Aysmptotics : Need help with residues

[saint_n]

Hey
Is there a method in calculating the residues.
Getting the poles is easy but i really don't know how my lecturer gets the residues
eg. 1/2(Zeta*Gamma[s/2]
where at s = 1 it is Sqrt[Pi]/2 etc
how does he do it?

 

[Tide]

The residue is just the coefficient of the 1/(z-z0) term in the Taylor series expansion of the function about the pole.

 

[shmoe]

If $$f(s)$$ has a simple pole at s=a, then it's residue at a is $$\lim_{s\rightarrow a}(s-a)f(s)$$

$$\zeta (s)$$ has a simple pole at s=1, actually $$\zeta (s)=\frac{1}{s-1}+O(1)$$ near s=1. $$\Gamma (s/2)$$ is analytic at s=1, so their product has a simple pole. To find their residue, just look at the above limit. You'll also need to know that $$\Gamma(1/2)=\sqrt{\pi}$$, which comes easily from the identity $$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin{\pi s}}$$.

 

[saint_n]

ok I am starting to get it,,but what's happens if its a double pole.Do you work out the residues for both functions and multiply the 2 residues together or does it depend eg GammaZeta[1-s] so the pole would be at 0.
Since the Gamma fn and the Zeta are multiplied together you multiply the residues together??
and if Gamma+Zeta[1-s] then we add the residues together??
Isnt there an expansion of the Gamma and Zeta Functions
because all i know,if you integrate e^(-x)*x^(s-1) with respect to x you get the gamma fn and Summing from 1 to infinity of k^(-s) equal the Zeta function.Thats why i never understood why the Zeta fn has poles at 1 and Gamma fn has poles at 0,-1,-2,-3,...

 

[Galileo]

There are different methods for calculating residues.
Let $z_0$ be a pole of $f$.

One way is to the coefficient of $z^{-1}$ in the Laurent expansion of $f(z)$ at $z_0$, like Tide said. The residue is equal to that coefficient.

Another way is:
$$Res(f,z_0)=\lim_{z\rightarrow z_0}\frac{[(z-z_0)^kf(z)]^{(k-1)}}{(k-1)!}$$
where $k$ is the order of the pole. (the (k-1) up in the numerator means taking the (k-1)th derivative of the numerator).
To find the order, use:
$$\lim_{z\rightarrow z_0}(z-z_0)^mf(z)=\left\{ \begin{array}{ll}b_k & m=k\\0 & m>k \\ \infty & m<k \end{array} \right.$$
Where $b_k$ is the coefficient of $z^{-k}$ in the Laurent expansion of f.
The second method is easy when the order of the pole is low (1 or 2).

 

[shmoe]

Thats why i never understood why the Zeta fn has poles at 1 and Gamma fn has poles at 0,-1,-2,-3,...

$$\zeta(s)=\sum_{n=1}^{\infty}n^{-s}$$
for real part s>1. Since the harmonic series,
$$\sum_{n=1}^{\infty}n^{-1}$$
diverges, zeta must have a pole at s=1. Have you seen any proofs of the analytic continuation of zeta? Anyone of them should make it clear that it has no other poles.


If you define the Gamma function the "integral way", you have
$$\Gamma(s)=\int_{0}^{\infty}e^{-x}x^{s-1}dx$$
valid for all complex s with real part >0, then by analytic continuation via
$$\frac{1}{s}\Gamma(s+1)=\Gamma(s)$$
The integral part shows handily you have no poles in the right half plane (it also diverges if you tried to stick s=0 in). Think about $$\Gamma(0)$$ you're going to try to evaluate
$$\frac{\Gamma(s+1)}{s}$$
at s=0, hence you get a pole, since $$\Gamma(1)=1\neq 0$$. This pole cascades through all the negative integers by the functional equation.