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Re: Re-defining Log over it's branch cut


Thanks guys.  I checked it on my machine (5.2) as well and seems Carl's technique works with 6.0.  Little involved though.  Maybe for now, it's not too bad to code it directly.  For example, when I integrate the Zeta integral  on the lower leg of the contour, I just enter:

NIntegrate[Exp[(s - 1)(Log[r] - Pi I)]/(Exp[r] - 1), {r, 1, Infinity}]

same on the upper leg with z=r Exp[Pi I] and on the route around the branch point, I can use the technique  with ContourNIntegrate that was described in another post.


> Hello.Carl Woll's solution is very smart.
> 
> Limit[z^(s - 1) /. z -> r Exp[-I (Pi -eps)], eps ->
> 0, Assumptions ->
> r > 0]
> E^((-1 + s)*((-I)*Pi + Log[r]))
> 
> But I guess he used Mathematica 6, since in 5.2 I get
> 
> In[17]:=
> Limit[z^(s - 1) /. z -> r*Exp[(-I)*(Pi - eps)], eps
> -> 0, Assumptions -
> > r > 0]
> 
> Out[17]=
> (-r)^(-1 + s)
> 
> So let's see what can I do (until upgrading!).
> 
> In[18]:=
> PowerExpand[Limit[z^(s - 1), z -> r*Exp[e*I]]]
> % /. E^(b_Times) :> ComplexExpand[E^b]
> TrigToExp[% /. e -> Pi]
> TrigToExp[%% /. e -> -Pi]
> 
> Out[18]=
> E^(I*e*(-1 + s))*r^(-1 + s)
> 
> Out[19]=
> r^(-1 + s)*(Cos[e*(-1 + s)] + I*Sin[e*(-1 + s)])
> 
> Out[20]=
> E^(I*Pi*(-1 + s))*r^(-1 + s)
> 
> Out[21]=
> r^(-1 + s)/E^(I*Pi*(-1 + s))
> 
> Everything is how you desired!
> 
> Regards
> Dimitris
> 
>


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