TITLE: Hints on calculating the integral $\int_0^1\frac{x^{19}-1}{\ln x}\,dx$
QUESTION [11 upvotes]: I would be happy to get some hints on the following integral:
$$
\int_0^1\frac{x^{19}-1}{\ln x}\,dx
$$

REPLY [3 votes]: You may like this method. Note
$$ \lim_{n\to\infty} n(x^{\frac{1}{n}}-1)=\ln x, \text{ for }x>0 $$
and hence, for $m\in\mathbb{N}$,
\begin{eqnarray}
\int_0^1\frac{x^m-1}{\ln x}dx&=&\int_0^1\sum_{i=0}^{m-1}\frac{x^i(x-1)}{\ln x}dx\\
&=&\int_0^1\sum_{i=0}^{m-1}\lim_{n\to\infty}\frac{x^i(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\
&=&\lim_{n\to\infty}\sum_{i=0}^{m-1}\int_0^1\frac{x^i(x-1)}{n(x^{\frac{1}{n}}-1)}dx\\
&=&\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{m-1}\int_0^1x^i\sum_{k=0}^{n-1}x^{\frac{k}{n}}dx\\
&=&\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{m-1}\int_0^1\sum_{k=0}^{n-1}x^{\frac{k}{n}+i}dx\\
&=&\lim_{n\to\infty}\sum_{i=0}^{m-1}\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\frac{k}{n}+i+1}\\
&=&\sum_{i=0}^{m-1}\int_0^1\frac{1}{x+i+1}dx\\
&=&\sum_{i=0}^{m-1}\ln\frac{i+2}{i+1}\\
&=&\ln(m+1).
\end{eqnarray}
So for $m=19$, we have
$$ \int_0^1\frac{x^{19}-1}{\ln x}dx=\ln20.$$