[seqfan] Re: Double zeta function derivative

[Olivier Oloa]

Hi,
It seems to me that there is a confusion between the multivariate function "zeta(s,1)=sum_{n>j>0} 1/(n^s*j)" and the Hurwitz zeta function "zeta(s,1)=sum_{n=0}^{\infty} 1/(n+1)^s=zeta(s)" also called the generalized zeta function.
We have: "zeta'(0)=-1/2*ln(2pi)" not to be confused with "zeta'(0,1)"...

Olivier.


> Message du 01/06/11 00:27
> De : "Maximilian Hasler" 
> A : "Sequence Fanatics Discussion list" 
> Copie à : olivier.oloa at wanadoo.fr
> Objet : Re: [seqfan] Double zeta function derivative
> 
> According to
> 
> http://www.wolframalpha.com/input/?i=zeta(s%2C1)
> the slope of s -> zeta(s,1) in s=0 is
> -log(2*Pi)/2 = -0.9189385332046727417803297364
> 
> Maximilian
> 
> 
> 
> 
> On Tue, May 31, 2011 at 6:41 AM, Richard Mathar
> wrote:
> >
> > There is a numerical evaluation of a derivative of a double zeta function
> > in http://oeis.org/A190643 which I do not understand:
> >
> > By definition
> > zeta(s1,s2) = sum_{j1>j2>0} 1/(j1^s1*j2^s2).
> > summing over positive integers j1 and j2. Taking s2=1 defines
> > zeta(s1,1) = sum_{j1>j2>0) 1/(j1^s1*j2).
> > The derivative with respect to the first variable is
> > (d/ds) zeta(s,1) = -sum_{j1>j2>0) log(j1)/(j1^s1*j2).
> > In the limit of s1->0 we get
> > zeta'(0,1) = -sum_{j1>j2>0) log(j1)/j2.
> > The following questions arise:
> > (i) Why is this value positive as claimed given that j1 and j2 are both
> > positive in the region of summation?
> > (ii) How does the double sum converge given that neither sum_{j1} log(j1)
> > nor sum_{j2}/j2 converge?
> > Is this some sort of renormalization/complex continuation?
> > Did I miss that summation and derivation do not commute somewhere?
> >
> > Richard Mathar
> >
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