SUMMARY
The integral $$\int^1_0 \frac{\log^3(1-x)}{x}\, dx$$ evaluates to $$-\frac{\pi^4}{15}$$ through a substitution of $$1-x=t$$, transforming the integral into $$I= \int_{0}^{1} \frac{\ln^{3} t}{1-t}\ dt$$. This result is confirmed by the relationship $$I = - 6\ \sum_{k=1}^{\infty} \frac{1}{k^{4}}$$, which equals $$-\frac{\pi^{4}}{15}$$. Additionally, the discussion introduces a generalization for $$n \geq 2$$, stating that $$\int^1_0 \frac{\log^{n-1}(1-x)}{x}\, dx = (-1)^{n-1} \Gamma(n) \zeta(n)$$, specifically for $$n = 4$$, reinforcing the established result.
PREREQUISITES
- Understanding of definite integrals and logarithmic functions
- Familiarity with the Gamma function and Riemann zeta function
- Knowledge of series convergence and summation techniques
- Experience with substitution methods in integral calculus
NEXT STEPS
- Study the properties of the Gamma function and its applications in integrals
- Explore the Riemann zeta function, particularly $$\zeta(4)$$ and its significance
- Learn advanced techniques in integral calculus, focusing on substitutions
- Investigate the convergence of series, specifically $$\sum_{k=1}^{\infty} \frac{1}{k^{4}}$$
USEFUL FOR
Mathematicians, students of advanced calculus, and researchers interested in integral evaluations and special functions.