How is the residue computed for a simple pole in the book on prime numbers?

[camilus]

my book on prime numbers has a line where it skims over a residue computation, and I am in dire need of clarification. It's rather simple, and I may very well be the one mistaken, but I am getting a extra factor of 1/2 in the residue whereas in the book it does not appear and apparently isn't a typo either.

We have $$\psi (z) = \sum_{n \in \textbf{N}} e^{-n^2\pi z} = {1 \over 2\pi i}\int_{c-i\infty}^{c+i\infty} {\xi (s) \over s(s-1)}z^{{-1 \over 2}s}ds$$ for c>1 and R(z)>0.

next, we move the line of integration left to R(s)=1/2 passing the simple pole at s=1 with residue 1/sqrt(z):

$$\psi (z) = {1 \over 2\pi i}\int_{1/2-i\infty}^{1/2+i\infty} {\xi (s) \over s(s-1)}z^{{-1 \over 2}s}ds + {1 \over \sqrt{z}}$$

and for some reason I am getting that the last term should be $${1 \over 2\sqrt{z}}$$The extra 1/2 is coming from the fact that $$\xi (1) = 1/2$$ when computing the residue of Res(f,1) when $$f(s) = {\xi (s) \over s(s-1)}z^{{-1 \over 2}s}$$

 

[camilus]

The book just states that passing the simple pole leaves a residue of 1/sqrt(z), just as posted above. Can someone see if I am doing this correctly..?

Basically I am computing the residue for $$f(s) = {\xi (s) \over s(s-1)}z^{{-1 \over 2}s}$$ at s=1 as such:

$$Res(f,1) = \lim_{s \rightarrow 1} (s-1)*{\xi (s) \over s(s-1)}z^{{-1 \over 2}s} = {\xi (1) \over 1}z^{{-1 \over 2}}={1 \over 2\sqrt{z}$$

the following link computes the same residue without the z^-s/2 factor. since there is no problems for letting s->1 in that factor so just multiply the link by z^-1/2...

http://www.wolframalpha.com/input/?i=residue+of+(xi+function)/(z(z-1))+at+z=1