1Hnxn,
Introduce a generating function for the generalized harmonic numbers for |x|<1
H(k)(x)=n=1H(k)nxn=Lik(x)1xH(x)=ln(1x)1x
and the following identity
H(k)n+1H(k)n=1(n+1)kHn+1Hn=1n+1

Let us integrating the indefinite form of the blue integral.
ln3xln(1x)1x dx=n=1Hnxnln3x dx=n=1Hnxnln3x dx=n=1Hn3n3[xn dx]=n=1Hn3n3[xn+1n+1]=n=1Hn[xn+1ln3xn+13xn+1ln2x(n+1)2+6xn+1lnx(n+1)36xn+1(n+1)4]=ln3xn=1Hn+1xn+1n+1+ln3xn=1xn+1(n+1)2+3ln2xn=1Hn+1xn+1(n+1)23ln2xn=1xn+1(n+1)36lnxn=1Hn+1xn+1(n+1)3+6lnxn=1xn+1(n+1)4+6n=1Hn+1xn+1(n+1)46n=1xn+1(n+1)5= n=1[Hnxnln3xnxnln3xn23Hnxnln2xn2+3xnln2xn3 +6Hnxnlnxn36xnlnxn46Hnxnn4+6xnn5]= H1(x)ln3x+Li2(x)ln3x+3H2(x)ln2x3Li3(x)ln2x 6H3(x)lnx+6Li4(x)lnx+6H4(x)6Li5(x).
Therefore
112ln3xln(1x)1x dx= 6H4(1)6Li5(1)[H1(12)ln32Li2(12)ln32 +3H2(12)ln223Li3(12)ln22+6H3(12)ln2 6Li4(x)ln2+6H4(x)6Li5(x)]= 12ζ(5)π2ζ(3)+38ζ(3)ln22π4120ln214ln52 6H4(12)+6Li4(12)ln2+6Li5(12).
Using the similar approach as calculating the blue integral, then
ln3xln(1x)x dx=n=1xn1nln3x dx=n=11nxn1ln3x dx=n=11n3n3[xn1 dx]=n=11n3n3[xnn]=n=11n[xnln3xn3xnln2xn2+6xnlnxn36xnn4]=n=1[xnln3xn2+3xnln2xn36xnlnxn4+6xnn5]=6Li5(x)6Li4(x)lnx+3Li3(x)ln2xLi2(x)ln3x.
Hence
112ln3xln(1x)x dx=π26ln32218ζ(3)ln226Li4(12)ln26Li5(12)+6ζ(5).
Combining altogether, we have

I= π4120ln2334ζ(3)ln22+π22ln321120ln52+6ζ(5)+π2ζ(3) +6H4(12)18Li4(12)ln224Li5(12).


Continuing my answer in: A sum containing harmonic numbers \displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}, we have
\begin{align}
\color{blue}{\mathbf{H}_{3}\left(x\right)}=&\frac12\zeta(3)\ln x-\frac18\ln^2x\ln^2(1-x)+\frac12\ln x\left[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\right]\\&+\operatorname{Li}_4(x)-\frac{\pi^2}{12}\operatorname{Li}_2(x)-\frac12\operatorname{Li}_3(1-x)\ln x+\frac{\pi^4}{60}.\tag1
\end{align}

Dividing (1) by x and then integrating yields
\small\begin{align}
\color{blue}{\mathbf{H}_{4}\left(x\right)}=&\frac14\zeta(3)\ln^2 x-\frac18\int\frac{\ln^2x\ln^2(1-x)}x\ dx+\frac12\int\frac{\ln x}x\bigg[\color{blue}{\mathbf{H}_{2}\left(x\right)}-\operatorname{Li}_3(x)\bigg]\ dx\\&+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac12\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx+\frac{\pi^4}{60}\ln x\\
=&\frac14\zeta(3)\ln^2 x+\frac{\pi^4}{60}\ln x+\operatorname{Li}_5(x)-\frac{\pi^2}{12}\operatorname{Li}_3(x)-\frac18\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}\\&+\frac12\left[\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}-\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}-\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}\right].\tag2
\end{align}

Evaluating the red integral using the same technique as the previous one yields
\begin{align}
\color{red}{\int\frac{\ln^2x\ln^2(1-x)}x\ dx}&=\frac13\ln^3x\ln^2(1-x)-\frac23\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}.
\end{align}

Evaluating the purple integral yields
\begin{align}
\color{purple}{\sum_{n=1}^\infty\frac{H_{n}}{n^2}\int x^{n-1}\ln x\ dx}&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\frac{\partial}{\partial n}\left[\int x^{n-1}\ dx\right]\\
&=\sum_{n=1}^\infty\frac{H_{n}}{n^2}\left[\frac{x^n\ln x}{n}-\frac{x^n}{n^2}\right]\\
&=\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 dv=\dfrac{\operatorname{Li}_3(x)}{x}\ dx yields
\begin{align}
\color{green}{\int\frac{\operatorname{Li}_3(x)\ln x}x\ dx}&=\operatorname{Li}_4(x)\ln x-\int\frac{\operatorname{Li}_4(x)}x\ 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 dv=\dfrac{\ln x}{x}\ dx yields
\begin{align}
\color{orange}{\int\frac{\operatorname{Li}_3(1-x)\ln x}x\ dx}&=\frac12\operatorname{Li}_3(1-x)\ln^2 x+\frac12\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}.
\end{align}

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

dv=\dfrac{\ln^2 x}{1-x}\ dx\quad\Rightarrow\quad
v=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x,

we have
\small{\begin{align}
\color{maroon}{\int\frac{\operatorname{Li}_2(1-x)\ln^2 x}{1-x}\ dx}=&\left[2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln x-\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)\\
&-2\int\frac{\operatorname{Li}_3(x)\ln x}{1-x}\ dx+2\int\frac{\operatorname{Li}_2(x)\ln x}{1-x}\ dx+\color{blue}{\int\frac{\ln(1-x)\ln^3 x}{1-x}\ dx}.
\end{align}}

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

\begin{align}
\int\frac{\operatorname{Li}_k(x)\ln x}{1-x}\ dx&=\sum_{n=1}^\infty H_{n}^{(k)}\int x^n\ln x\ dx\\
&=\sum_{n=1}^\infty H_{n}^{(k)}\frac{\partial}{\partial n}\left[\int x^n\ dx\right]\\
&=\sum_{n=1}^\infty H_{n}^{(k)}\left[\frac{x^{n+1}\ln x}{n+1}-\frac{x^{n+1}}{(n+1)^2}\right]\\
&=\sum_{n=1}^\infty\left[\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}}\right]\\
&=\sum_{n=1}^\infty\left[\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}}\right]\\
&=\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

\begin{align}
\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n}&=\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\
\color{blue}{\mathbf{H}_{1}^{(k)}(x)}&=\int\frac{\operatorname{Li}_k(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx\\
&=\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx.
\end{align}

Repeating the process above yields

\begin{align}
\sum_{n=1}^\infty \frac{H_{n}^{(k)}x^n}{n^2}
&=\int\frac{\operatorname{Li}_{k+1}(x)}{x}\ dx+\int\frac{\operatorname{Li}_k(x)}{x(1-x)}\ dx\\
\color{blue}{\mathbf{H}_{2}^{(k)}(x)}&=\operatorname{Li}_{k+2}(x)+\operatorname{Li}_{k+1}(x)+\int\frac{\operatorname{Li}_k(x)}{1-x}\ dx,
\end{align}

where it is easy to show by using IBP that

\begin{align}
\int\frac{\operatorname{Li}_2(x)}{1-x}\ dx&=-\int\frac{\operatorname{Li}_2(1-x)}{x}\ dx\\
&=2\operatorname{Li}_3(x)-2\operatorname{Li}_2(x)\ln(x)-\operatorname{Li}_2(1-x)\ln x-\ln (1-x)\ln^2x
\end{align}

and


\int\frac{\operatorname{Li}_3(x)}{1-x}\ dx=-\int\frac{\operatorname{Li}_3(1-x)}{x}\ dx=-\frac12\operatorname{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
\small{\begin{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^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{blue}{\mathbf{H}_{2}(x)}\ln^2x+\frac1{15}\color{blue}{\mathbf{H}_{1}(x)}\ln^3x+C.\tag3
\end{align}}

The next step is finding the constant of integration. Setting x=1 to (3) yields
\small{\begin{align}
\color{blue}{\mathbf{H}_{4}(1)}
&=-\frac{\pi^2}{30}\operatorname{Li}_3(1)+\frac65\operatorname{Li}_5(1)-\frac15\operatorname{Li}_4(1)-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(1)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(1)}+C\\
3\zeta(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)-\frac35\zeta(3)+3\zeta(5).
\end{align}}

Thus
\small{\begin{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^3x\ln^2(1-x)+\frac65\operatorname{Li}_5(x)\\&-\frac15\left[\operatorname{Li}_3(x)-\operatorname{Li}_2(x)\ln x-\frac12\ln(1-x)\ln^2x\right]\operatorname{Li}_2(1-x)-\frac15\operatorname{Li}_4(x)\\&-\frac35\operatorname{Li}_4(x)\ln x+\frac15\operatorname{Li}_3(x)\ln x+\frac15\operatorname{Li}_3(x)\ln^2x-\frac1{10}\operatorname{Li}_3(1-x)\ln^2 x\\&-\frac1{15}\operatorname{Li}_2(x)\ln^3x-\frac15\color{blue}{\mathbf{H}_{2}^{(3)}(x)}+\frac15\color{blue}{\mathbf{H}_{2}^{(2)}(x)}
+\frac15\color{blue}{\mathbf{H}_{1}^{(3)}(x)}\ln x\\&-\frac15\color{blue}{\mathbf{H}_{1}^{(2)}(x)}\ln x+\frac25\color{blue}{\mathbf{H}_{3}(x)}\ln x-\frac15\color{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)-\frac35\zeta(3)+3\zeta(5)\tag4
\end{align}}

and setting x=\frac12 to (4) yields
\begin{align}
\color{blue}{\mathbf{H}_{4}\left(\frac12\right)}=&\ \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}_4\left(\frac12\right)\ln2+2\operatorname{Li}_5\left(\frac12\right).\tag5
\end{align}


Finally, we obtain

\begin{align}
\int_0^1\frac{\ln^3(1+x)\ln x}x\ dx=&\ \color{blue}{\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)-\frac25\ln^52+\frac{\pi^2}3\ln^32-\frac{21}4\zeta(3)\ln^22}\\&\color{blue}{-12\operatorname{Li}_4\left(\frac12\right)\ln2-12\operatorname{Li}_5\left(\frac12\right)},
\end{align}

which again matches @Cleo's answer.


References :

[1]\ Harmonic number

[2]\ Polylogarithm

Attribution
Source : Link , Question Author : Oksana Gimmel , Answer Author : Community

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