Is there a closed form of the integral with $s>1$?
$$ \int_{0}^{1} \frac{x^s -1}{\ln x} dx$$
I used the expansion $$x^s = 1 + \sum_{n=1}^{\infty}s^{n}\ln^{n}(x)/n!.$$
Is there a closed form of the integral with $s>1$?
$$ \int_{0}^{1} \frac{x^s -1}{\ln x} dx$$
I used the expansion $$x^s = 1 + \sum_{n=1}^{\infty}s^{n}\ln^{n}(x)/n!.$$