# I Limits and Laurent series

1. Nov 9, 2016

### MAGNIBORO

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. Nov 10, 2016

### 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.

3. Nov 10, 2016

### MAGNIBORO

I do understand that, but what does it mean "expansion at $n=\infty$". and how wolfram alpha obtained for complicated expresions like
$$\lim_{n\rightarrow \infty } \frac{H_{n}}{2} + n + log(\Gamma (n+1)) - (n+1)(log(n+1))$$

4. Nov 10, 2016

### Staff: Mentor

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.

5. Nov 13, 2016

### MAGNIBORO

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?

6. Nov 14, 2016

### Staff: Mentor

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.

7. Nov 14, 2016

thanks =D