Let $s$ be a complex number.

$$T(s)=\sum_{n>0} (n^s + n^{-s})^{-1} $$

This is well defined for $Re(s) > 1$.

It seems $T(s) = T(-s)$ but then again we have it only defined for $Re(s)>1$ for now.

We try analytic continuation by

$$\sum_{n>0} (n^s + n^{-s})^{-1} - n^{-s} = - \sum_{n>0} (n^{3s} + n^{s})^{-1}$$

Then assuming $sum_{n>0} n^{-s}$ has interpretation as $\zeta(s)$ then :

$T(s)=\sum_{n>0} (n^{3s} + n^s)^{-1} + \zeta(s)$

This now is suppose to be analytic for $Re(s) >1/3$.

We still assume $T(s) = T(-s)$ here.

We can continue this proces by noticing

$$g(x) = 1/(x + 1/x) = x/(1+x^2) = x - x^3 + x^5 - x^7 + ... = 1/x - 1/x^3 + 1/x^5 - 1/x^7 + ...$$

and $g(1/x) = g(x)$.


$$g(x) - 1/x = - 1/(x^3 + x)$$

$$g(x) - 1/x + 1/x^3 = 1/(x^5 + x^3)$$


$$g(x) - x = - 1/(x^{-3} + x)$$

$$g(x) - x + x^3 = 1/(x^{-5} + x^{-3})$$


So we have error terms too.

letting $x = n^{-s}$ we can use this to get the idea for a zeta series expansion :

$$T(s) = \zeta(s) - \zeta(3s) + \zeta(5s) - \zeta(7s) + ...$$

And the error term is suppose to go to zero.

So we consider the infinite sum and use $T(s) = T(-s)$ :

$$T(s) = \zeta(s) - \zeta(3s) + \zeta(5s) - \zeta(7s) + ... = \ T(-s) = \zeta(-s) - \zeta(-3s) + \zeta(-5s) - \zeta(-7s) + ...$$

by symmetry.


$$E(s)=\sum_{n>0} (n^{ks} + n^{(k-2)s})^{-1}$$

seems to be well defined and meromorphic for $Re(s) > 1/k$.

We get the feeling since it appears by the above logic that $T(s)$ is defined for $Re(s) > 0$ ; the equation $T(s) = T(-s)$ seems justified.

But all of these steps are possibly dubious.

changing order of summation , ignoring radius of taylors ($g(x)$ is expanded in a taylor without mentioning the radius !? ) etc

Then again it is very similar to what we do with the Riemann zeta function.

Changing the taylor expansion points slowly towards values s with small or negative real part might be more formal and better, but very complicated.

I prefer $T(s) = \zeta(s) - \zeta(3s) + ...$ truncated at stopping at substraction.

But maybe that zeta expansion is not valid or not so good.

What is remarkable is that this $T(s)$ is close to the zeta function for $Re(s) >>1$.

Computing zero's, poles, branches, analytic continuation etc is the main goal.

But things are tricky; changing order of summation , fubini like ideas and such might not be justified.

Plots are appreciated too :)

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    $\begingroup$ You might be interested in my question at math.stackexchange.com/q/4470369 which seems to be at least loosely related. $\endgroup$ Jan 20 at 22:57
  • $\begingroup$ @StevenClark thank you ! Very interesting and inspirational. I upvoted and favo your post and comment. Hope you get answers. $\endgroup$
    – mick
    Jan 20 at 23:01
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    $\begingroup$ your arguments clearly show that $T$ has an analytic continuation to $\Re s >0$ outside $1,1/3,1/5..$ where it has simple poles; then I suspect (and I do not think it's that hard to prove though I may be wrong) that $\Re s=0$ is a natural boundary, so $T$ cannot be continued at any point there $\endgroup$
    – Conrad
    Jan 20 at 23:06
  • $\begingroup$ @StevenClark I got my function from my mentor. He mentioned that sometimes considering a power of your dirichlet series has the same zero's could help. You might also be interested in this generalization : math.stackexchange.com/questions/3805558/… $\endgroup$
    – mick
    Jan 20 at 23:07
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    $\begingroup$ No I am talking about $ia$ (eg $2ia\log n =i\pi+2k \pi$) $\endgroup$
    – Conrad
    Jan 21 at 23:59

1 Answer 1


It is useful to note that $$T(s)=\sum_{n=1}^\infty \frac{1}{n^s + n^{-s}} = \sum_{n=1}^\infty \frac{n^s}{n^{2s} + 1}=\sum_{n=1}^\infty \sum_{k=0}^\infty (-1)^k n^{(2k+1)s}= \sum_{k=0}^\infty (-1)^k \zeta(-(2k+1)s)$$ However, we could have also used the fact that $\frac{1}{1+x} = -\sum_{n=1}^\infty (-1)^{-n} x^{-n}$ in which case we would have obtained $$T(x)= - \sum_{k=1}^\infty (-1)^k \zeta\left(-(-2k+1)x\right)$$

However, the previous steps have involved replacing a divergent series with its analytical continuation, and this is not generally a valid step on its own. Instead, when we do this, we may have to pick up the new residues introduced by the inner function (in this case, the $\zeta$ function). And, in this case, there is a residue we need to pick up.

Before I go into that, we should first briefly look at the relationship between $\sum_{k=0}^\infty (-1)^k \zeta(-(2k+1)x)$ and $T(s)$ when $\mathfrak{Re}(x)<-1$. Notice that as $\lim_{x \to \infty} \zeta(x) = 1$, so the series doesn't converge. This is not a problem, since this type of divergence is very tame-- we have lots of options here, such as Cesaro summation, or applying a smooth cutoff function. However, since we 'morally' know that $1-1+1-1+1+\dots = \frac{1}{2}$, then the easiest computational option is to regularize the sum to $$\sum_{k=0}^\infty (-1)^k \zeta(-(2k+1)x) \to \lim_{N \to \infty} \left(\frac{1}{2} \pm \sum_{k=0}^N (-1)^k \zeta(-(2k+1)x) \right)$$ where the plus or minus is determined by whether $N$ is even or odd. One can numerically check that for $\mathfrak{Re}(x)<-1$ this series agrees with $T(s)$. An analogous argument can be made for $- \sum_{k=1}^\infty (-1)^k \zeta(-(-2k+1)x)$ where $\mathfrak{Re}(x)>1$. The series both define analytical continuations up to $\mathfrak{Re}(x)=0$ (and beyond).

To understand why this function has cannot be extended to a single compelx function, let us look in the complex plane. The function for us to analyze is $\zeta(-(2z+1)x) \frac{\csc(\pi z)}{2i}$. At the integers, this function has the residue $(-1)^k \zeta(-(2k+1)x)$. If these residues were all of the residue of the function, we would have a function that could be analytically continued to the whole plane. However, $\zeta(-(2z+1)x)$ contributes its own residue because $\zeta(1) = \infty$ and moreover the position of this residue depends on $x$. Indeed, the residue is located at $-(2z+1)x =1 \implies z = -\frac{x+1}{2x}$ and has value $\frac{\pi \csc\left(\pi \frac{1+x}{2x}\right)}{2x}$.

Now, observe the following contour integral $$\int_{c - i \infty}^{c + i \infty} \zeta(-(2z+1)x)\frac{\csc(\pi z)}{2i}dz $$ Here is a graph of what the contour integral looks like when $\mathfrak{Re}(x)<0$ enter image description here Notice that when $\mathfrak{Re}(s)<0$ the $\zeta$ residue is on the LHS of the contour, so we obtain that $$\int_{c - i \infty}^{c + i \infty} \zeta(-(2z+1)x)\frac{\csc(\pi z)}{2i} = \sum_{k=0}^\infty (-1)^k \zeta(-(2k+1)x) = T(x)$$ However, and this is crucial, we can switch the orientation direction of the contour (i.e. go from $-\varepsilon + i \infty$ down towards $-\varepsilon - i \infty$ instead of traveling upwards) and now the integral is also equal to $$T(x) = \int_{c - i \infty}^{c + i \infty} \zeta(-(2z+1)x)\frac{\csc(\pi z)}{2i} = -\sum_{k=1}^\infty (-1)^k \zeta(-(-2k+1)x) + \mathbf{\frac{\boldsymbol{\pi} \boldsymbol{\csc}\left(\boldsymbol{\pi} \frac{1+x}{2x}\right)}{2x}}$$ Now, notice what happens when we switch the sign of $x$ so that $\mathfrak{Re}(x) > 0$. Then we obtain that $$\int_{c - i \infty}^{c + i \infty} \zeta(-(2z+1)x)\frac{\csc(\pi z)}{2i} = -\sum_{k=1}^\infty (-1)^k \zeta(-(-2k+1)x) = T(x)$$ and as a consequence $$T(x) = \int_{c - i \infty}^{c + i \infty} \zeta(-(2z+1)x)\frac{\csc(\pi z)}{2i} = \sum_{k=0}^\infty (-1)^k \zeta(-(2k+1)x) + \mathbf{\frac{\boldsymbol{\pi} \boldsymbol{\csc}\left(\boldsymbol{\pi} \frac{1+x}{2x}\right)}{2x}} $$

Now, altogether, this tells us that $T(x)$ has a branch cut on the line $\Re(s) = 0$ since we have that $$T(x) = \sum_{k=0}^\infty (-1)^k \zeta(-(2k+1)x) + \begin{cases} \frac{\pi \csc\left(\pi \frac{1+x}{2x}\right)}{2x} & \mathfrak{Re}(x)<0 \\ 0 & \mathfrak{Re}(x)>0 \\ \end{cases}$$ And so $T(x)$ is part of two different branches of the same Reimann surface.

Now, let us analyze more closely the behavior of the contour needed to make our sum converge. $\zeta(-z)$ becomes large when $\Re{z}>0$. Thus $\zeta(-e^{i \theta}z)$ is large between angles $-\frac{\pi}{2}-\theta<\alpha<\frac{\pi}{2} - \theta$. Therefore, our contour should be integrated on the following path $$\int_{-\infty e^{(-\frac{\pi}{2}-\theta)i}}^{\infty e^{(\frac{\pi}{2}-\theta)i}} f(z)dz$$ However, this contour breaks down when $\theta = \pm \frac{\pi}{2}$, i.e. exactly when $\theta$ is purely imaginary. In that case, we need to integrate along the real line, but this is where all the poles are, and we can't integrate over poles (the integration doesn't even converge in the principal value sense, since the series doesn't converge). Therefore, we are forced to analyze the sum. When $s$ is purely imaginary then $$\sum \frac{1}{n^s + n^{-s}} = \sum \frac{1}{e^{\ln(n) i t} + e^{-\ln(n) i t}} = \sum \frac{1}{\cos(\ln(n)t)}$$ Which suggests there are poles at $t = \frac{\pi + 2\pi k}{2 \ln(n)}, n,k \in \mathbb{N}$. Though, at the moment, I can't think of a good way to show there are actually poles there.


Your function obviously shares some resemblance with the usual zeta function. Therefore, we might wish to find the number-theoretic significance of the function. One way we could go about this is to try to represent it as a Dirichlet series. Let us simplify a different way at the point where our series is of this form. $$T(s)=\sum_{n=1}^\infty \sum_{k=0}^\infty (-1)^k n^{(2k+1)s}$$ We turn this into a Dirichlet series in the following way $$T(-s) = \sum_{m=1}^\infty \chi(m) m^{-s} \quad \chi(m) = \sum_{m = n^{2k+1}\ \ \ n,k} (-1)^k $$ What we are doing is collecting common powers of $m^s$ by finding all the ways to write $m^s$ as $n^{2k+1}$ for some $n$ and $k$. For instance, $\chi(8) = 0$, since $8^{2(0)+1} = 8$ but also $2^{2(1)+1} = 8$. Similarly, $\chi(32)=2$ since $32 = (2^{1})^5 = (2^{5})^1$. The first time $\chi(m) = 3$ is at $2^{25}$ since $25 = 1\cdot 25, 5 \cdot 5, 25 \cdot 1$, and all factors here $\equiv 1 \mod 4$. I'm not all that knowledgeable in number theory myself, but perhaps this gives you a starting place to look if you're interested in analyzing the number theoretic properties of your function.

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    $\begingroup$ I am not sure about your claim that changing to $(-1)^n$ changes much since imho the singularities on $\Re s =0$ come from the roots of $n^{2s}+1=0$ and those don't change if you just put a minus sign as above on $n$ odd; this being said I didn't prove that rigouroously - since it is definitely possible that if one is close to such a root $ia$ for some fixed $m$, there is interference from $n^{2a}+1$ for other $n$, but again I do not see any reason for that interference to depend on the parity of $n$ $\endgroup$
    – Conrad
    Jan 22 at 17:24
  • $\begingroup$ Thank you ! Very instructive. $\endgroup$
    – mick
    Jan 28 at 23:29
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    $\begingroup$ @mick I noticed that you have accepted this as an answer. Just so you know, the two graphs at the end of the answer are incorrect. I had lazily just assumed that there wouldn't be much problem in just choosing the same contour everywhere, but Conrad's comment made me realize there is some non-obvious behaviour as a result of being forced to change the angle of the contour when $s$ is complex. I plan to update the answer later with the correct arguments at some point, but I have been busy for the past few days. In particular, there actually are singularities at $n^{2s}+1=0$ as Conrad predicts. $\endgroup$ Jan 29 at 2:06
  • $\begingroup$ @CalebBriggs thank you. Im looking forward. I hesitated not to accept because it was mainly solved and rude not too. So eventually I did. Nice to see you will put in more effort ! $\endgroup$
    – mick
    Jan 29 at 15:20

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