Question on a form of the Euler constant

1. Aug 21, 2007

themandotcom

I was wondering how I would go about proving this equation:
$$\int_{1}^{\infty}\frac{u-}{u^2}du=1-\gamma$$ where $$\gamma$$ is the Euler Constant, and  is the floor function

Last edited: Aug 22, 2007
2. Aug 22, 2007

dodo

Maybe, turning the left-hand side into
$$\sum_ {i=1} ^ \infty {\int_{i}^{i+1}\frac{u-i}{u^2}du }$$