I think it should be -i*pi, and so does Mathematica.
Commercial Macsyma thinks i*pi.
Numerical evaluation as x->0 from real positive values confirms
-i*pi is consistent with Maxima's numerics.
RJF
Raymond Toy wrote:
>>>>>>"gosei" == go furuya writes:
>>>>>>
>>>>>>
>
> gosei> Hi Raimond
> >> I've looked a bit more at this. A simpler example is
> >>
> >> limit(log(-1-%i*x),x,0,plus);
> >>
> >> Maxima returns log(-1) = %i*%pi. However, the correct answer is
> >> -%i*%pi because we're always in the third quadrant.
>
> gosei> You may confused calculating limit in 2D with 1D.
> gosei> In case of uniformization,both %i*%pi and -%i*%pi shoud be correct.
> gosei> All that is required is to select domain of uniformization
> gosei> ,such that principal value.
>
>What does uniformization mean? And why are they both correct? What
>do you mean limit in 2D vs 1D?
>
>Perhaps I'm wrong in assuming log means principal log so the branch
>cut is the negative real axis.
>
>Anyway, my reasoning is this:
>
>log(-1-%i*x) = 1/2*log(x^2+1) + %i*arg(-1-%i*x)
>
>As x -> 0 from above, the realpart has limit 0. For the imaginary part
>arg(-1-%i*x) is -pi + eps for small x > 0, so the limit is -pi.
>
>Thus the correct answer should be -%i*%pi.
>
>Ray
>
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