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I Limits and Laurent series

  1. Nov 9, 2016 #1
    hi, I try to calculate the integral
    $$\int_{0}^{1}log(\Gamma (x))dx$$

    and the last step To solve the problem is:
    $$1 -\frac{\gamma }{2} + \lim_{n\rightarrow \infty } \frac{H_{n}}{2} + n + log(\Gamma (n+1)) - (n+1)(log(n+1))$$
    and wolfram alpha tells me something about series expansion at ##n=\infty## of laurent series
    http://www.wolframalpha.com/input/?...rmassumption={"C",+"limit"}+->+{"Calculator"}

    I know a little about series of laurent, but I do not understand how they serve to solve limits
    and expansion at ##n=\infty##.
     
  2. jcsd
  3. Nov 10, 2016 #2

    mfb

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    Staff: Mentor

    If the Laurent series does not include positive powers of n (as in this problem), then the absolute term is your limit (the negative powers vanish for n->infinity).

    If it has a finite number of positive powers of n, then your expression does not have a limit. If none of those cases apply, it gets complicated.
     
  4. Nov 10, 2016 #3
    I do understand that, but what does it mean "expansion at ##n=\infty##". and how wolfram alpha obtained for complicated expresions like
    $$\lim_{n\rightarrow \infty } \frac{H_{n}}{2} + n + log(\Gamma (n+1)) - (n+1)(log(n+1))$$
     
  5. Nov 10, 2016 #4

    mfb

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    It is the same as an expansion around u=0 for u=1/n.

    How: Sum all terms of the expansion of the summands in the usual way.
     
  6. Nov 13, 2016 #5
    but expancion at u=0 of n=1/u Is not it just 1/u?
    like expancion at z=0 of y= z^2 - 3z +1/z is not just z^2 - 3z +1/z?

    and other question:
    the expancion of 1/z at z=n is
    $$\frac{1}{z}=\frac{1}{u+n}=\frac{1}{n}\, \frac{1}{1+\frac{u}{n}}=\sum_{k=0}^{\infty }\frac{(-1)^{k}\, (z-n)^{k}}{n^{k+1}}$$

    for n tends to infinity Would be
    $$\lim_{n\rightarrow \infty }+\frac{1}{n}-\frac{z-n}{n^{2}}+... = 0$$

    so... what mistake I have?
     
  7. Nov 14, 2016 #6

    mfb

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    The limit of 1/z for z->infinity is 0. Where is the problem?
    If you expand around n, letting the expansion point go to infinity looks odd.
    Those are the correct expansions.
     
  8. Nov 14, 2016 #7
    thanks =D
     
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