I have the following sum
$\sum\_{j=1}^K {K \choose j} (-1)^{j+1}/j$. Now I can write this as the integral $\int\_{-1}^0 \frac{(1+x)^K - 1}{x} dx$. However, I wonder whether there is a closed form expression for that integral? Thanks.



Sum involving binomial coefficients
The closed form of the integral is


$$
\int\_{-1}^{0}\frac{(1+x)^k - 1}{x} = \frac{s(k+1,2)}{k!}
$$


where $s(k+1,2)$ denote the [Stirling number](http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind) of the first kind.


