# Tag Archives: logarithms

## General formula for an integral involving powers of logarithms

$$\int\limits_0^1 \dfrac{\log^{m} (1+x)\log^n x}{x}\; dx = (-1)^{n+1}(n!) (m)! \sum_{\{m-1\}} \sum_{k=1}^\infty \frac{(-1)^k}{k^{n+2}} \prod^{l’}_{j=1}\frac{(-1)^{i_j}}{(i_j)!} \left( \frac{H_{k-1}^{(r_j)}}{r_j}\right)^{i_j}$$ $$\textit{solution}$$ Stirling numbers of the first kind might be useful here, Consider $$m! \sum_{k=m}^\infty (-1)^{k-m} \left[k\atop m\right] \frac{x^k}{k!} = \log^m(1+x)$$ \int\limits_0^1 \dfrac{\log^m (1+x)\log^n x}{x}\; dx = … Continue reading