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Mathematica knows that:

$$\log (n)=\lim_{s\to 1} \, \left(1-\frac{1}{n^{s-1}}\right) \zeta (s)$$

Kind of tautological starting with logarithms, but I would like to know better why this limit works:

$${\Large \log (n)=\lim_{s\to 1+\frac{2 i \pi k}{\log (n)}} \, \left(1-\frac{1}{n^{s-1}}\right) \zeta (s-i \Im(s))}$$

for $k$ an integer.

Solving it symbolically for some integer $k$ while leaving $n$ as a variable Mathematica says it is equal to zero. But setting $n$ to any value I get $\log (n)$.


Table[Limit[Zeta[s - I*Im[s]]* Total[{1 - 1/(2)^(s - 1)}], 
  s -> (1 + (2*I \[Pi]*k/Log[2]))], {k, 1, 12}]
N[%, 12]


  Limit[Zeta[s - I*Im[s]]*Total[{1 - 1/(n)^(s - 1)}], 
   s -> (1 + (2*I \[Pi]*k/Log[n]))], {k, 1, 6}], {n, 1, 12}]
 N[%, 12]
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Could you post your code for the second problem ? Whatever I do, I always get $0$. – Claude Leibovici Mar 1 '14 at 13:00
I will add some Mathematica code. – Mats Granvik Mar 1 '14 at 13:03
This is really strange, indeed ! – Claude Leibovici Mar 1 '14 at 13:09
up vote 1 down vote accepted

I have no idea what Mathematica is up to, but if we write $s = \sigma + \frac{2i\pi k}{\log n} + it$, we have

$$\begin{align} \frac{1}{n^{s-1}} &= \exp \left((1-s)\log n\right)\\ &= \exp \left(\left(1-\sigma - \frac{2i\pi k}{\log n} - it\right)\log n\right)\\ &= \exp \left((1-\sigma-it)\log n\right)\cdot e^{-2i\pi k}\\ &= e^{(1-\sigma-it)\log n}\\ &= 1 - (\sigma-1+it)\log n + O\left((\sigma-1+it)^2\right). \end{align}$$


$$\zeta(s-i\Im s) = \zeta(\sigma) = \frac{1}{\sigma-1} + O(1),$$

we therefore have

$$\begin{align} \left(1 - \frac{1}{n^{s-1}}\right)\zeta(s-i\Im s) &= \left((\sigma-1+it)\log n + O\left((\sigma-1+it)^2\right)\right)\left(\frac{1}{\sigma-1} + O(1)\right)\\ &= \log n + \frac{it\log n}{\sigma-1} + O\left(\frac{t^2}{\sigma-1}\right) + o(1), \end{align}$$

and the limit does not generally exist. The limit exists and equals $\log n$ if $s$ approaches $1+ \frac{2i\pi k}{\log n}$ in such a way that

$$\lim \frac{t}{\sigma-1} = 0,$$

but along the ray $t = c(\sigma-1)$, the limit is $(1+ic)\log n$, and if $t$ approaches $0$ slower than $\sigma-1$, the expression is unbounded.

Apparently, for the nonexisiting limit, Mathematica just returned $0$ in the general case, but what made it produce the limit $\log n$ when given a specific $n$, I cannot guess.

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