camilus
May31-11, 08:25 PM
my book on prime numbers has a line where it skims over a residue computation, and Im in dire need of clarification. It's rather simple, and I may very well be the one mistaken, but Im 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 Im 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}
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 Im 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}