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Jan 3, 2012 at 19:18 comment added user18237 Note that this works if $n$ and $\chi$ have the same parity (i.e. $n$ is even if $\chi(-1)=1$ and $n$ is odd if $\chi(-1)=-1$.) Otherwise $L(1-n,\chi)$ vanishes while the corresponding $\Gamma$ factor in the functional equation has a pole.
Jan 3, 2012 at 17:39 history answered Stopple CC BY-SA 3.0