Question on a form of the Euler constant

[themandotcom]

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

 

[dodo]

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