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 $\dfrac{\pi^4}{15}$

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 $6.38423$.
