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While I was working on the evaluation of a certain series, the following limit came up:

\begin{align} \lim_{n \to 1} \frac{d(n)-1}{n(n-1)} &= \lim_{n \to 1} \frac{d'(n)}{2n-1} \\ &= d'(1) .\end{align}

Here, I used l'Hôpitals rule, and $d(\cdot)$ denotes the divisor function. In order to compute the derivative of a function, one must know how it is defined on the real numbers. Unfortunately, I have not found any proposals that described the notion of the analytic continuation for the divisor function so far. It seems that it is defined on $\mathbb{Z}$ only.

I did find identities for $\sigma_{\alpha} (x)$ for any $\alpha \in \mathbb{C}$ and $x \in \mathbb{Z}_{\geq 1}$. However, this is not what I am looking for. I seek to find a suitable expression for $\sigma_{\alpha} (x)$ when $\alpha = 0$, and real or complex arguments $x$.

I believe it may be possible to find such an extension of the divisor function, in part because such extensions have been found for the lowercase prime omega function. This arithmetic function is related to the divisor function. The continuation of the function is as follows: $$ \DeclareMathOperator{\sinc}{sinc} \omega(z) = \log_{2} \Bigg{(} \sum_{x=1}^{\lceil Re(z) \rceil} \sinc \Bigg{(} \prod_{y=1}^{\lceil Re(z) \rceil + 1 } (x^{2} + x - yz) \Bigg{)} \Bigg{)} ,$$ where $\sinc(\cdot)$ is the normalized sinc function.

Question: have any proposals been advanced for the analytic continuation of the divisor function, thereby extending the domain to $\mathbb{R}$ or $\mathbb{C}$ ?

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    $\begingroup$ The formula for $\omega(z)$ seems to be more of an approximation than an exact formula as it doesn't seem to exactly match $\omega(z)$ at most integer values of $z$ (with the exceptions of $z=0$ and $z=1$). The errors at $z=2$ and $z=3$ are $-0.000999437$ and $-0.000233918$ respectively. The error does seem to decrease in magnitude as $z$ increases in magnitude. $\endgroup$ May 12, 2021 at 19:35
  • $\begingroup$ The Harmonic number function $H_z=\gamma+\psi^{(0)}(z+1)$ and Gamma function $\Gamma(z)$ might be better examples. $\endgroup$ May 12, 2021 at 19:41
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    $\begingroup$ You mention evaluation of a certain series: which one, and doesn't it give you some idea of the value of $d'(1)$ ? I love heuristic guesses, let me guess (and don't ask me why) $d'(1)=1.56061785204...$ $\endgroup$ May 12, 2021 at 20:46
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    $\begingroup$ Is section 3.1 of the paper at arxiv.org/pdf/1905.09818.pdf close to what you're looking for? $\endgroup$ May 12, 2021 at 21:56
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    $\begingroup$ @მამუკაჯიბლაძე Interesting. I have a simpler analytic formula for $\mu(x)$ valid for $x\in \mathbb{R}$ and I just noticed that my formula seems to evaluate almost exactly the same as the figure in the paper. I'm now wondering if my formula is related or somehow equivalent, so I think I'll ask a question on this topic. $\endgroup$ May 23, 2021 at 22:39

2 Answers 2

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I believe the divisor function $d(n)=\sigma_0(n)$ can be analytically continued at least for $n\in\mathbb{R}$, but I'm not sure about $n\in\mathbb{C}$.


Consider the divisor summatory function defined in formula (1) below.

$$D(x)=\sum\limits_{n=1}^x\sigma_0(n)\tag{1}$$


Now consider the analytic representation of $D_o(x)=\underset{\epsilon\to 0}{\text{lim}}\frac{D(x-\epsilon)+D(x+\epsilon)}{2}$ and it's first order derivative $D_o'(x)$ defined in formulas (2) and (3) below where the evaluation frequency $f$ is assumed to be a positive integer.

$$D_o(x)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \left(\frac{x}{n}-\left(\frac{1}{2}-\frac{1}{\pi}\sum\limits_{k=1}^{f\,n}\frac{\sin\left(\frac{2 \pi k x}{n}\right)}{k}\right)\right)\right),\quad x>0\tag{2}$$

$$D_o'(x)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{1}{n}\left(1+2\sum\limits_{k=1}^{f\,n}\cos\left(\frac{2 \pi k x}{n}\right)\right)\right),\quad x>0\tag{3}$$


Finally consider the function $f'(x)$ defined in formula (4) below which is a subset of the function $D_o'(x)$ defined in formula (3) above.

$$f'(x)=\underset{N,f\to\infty}{\text{lim}}\left(2\sum\limits_{n=1}^N\frac{1}{n}\sum\limits_{k=1}^{f\,n}\cos\left(\frac{2 \pi k x}{n}\right)\right),\quad x>0\tag{4}$$


The function $f'(x)$ defined in formula (4) above evaluates exactly to $2\,f \sigma_0(n)$ when $x=n$ and $n\in\mathbb{Z}\land|n|\le N\land n\ne 0$ which leads to the following analytic formula for $\sigma_0(x)$ where the evaluation frequency $f$ may be chosen to be any positive integer.

$$\sigma_0(x)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{f}\sum\limits_{n=1}^N\frac{1}{n}\sum\limits_{k=1}^{f\,n}\cos\left(\frac{2 \pi k x}{n}\right)\right)\tag{5}$$


The following two figures illustrate formula (5) for $\sigma_0(x)$ above where Figure (1) is evaluated at $f=1$, Figure (2) is evaluated at $f=2$, and both figures are evaluated at $N=5$. The red discrete portions of the figures illustrate the value of $\sigma_0(x)$ at non-zero integer values of $x$. Note formula (5) for $\sigma_0(x)$ evaluates exactly correct when $x=n$ and $n\in\mathbb{Z}\land|n|\le N\land n\ne 0$. Also note that formula (5) evaluates to $N$ at $x=0$, and therefore the evaluation of formula (5) at $x=0$ diverges to $\infty$ as $N\to\infty$ which is consistent with the fact that zero has an infinite number of divisors. I'll also note that when evaluated at $f=2$, formula (5) for $\sigma_0(x)$ evaluates exactly to zero when evaluated at half-integer values of $x$ which is illustrated in Figure (2) below.


Illustration of formula (5) for sigma_0(x) evaluated at N=5 and f=1

Figure (1): Illustration of formula (5) for $\sigma_0(x)$ evaluated at $N=5$ and $f=1$


Illustration of formula (5) for sigma_0(x) evaluated at N=5 and f=2

Figure (2): Illustration of formula (5) for $\sigma_0(x)$ evaluated at $N=5$ and $f=2$


The evaluation limit $N=5$ was used in Figures (1) and (2) above to illustrate that formula (5) for $\sigma_0(x)$ evaluates exactly correct when $x=n$ and $n\in\mathbb{Z}\land|n|\le N\land n\ne 0$. I usually select a value of $N$ much greater than the largest magnitude of $x$ in the evaluation range which I think is generally desirable. Figure (3) below illustrates formula (5) for $\sigma_0(x)$ evaluated at $f=1$ and $N=100$ in the range $0<x<20.5$.


Illustration of formula (5) for sigma_0(x) evaluated at f=1 and N=100

Figure (3): Illustration of formula (5) for $\sigma_0(x)$ evaluated at $f=1$ and $N=100$


The derivative $\sigma_0'(x)$ of formula (5) for $\sigma_0(x)$ above is illustrated in formula (6) below.

$$\sigma_0'(x)=\underset{N\to\infty}{\text{lim}}\left(-\frac{2\pi}{f}\sum\limits_{n=1}^N\frac{1}{n^2}\sum\limits_{k=1}^{f\,n} k \sin\left(\frac{2 \pi k x}{n}\right)\right)\tag{6}$$


Formula (6) for $\sigma_0'(x)$ above seems to be independent of the value of $f$ when evaluated at $x=1$ (see my related Math StackExchange question), so Figure (4) below just illustrates formula (6) above for $\sigma_0'(1)$ evaluated at $f=1$ as a function of $N$. Note as $N$ increases $\sigma_0'(1)$ also increases in an almost linear manner implying $\underset{N\to\infty}{\text{lim}}\sigma_0'(1)=\infty$.


Illustration of formula (6) for sigma_0'(1) evaluated at f=1 as a function of N

Figure (4): Illustration of formula (6) for $\sigma_0'(1)$ evaluated at $f=1$ as a function of $N$


The following table illustrates the trend illustrated in Figure (4) above continues as the magnitude of $N$ increases.


$$\begin{array}{cc} N & \sigma_0'(1) \\ 10 & 6.96764 \\ 100 & 96.6867 \\ 1000 & 996.657 \\ 10000 & 9996.65 \\ \end{array}$$

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  • $\begingroup$ interesting answer, thank you. The value you obtain for $\sigma_{0}'(1) $ does not correspond with the value that I was anticipating though based on numerical calculations. In my calculations based on partial summations of a rational multiple zeta series, it should amount to approximately $0,0002$. Maybe your analytic continuation can or should be normalized somewhere? $\endgroup$
    – Max Muller
    May 25, 2021 at 20:25
  • $\begingroup$ The trigonometric sum is reminiscent of a Fourier series, perhaps one can consider the continuous analogue of the sum-based formulation for Fourier series by formulating it as an integral (see e.g. lectures 5 and 6 of: see.stanford.edu/Course/EE261 ) $\endgroup$
    – Max Muller
    May 25, 2021 at 20:27
  • $\begingroup$ Is it clear that the $N$ limit is defined on non-integers and will be real analytic? I am in great doubt about the complex case as $\cos$ and $\sin$ grow exponentially in the imaginary direction. Also, what is the status of $f$? Does increasing it improve convergence? $\endgroup$ May 26, 2021 at 7:00
  • $\begingroup$ After some experimenting - seems like your limit is infinite on any non-integer, regardless of $f$ $\endgroup$ May 26, 2021 at 7:14
  • $\begingroup$ @მამუკაჯიბლაძე I don't believe the evaluation limit $N$ affects the evaluation at a non-zero integer $x=n$ which is less than or equal in magnitude to $N$ ($0<|x|\le N$), but you're correct that the evaluation limit $N$ affects the evaluation at non-integer values of $x$ as well as the evaluation at $x=0$. This is most noticeable for $|x|<1$ since using formula (5) $\mu(0)\to\infty$ as $N\to\infty$. $\endgroup$ May 26, 2021 at 12:17
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Since the formula I posted in my previous answer only converges as $N\to\infty$ at $x\in\mathbb{Z}_{\ne 0}$, I decided to post another formula which I believe converges as $N\to\infty$ for $x\in\mathbb{C}$. The previous answer I posted is consistent with the fact that zero has an infinite number of divisors, but this answer assumes the definition $\sigma_0(0)=0$.


Consider the following definitions of the analytic extension $\tilde{\sigma}_0(x)$ of $\sigma_0(n)$ and it's first order derivative $\tilde{\sigma}_0'(x)$ which are based on partial evaluations of real analytic formulas for $\tilde{f_{\sigma_0}}'(x)=\sum\limits_n \sigma_0(n)\,\delta(x-n)$ and $\tilde{f_{\sigma_0}}''(x)=\sum\limits_n \sigma_0(n)\,\delta'(x-n)$ (see this answer I posted to a related question on Math StackExchange).


$$\tilde{\sigma}_0(x)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N\left(-\frac{\sin(2 \pi x)}{\pi x}+\frac{1}{n}\sum\limits_{k=1}^n\left(\cos\left(\frac{2 \pi k x}{n}\right)+\cos\left(\frac{2 \pi (k-1) x}{n}\right)\right)\right)\right)\tag{1}$$

$$\tilde{\sigma}_0'(x)=\underset{N\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N\left(\frac{\sin(2 \pi x)-2 \pi x \cos(2 \pi x)}{\pi x^2}-\frac{2 \pi}{n^2}\sum\limits_{k=1}^n\left(k \sin\left(\frac{2 \pi k x}{n}\right)+(k-1) \sin\left(\frac{2 \pi (k-1) x}{n}\right)\right)\right)\right)\tag{2}$$


Figure (1) below illustrates formula (1) for $\tilde{\sigma}_0(x)$ evaluated at $N=100$ in blue where the red discrete evaluation points illustrate $\sigma_0(x)$ for $x=n\in\mathbb{Z}$.


Illustration of formula (1)

Figure (1): Illustration of formula (1) for $\tilde{\sigma}_0(x)$


The following table illustrates formula (2) for $\tilde{\sigma}_0'(x)$ evaluated at $x=1$ seems to converge to approximately $-3.35$ as $N\to\infty$.


$$\begin{array}{cc} N & \tilde{\sigma}_0'(1) \\ 100 & -3.31334 \\ 1000 & -3.34278 \\ 10000 & -3.34574 \\ \end{array}$$

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