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Question on a form of the Euler constant

  1. Aug 21, 2007 #1
    I was wondering how I would go about proving this equation:
    [tex]\int_{1}^{\infty}\frac{u-}{u^2}du=1-\gamma[/tex] where [tex]\gamma[/tex] is the Euler Constant, and [tex] [/tex] is the floor function
    Last edited: Aug 22, 2007
  2. jcsd
  3. Aug 22, 2007 #2
    Maybe, turning the left-hand side into
    [tex]\sum_ {i=1} ^ \infty {\int_{i}^{i+1}\frac{u-i}{u^2}du }[/tex]
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