I Limits and Laurent series

  • Thread starter MAGNIBORO
  • Start date
107
26
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/?i=limit&rawformassumption={"F",+"Limit",+"limitfunction"}+->"H_n/2+++n+++ln(n!)+-+(n+1)ln(n+1)"&rawformassumption={"F",+"Limit",+"limitvariable"}+->"n"&rawformassumption={"F",+"Limit",+"limit"}+->"infinity"&rawformassumption={"FVarOpt"}+->+{{"Limit",+"direction"},+{"Limit",+"limitvariable"}}&rawformassumption={"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##.
 
33,001
8,779
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.
 
107
26
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.
I do understand that, but what does it mean "expansion at ##n=\infty##". and how wolfram alpha obtained for complicated expresions like
$$\lim_{n\rightarrow \infty } \frac{H_{n}}{2} + n + log(\Gamma (n+1)) - (n+1)(log(n+1))$$
 
33,001
8,779
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.
 
107
26
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.
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?
 
33,001
8,779
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.
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?
Those are the correct expansions.
 
107
26
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.
thanks =D
 

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