An entire function interpolating $\mu(n)$

This answer is related to analytic formulas for $a(n)$ where $f_a(x)$ and $F_a(s)$ defined in formulas (1) and (2) below are the summatory function and Dirichlet series associated with $a(n)$.

$$f_a(x)=\sum\limits_{n=1}^x a(n)\tag{1}$$

$$F_a(s)=s\int\limits_0^\infty f_a(x)\,x^{-s-1}\,dx=\sum\limits_{n=1}^\infty a(n)\ n^{-s}\tag{2}$$

In the remainder of this answer $\tilde{a}(x)$ is used to refer to an analytic representation of the arithmetic function $a(n)$. The analytic representation $\tilde{a}(x)$ typically converges to $a(n)$ when $|x|=n\in\mathbb{Z}_{\ne 0}$, but $\tilde{a}(0)$ may or may not converge to $a(0)$ when $a(0)$ is defined and meaningful.

The following two analytic formulas for $\tilde{a}(x)$ are based on the answer above and An Exact Formula for the Prime Counting Function which was referenced in the answer above. Formula (4) below is indeterminate at $x=0$ (because $0^0$ is indeterminate) but it converges at $x=0$ in a limit sense. I'll note that formulas (3) and (4) are both somewhat slow to converge and somewhat sensitive to evaluation precision.

$\tilde{a}(x)=\underset{K\to\infty}{\text{lim}}\left(-2\sum\limits_{k=2}^K x^k \sum\limits_{m=0}^{\frac{k}{2}-1} \frac{(2 i \pi )^{k-2 m-2}\,F_a(2 m+2)}{(k-2 m-1)!}\right)\tag{3}$

$\tilde{a}(x)=\underset{I\to\infty}{\text{lim}}\left(-2\sum\limits_{i=0}^I (-1)^i (2 \pi n)^{2 i} \sum\limits_{j=1}^i \frac{(-1)^j (2 \pi)^{-2 j} F_a(2 j)}{(2 i-2 j+1)!}\right)\tag{4}$

For $x\in\mathbb{R}$ formula (4) above seems to evaluate very similar to the real part of formula (3) above for every definition of $a(n)$ which I've tested.

For some definitions of $a(n)$ and under certain conditions, the analytic formula for $\tilde{a}(n)$ defined in formula (5) below seems to evaluate very similar to formula (3) above. The analytic function $\tilde{a}(x)$ defined in formula (5) evaluates exactly to $a(n)$ when $x=n\in\mathbb{Z}\land0<|n|\le N$. When formula (3) above converges for a particular definition of $a(n)$, there are at least two conditions related to the evaluation limit $N$ in formula (5) below that are necessary, but perhaps not sufficient, to achieve this similarity which are discussed following formula (5) below.

$$\tilde{a}(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{b(n)}{n}\sum\limits_{k=1}^n e^{\frac{2 \pi i k x}{n}}\right)\quad\text{where}\quad b(n)=\sum\limits_{d|n}a(d)\,\mu\left(\frac{n}{d}\right)\tag{5}$$

Condition (1): The first condition necessary for formula (5) to evaluate similar to formula (3) is the limit defined in formula (6) below must converge, and more specifically I believe it must converge to zero. If this limit diverges, formula (4) will still evaluate exactly to $a(n)$ when $x=n\in\mathbb{Z}\land0<|n|\le N$, but will diverge at non-integer values of $x$.

$$\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{b(n)}{n}\right)=0\tag{6}$$

Condition (2): The second condition necessary for formula (5) to evaluate similar to formula (3) is it must be possible to evaluate $\tilde{a}(0)$ defined in formula (7) below to a particular value for arbitrarily large magnitudes of the evaluation limit $N$. This implies $\tilde{a}(0)$ defined in formula (7) below can be evaluated to this specific value for an infinite number of values of $N$.

$$\tilde{a}(0)=\sum\limits_{n=1}^N b(n)\tag{7}$$

In all of the figures below, the red discrete portion of the plots represents the evaluation of the arithmetic function $a(n)$ at integer values. Formula (3) is illustrated in blue and formula (5) is illustrated in orange.

In the simplest case where $a(n)=\delta_{n-1}$ (Kronecker delta function), $f_a(x)=\theta(x-1)$, $F_a(s)=1$, and $b(n)=\mu(n)$. Note conditions (1) and (2) specified above are both met since $\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\mu(n)}{n}\right)=0$ and $\tilde{a}(0)=\sum\limits_{n=1}^N\mu(n)$ is the Mertens function $M(N)$ which evaluates to every integer an infinite number of times.

In the case of $a(n)=\delta_{n-1}$, formula (8) below is equivalent to formula (3) above where $_1\tilde{F}_2()$ is the Hypergeometric PFQ Regularized function.

$\tilde{a}(x)=\underset{K\to\infty}{\text{lim}}\left(-2\sum\limits_{k=2}^K -\frac{1}{2} i^k \pi ^{k+\frac{1}{2}} \, _1\tilde{F}_2\left(1;\frac{k+2}{2},\frac{k+3}{2};-\pi ^2\right) x^k\right)\tag{8}$

Figures (1) and (2) below illustrate the real and imaginary parts of formulas (3) and (5) for $\tilde{a}(x)$ (blue and orange) where $a(n)=\delta_{n-1}$. Formula (3) is evaluated at $K=100$, and formula (5) is evaluated at $N=101$ which was selected to achieve the correct evaluation at $x=0$.

Figure (1): Illustration of real part of formulas (3) and (5) for $\tilde{a}(x)$ (blue and orange) where $a(n)=\delta_{n-1}$

Figure (2): Illustration of imaginary part of formulas (3) and (5) for $\tilde{a}(x)$ (blue and orange) where $a(n)=\delta_{n-1}$

In the case where $a(n)=\mu(n)$ (Moebius function), $f_a(x)=M(x)$ which is the Mertens function, $F_a(s)=\frac{1}{\zeta(s)}$, and $b(n)=\sum\limits_{d|n}\mu(d)\,\mu\left(\frac{n}{d}\right)$. Note condition (1) specified above is met since $\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{b(n)}{n}\right)=0$ (see this answer to my related question on Math StackExchange). The second necessary condition for equivalence of formula (5) to formulas (3) in the case of $a(n)=\mu(n)$ is it must be possible to evaluate formula (7) above for $\tilde{a}(0)$ to $0$ for arbitrarily large magnitudes of the limit $N$. I believe this condition is met but this is still an open issue (see my related question on Math StackExchange).

Figures (3) and (4) below illustrate the real and imaginary parts of formulas (3) and (5) for $\tilde{a}(x)$ (blue and orange) where $a(n)=\mu(n)$. Formula (3) is evaluated at $K=100$, and formula (5) is evaluated at $N=140$ which was selected to achieve the correct evaluation at $x=0$.

Figure (3): Illustration of real part of formulas (4) and (5) for $\tilde{a}(x)$ (blue and orange) where $a(n)=\mu(n)$

Figure (4): Illustration of imaginary part of formulas (4) and (5) for $\tilde{a}(x)$ (blue and orange) where $a(n)=\mu(n)$

The analytic representation for $\tilde{a}(x)$ defined in formula (5) above is based on the first-order derivative $\tilde{f}_a'(x)$ of an analytic representation of $\tilde{f}_a(x)=\underset{\epsilon\to 0}{\text{lim}}\frac{f_a(x-\epsilon)+f_a(x+\epsilon)}{2}$ where $\cos(\frac{2 \pi k x}{n})$ was replaced with $e^{\frac{2 \pi i k x}{n}}$ in formula (5) above. This answer I posted to my own related question on MathOverflow illustrates a much deeper connection between the analytic representation for $\tilde{f}_a'(x)$ and the analytic representation for $\tilde{a}(x)$ defined in formula (4) above.

This connection led to the following analytic representations of $f_a(x)=\sum\limits_n a(n)\,\theta(x-n)$ and it's first order derivative $f_a'(x)=\sum\limits_n a(n)\,\delta(x-n)$ where $F_a(s)=\sum\limits_n a(n)\ n^{-s}$ and the evaluation frequency $f$ is assumed to be a positive integer. The MathOverflow answer I linked to above illustrates formula (9) for $\tilde{f}_a(x)$ below for a variety of arithmetic functions.

$$\tilde{f}_a(x)=\underset{\substack{K,f\to\infty \\ K\gg f\,x}}{\text{lim}}\left(-4 f \sum\limits_{k=1}^K \frac{(-1)^k\ x^{2 k+1}}{2 k+1} \sum\limits_{j=1}^k \frac{(-1)^j\ (2 \pi f)^{2(k-j)}\ F_a(2 j)}{(2 k-2 j+1)!}\right)\tag{9}$$

$$\tilde{f}_a'(x)=\underset{\substack{K,f\to\infty \\ K\gg f\,x}}{\text{lim}}\left(-4 f\sum\limits_{k=1}^K (-1)^k\ x^{2 k}\sum\limits_{j=1}^k \frac{(-1)^j\ (2 \pi f)^{2(k-j)}\ F_a(2 j)}{(2 k-2 j+1)!}\right)\tag{10}$$

I believe formulas (11) and (12) below may perhaps be equivalent to formulas (9) and (10) above.

$$\tilde{f}_a(x)=\underset{\substack{K,f\to\infty \\ K\gg f\,x}}{\text{lim}}\Re\left(-4 f \sum\limits_{k=2}^K \frac{x^{k+1}}{k+1} \sum\limits_{m=0}^{\frac{k}{2}-1} \frac{(2 \pi f i)^{k-2 m-2}\ F_a(2 m+2)}{(k-2 m-1)!}\right)\tag{11}$$

$$\tilde{f}_a'(x)=\underset{\substack{K,f\to\infty \\ K\gg f\,x}}{\text{lim}}\Re\left(-4 f \sum\limits_{k=2}^K x^k \sum\limits_{m=0}^{\frac{k}{2}-1} \frac{(2 \pi f i)^{k-2 m-2}\ F_a(2 m+2)}{(k-2 m-1)!}\right)\tag{12}$$

Consider the conjectured analytic formulas defined in (13) and (14) below for $\tilde{M}(x)=\sum\limits_n\mu(n)\,\theta(x-n)=\underset{\epsilon\to 0}{\text{lim}}\frac{M(x-\epsilon)+M(x+\epsilon)}{2}$ where $M(x)=\sum\limits_{n\le x}\mu(n)$ is the Mertens function and $b(n)$ in formula (14) below is the Moebius transform of $\mu(n)$ (see see OEIS entry A007427).

$$\tilde{M}(x)=\underset{\substack{K,f\to\infty \\ K\gg f\,x}}{\text{lim}}\left(-4 f \sum\limits_{k=1}^K \frac{(-1)^k\ x^{2 k+1}}{2 k+1} \sum\limits_{j=1}^k \frac{(-1)^j\ (2 \pi f)^{2(k-j)}}{(2 k-2 j+1)!\ \zeta(2 j)}\right),\quad x\ge 0\tag{13}$$

$$\tilde{M}(x)=\underset{\substack{N,f\to\infty \\ \sum\limits_{n=1}^N b(n)=0}}{\text{lim}}\left(\sum\limits_{n=1}^N b(n) \left(\frac{x}{n}+\frac{1}{\pi }\sum\limits_{k=1}^{f\,n} \frac{\sin\left(\frac{2 \pi k x}{n}\right)}{k}\right)\right),\quad x\ge 0\tag{14}$$

In formula (14) above, the evaluation limit $N$ must be selected such that $\sum\limits_{n=1}^N b(n)=0$ where $b(n)=\sum\limits_{d|n}\mu(d)\,\mu\left(\frac{n}{d}\right)$, and the inner sum over $k$ must be evaluated with the upper limit $f\,n$ in order to satisfy a conditional convergence requirement when formula (14) is evaluated at finite integer values of $f$. The $\frac{x}{n}$ term was included in formula (14) above to accelerate convergence, but this term can be dropped when $N\gg x$ since $\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{b(n)}{n}\right)=0$.

Figures (5) to (8) below illustrate formulas (13) and (14) (orange and green) for $\tilde{M}(x)$ (blue) evaluated at $f=1$, $f=2$, $f=3$, and $f=4$. All four figures below were evaluated at the limits $K=200$ for formula (13) above and $N=140$ for formula (14) above. Each time the evaluation frequency $f$ increases, the convergence range of formula (13) with respect to $x$ decreases, and therefore the evaluation in each figure below was terminated slightly after formula (13) started to diverge. Note in all four figures below, the evaluations of formulas (13) and (14) were nearly identical up to the point where formula (13) starts to diverge, and to me this seems to suggest an equivalence between formulas (13) and (14) above.

Figure (5): Illustration of formulas (13) and (14) (orange and green) for $\tilde{M}(x)$ (blue) evaluated at $f=1$

Figure (6): Illustration of formulas (13) and (14) (orange and green) for $\tilde{M}(x)$ (blue) evaluated at $f=2$

Figure (7): Illustration of formulas (13) and (14) (orange and green) for $\tilde{M}(x)$ (blue) evaluated at $f=3$

Figure (8): Illustration of formulas (13) and (14) (orange and green) for $\tilde{M}(x)$ (blue) evaluated at $f=4$

For the cases $a(n)=1$, $a(n)=(-1)^{n-1}$ and $a(n)=\delta_{n-1}$ where $F_a(s)=\zeta(s)$, $F_a(s)=\eta(s)$, and $F_a(s)=1$, I've determined formulas (9) and (10) above are simply the power series for the functions defined in formulas (15) to (20) below. This proves the validity of formulas (9) and (10) above for these particular cases, but I believe formulas (9) and (10) are more generally applicable to any definition of $a(n)$ for which the Dirichlet series $F_a(s)=\sum\limits_n\frac{a(n)}{n^s}$ converges for $\Re(s)\ge 2$. All formulas below are for $x\ge 0$, but $\tilde{f}_a'(x)$ and $\tilde{f}_a(x)$ are actually even and odd functions respectively.

$\quad a(n)=1 \text{ where } F_a(s)=\zeta(s)$:

$$\tilde{f}_a'(x)=\sum\limits_n\delta(x-n)=\underset{f\to\infty}{\text{lim}}\left(-\frac{\sin (2 f \pi x)}{\pi x}+\sum\limits_{n=1}^f (\cos(2 n \pi x)+\cos(2 (n-1) \pi x))\right)\tag{15}$$

$$\tilde{f}_a(x)=\sum\limits_n\theta(x-n)=\underset{f\to\infty}{\text{lim}}\left(-\frac{\text{Si}(2 f \pi x)}{\pi}+\sum\limits_{n=1}^f \left(\frac{\sin(2 n \pi x)}{2 n \pi}+x\ \text{sinc}(2 (n-1) \pi x)\right)\right)\tag{16}$$

$\quad a(n)=(-1)^{n-1} \text{ where } F_a(s)=\eta(s)$:

$$\tilde{f}_a'(x)=\sum\limits_n (-1)^{n-1}\delta(x-n)=\underset{f\to\infty}{\text{lim}}\left(\frac{\sin(2 f \pi x)}{\pi x}-2 \sum\limits_{n=1}^f \cos((2 n-1) \pi x)\right)\tag{17}$$

$$\tilde{f}_a(x)=\sum\limits_n (-1)^{n-1}\theta(x-n)=\underset{f\to\infty}{\text{lim}}\left(\frac{\text{Si}(2 f \pi x)}{\pi }-\frac{2}{\pi}\sum\limits _{n=1}^f \frac{\sin ((2 n-1) \pi x)}{2 n-1}\right)\tag{18}$$

$\quad a(n)=\delta_{n-1} \text{ where } F_a(s)=1$:

$$\tilde{f}_a'(x)=\delta(x-1)=\underset{f\to\infty}{\text{lim}}\left(\frac{\sin(2 f \pi (x+1))}{\pi (x+1)}+\frac{\sin(2 f \pi (x-1))}{\pi (x-1)}\right)\tag{19}$$

$$\tilde{f}_a(x)=\theta(x-1)=\underset{f\to\infty}{\text{lim}}\left(\frac{\text{Si}(2 f \pi (x+1))+\text{Si}(2 f \pi (x-1))}{\pi }\right)\tag{20}$$

Formulas (15) to (18) above lead to the following formulas for the Riemann zeta function $\zeta(s)$ and Dirichlet eta function $\eta(s)$. Formulas (21) and (22) for $\zeta(s)$ and $\eta(s)$ can also be used to derive formulas for $\zeta(s)$ and $\eta(s)$ which converge for $\Re(s)>-1$.

$$\zeta(s)=\underset{f\to\infty}{\text{lim}}\left(2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s) \left(-\frac{f^s}{s}+\frac{1}{2} \left(1+\sum\limits_{n=2}^f \left(n^{s-1}+(n-1)^{s-1}\right)\right)\right)\right),\ \Re(s)<2\tag{21}$$

$$\eta(s)=\underset{f\to\infty}{\text{lim}}\left(2 \pi^{s-1} \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s) \left(\frac{2^{s-1} f^s}{s}-\sum\limits_{n=1}^f (2 n-1)^{s-1}\right)\right),\ \Re(s)<2\tag{22}$$

Formulas (16), (18), (20), and (22) above are illustrated in this answer I recently posted to my own question on MathOverflow.