Integral $\int_{1}^{\infty} \frac{\log^3 x}{x(x-1)} dx$ How do I arrive at the closed form expression of the integral $$\displaystyle\int_{1}^{\infty} \dfrac{\log^3 x}{x(x-1)}dx$$
Most probably the closed form is $\dfrac{\pi^4}{15}$
 A: Using a CAS, $$I_1=\displaystyle\int\dfrac{\log^3 x}{x(x-1)}dx=$$ $$6 \text{Li}_4(x)+3 \text{Li}_2(x) \log ^2(x)-6 \text{Li}_3(x) \log (x)-\frac{1}{4}
   \log ^4(x)+\log (1-x) \log ^3(x)$$ $$I_2=\displaystyle\int_{1}^{a} \dfrac{\log^3 x}{x(x-1)}dx=$$ $$6 \text{Li}_4(a)+3 \text{Li}_2(a) \log ^2(a)-6 \text{Li}_3(a) \log (a)-\frac{1}{4}
   \log ^4(a)+\log (1-a) \log ^3(a)-\frac{\pi ^4}{15}$$and, going to limit $$I_3=\displaystyle\int_{1}^{\infty} \dfrac{\log^3 x}{x(x-1)}dx=\frac{\pi ^4}{15}$$
For large values of $a$ $$I_2\simeq \frac{\pi ^4}{15}-\frac{\log ^3(a)+3 \log ^2(a)+6 \log (a)+6}{a}$$ For $a=10^3$, the value of the integral is $5.97352$ while the approximation leads to $5.97372$. For $a=10^4$, the value of the integral is $6.38423$ while the approximation leads to $6.38423$.
A: First, make the substitutions  $x=\frac{1}{u}$ followed by $u=1-z$, and then integrate by parts twice. Substitute $z=1-w$, then integrate by parts again:
$$\begin{align}
\int_{1}^{\infty}\frac{\log^3{x}}{x(1-x)}\mathrm{d}x
&=\int_{1}^{0}\frac{\log^3{\frac{1}{u}}}{\frac{1}{u}(1-\frac{1}{u})}\cdot\frac{(-\mathrm{d}u)}{u^2}\\
&=-\int_{0}^{1}\frac{\log^3{u}}{1-u}\mathrm{d}u\\
&=-\int_{0}^{1}\frac{\log^3{\left(1-z\right)}}{z}\mathrm{d}z\\
&=-\left[\ln{\left(z\right)}\log^3{\left(1-z\right)}\right]_{0}^{1}+\int_{0}^{1}\frac{3\ln{\left(z\right)}\ln^2{\left(1-z\right)}}{z-1}\mathrm{d}z\\
&=-3\int_{0}^{1}\frac{\ln{\left(z\right)}\ln^2{\left(1-z\right)}}{1-z}\mathrm{d}z\\
&=-3\left[\ln^2{\left(1-z\right)}\operatorname{Li}_{2}{\left(1-z\right)}\right]_{0}^{1}+3\int_{0}^{1}\frac{2\log{\left(1-z\right)}\operatorname{Li}_{2}{\left(1-z\right)}}{z-1}\mathrm{d}z\\
&=-6\int_{0}^{1}\frac{\log{\left(1-z\right)}\operatorname{Li}_{2}{\left(1-z\right)}}{1-z}\mathrm{d}z\\
&=-6\int_{0}^{1}\frac{\log{\left(w\right)}\operatorname{Li}_{2}{\left(w\right)}}{w}\mathrm{d}w\\
&=-6\left[\log{\left(w\right)}\operatorname{Li}_{3}{\left(w\right)}\right]_{0}^{1}+6\int_{0}^{1}\frac{\operatorname{Li}_{3}{\left(w\right)}}{w}\mathrm{d}w\\
&=6\int_{0}^{1}\frac{\operatorname{Li}_{3}{\left(w\right)}}{w}\mathrm{d}w\\
&=6\operatorname{Li}_{4}{\left(1\right)}\\
&=\frac{\pi^4}{15}.
\end{align}$$
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$\ds{\int_{1}^{\infty}{\ln^{3}\pars{x} \over x\pars{x - 1}}\,\dd x:\ {\large ?}}$.

\begin{align}&\color{#66f}{\Large\int_{1}^{\infty}%
{\ln^{3}\pars{x} \over x\pars{x - 1}}\,\dd x}
=\int_{1}^{0}{\ln^{3}\pars{1/x} \over \pars{1/x - 1}/x}\,
\pars{-\,{\dd x \over x^{2}}}
=-\int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x
\\[3mm]&=-\int_{0}^{1}\ln\pars{1 - x}\bracks{3\ln^{2}\pars{x}\,{1 \over x}}\,\dd x
=3\int_{0}^{1}{\rm Li}_{2}'\pars{x}\ln^{2}\pars{x}\,\dd x
\\[3mm]&=-3\int_{0}^{1}{\rm Li}_{2}\pars{x}\bracks{2\ln\pars{x}\,{1 \over x}}
\,\dd x
=-6\int_{0}^{1}{\rm Li}_{3}'\pars{x}\ln\pars{x}\,\dd x
=6\int_{0}^{1}{\rm Li}_{3}\pars{x}\,{1 \over x}\,\dd x
\\[3mm]&=6\int_{0}^{1}{\rm Li}_{4}'\pars{x}\,\dd x
=6\,{\rm Li}_{4}\pars{1}=6\
\underbrace{\zeta\pars{4}}_{\ds{=\ \color{#c00000}{\pi^{4} \over 90}}}\
=\ 6\,{\pi^{4} \over 90}=\color{#66f}{\Large{\pi^{4} \over 15}} \approx {\tt 6.4939}
\end{align}

A: The change of variables $x\leftarrow 1/x~$ shows that
$$\eqalign{
I&=\int_1^\infty\frac{\log^3x}{x(x-1)}dx\cr
&=\int_0^1\frac{-\log^3x}{1-x}dx
=-\sum_{n=0}^\infty \int_{0}^1x^n \log^3 x\,dx\cr
&=\sum_{n=0}^\infty \frac{6}{(n+1)^4}=6\zeta(4)=\frac{\pi^4}{15}
}
$$
which is the desired answer.
Edit. Indeed, generally, the change of variables $x=e^{-t}$ shows that
$$\eqalign{
\int_0^1x^n\log^p(1/x)\,dx&=\int_0^\infty e^{-(n+1)t}t^pdt\cr
&=\frac{1}{(n+1)^{p+1}}\int_0^\infty e^{-u}u^pdu\cr
&=
\frac{p!}{(n+1)^{p+1}}}
$$
