Consider the following formula which defines a piece-wise function which I believe corresponds to a series representation for the Dirac delta function $\delta(x)$. The parameter $f$ is the evaluation frequency and assumed to be a positive integer, and the evaluation limit $N$ must be selected such that $M(N)=0$ where $M(x)=\sum\limits_{n\le x}\mu(n)$ is the Mertens function.
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**(1)** $\quad\delta(x)=\underset{N,f\to\infty}{\text{lim}}\ 2\left.\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\ n}\ \left(\left\{
\begin{array}{cc}
\begin{array}{cc}
\cos \left(\frac{2 k \pi (x+1)}{n}\right) & x\geq 0 \\
\cos \left(\frac{2 k \pi (x-1)}{n}\right) & x<0 \\
\end{array}
\\
\end{array}
\right.\right.\right),\quad M(N)=0$
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The following figure illustrates formula (1) above evaluated at $N=39$ and $f=4$. The red discrete dots in figure (1) below illustrate the evaluation of formula (1) at integer values of $x$. I believe formula (1) always evaluates to exactly $2\ f$ at $x=0$ and exactly to zero at other integer values of $x$.
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[![Illustration of formula (1) for delta(x)][1]][1]
**Figure (1)**: Illustration of formula (1) for $\delta(x)$
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Now consider formula (2) below derived from the integral $f(0)=\int_0^{\infty}\delta(x)\ f(x)\, dx$ where $f(x)=e^{-\left| x\right|}$ and formula (1) above for $\delta(x)$ was used to evaluate the integral. Formula (2) below can also be evaluated as illustrated in formula (3) below.
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**(2)** $\quad e^{-\left| 0\right|}=1=\underset{N,f\to\infty}{\text{lim}}\ 4\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\ n}\frac{n\ \cos\left(\frac{2\ \pi\ k}{n}\right)-2\ \pi\ k\ \sin\left(\frac{2\ \pi\ k}{n}\right)}{4\ \pi^2\ k^2+n^2}\,,\quad M(N)=0$
**(3)** $\quad e^{-\left| 0\right|}=1=\underset{N\to\infty}{\text{lim}}\ \mu(1)\left(\coth\left(\frac{1}{2}\right)-2\right)+4\sum\limits_{n=2}^N\frac{\mu(n)}{4 e \left(e^n-1\right) n}\\$ $\left(-2 e^{n+1}+e^n n+e^2 n-e \left(e^n-1\right) \left(e^{-\frac{2 i \pi }{n}}\right)^{\frac{i n}{2 \pi }} B_{e^{-\frac{2 i \pi }{n}}}\left(1-\frac{i n}{2 \pi },-1\right)+e \left(e^n-1\right) \left(e^{-\frac{2 i \pi }{n}}\right)^{-\frac{i n}{2 \pi }} B_{e^{-\frac{2 i \pi }{n}}}\left(\frac{i n}{2 \pi }+1,-1\right)+\left(e^n-1\right) \left(B_{e^{\frac{2 i \pi }{n}}}\left(1-\frac{i n}{2 \pi },-1\right)-e^2 B_{e^{\frac{2 i \pi }{n}}}\left(\frac{i n}{2 \pi }+1,-1\right)\right)+2 e\right),\quad M(N)=0$
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The following table illustrates formula (3) above evaluated for several values of $N$ corresponding to zeros of the Mertens function $M(x)$. Note formula (3) above seems to converge to $e^{-\left| 0\right|}=1$ as the magnitude of the evaluation limit $N$ increases.
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$$\begin{array}{ccc}
n & \text{N=$n^{th}$ zero of $M(x)$} & \text{Evaluation of formula (3) for $e^{-\left| 0\right|}$} \\
10 & 150 & 0.973479\, +\ i\ \text{5.498812269991985$\grave{ }$*${}^{\wedge}$-17} \\
20 & 236 & 0.982236\, -\ i\ \text{5.786047752866836$\grave{ }$*${}^{\wedge}$-17} \\
30 & 358 & 0.988729\, -\ i\ \text{6.577233629689039$\grave{ }$*${}^{\wedge}$-17} \\
40 & 407 & 0.989363\, +\ i\ \text{2.6889189402888207$\grave{ }$*${}^{\wedge}$-17} \\
50 & 427 & 0.989387\, +\ i\ \text{4.472005325912989$\grave{ }$*${}^{\wedge}$-17} \\
60 & 785 & 0.995546\, +\ i\ \text{6.227857765313369$\grave{ }$*${}^{\wedge}$-18} \\
70 & 825 & 0.995466\, -\ i\ \text{1.6606923419056456$\grave{ }$*${}^{\wedge}$-17} \\
80 & 893 & 0.995653\, -\ i\ \text{1.1882293286557667$\grave{ }$*${}^{\wedge}$-17} \\
90 & 916 & 0.995653\, -\ i\ \text{3.521050901644269$\grave{ }$*${}^{\wedge}$-17} \\
100 & 1220 & 0.997431\, -\ i\ \text{1.2549006768893629$\grave{ }$*${}^{\wedge}$-16} \\
\end{array}$$
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Finally consider the following three formulas derived from the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ where all three convolutions were evaluated using formula (1) above for $\delta(x)$.
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**(4)** $\quad e^{-\left|y\right|}=\underset{N,f\to\infty}{\text{lim}}\ 4\sum\limits_{n=1}^N\mu(n)\sum\limits_{k=1}^{f\ n}\frac{1}{4\ \pi^2\ k^2+n^2}\ \left(\left\{
\begin{array}{cc}
\begin{array}{cc}
n \cos\left(\frac{2\ k\ \pi\ (y+1)}{n}\right)-2\ k\ \pi\ e^{-y} \sin\left(\frac{2\ k\ \pi}{n}\right) & y\geq 0 \\
n \cos\left(\frac{2\ k\ \pi\ (y-1)}{n}\right)-2\ k\ \pi\ e^y \sin\left(\frac{2\ k\ \pi}{n}\right) & y<0 \\
\end{array}
\\
\end{array}\right.\right),\ M(N)=0$
**(5)** $\quad e^{-y^2}=\underset{N,f\to\infty}{\text{lim}}\ \sqrt{\pi}\sum\limits_{n=1}^N\frac{\mu(n)}{n}\\$ $\ \sum\limits_{k=1}^{f\ n}e^{-\frac{\pi\ k\ (\pi\ k+2\ i\ n\ y)}{n^2}}\ \left(\left(1+e^{\frac{4\ i\ \pi\ k\ y}{n}}\right) \cos\left(\frac{2\ \pi\ k}{n}\right)-\sin\left(\frac{2\ \pi\ k}{n}\right) \left(\text{erfi}\left(\frac{\pi\ k}{n}+i\ y\right)+e^{\frac{4\ i\ \pi\ k\ y}{n}} \text{erfi}\left(\frac{\pi\ k}{n}-i\ y\right)\right)\right),\ M(N)=0$
**(6)** $\quad\sin(y)\ e^{-y^2}=\underset{N,f\to\infty}{\text{lim}}\ \frac{1}{2} \left(i \sqrt{\pi }\right)\sum\limits _{n=1}^{\text{nMax}} \frac{\mu(n)}{n}\sum\limits_{k=1}^{f n} e^{-\frac{(2 \pi k+n)^2+8 i \pi k n y}{4 n^2}} \left(-\left(e^{\frac{2 \pi k}{n}}-1\right) \left(-1+e^{\frac{4 i \pi k y}{n}}\right) \cos\left(\frac{2 \pi k}{n}\right)+\right.\\$ $\left.\sin\left(\frac{2 \pi k}{n}\right) \left(\text{erfi}\left(\frac{\pi k}{n}+i y+\frac{1}{2}\right)-e^{\frac{4 i \pi k y}{n}} \left(e^{\frac{2 \pi k}{n}} \text{erfi}\left(-\frac{\pi k}{n}+i y+\frac{1}{2}\right)+\text{erfi}\left(\frac{\pi k}{n}-i y+\frac{1}{2}\right)\right)+e^{\frac{2 \pi k}{n}} \text{erfi}\left(-\frac{\pi k}{n}-i y+\frac{1}{2}\right)\right)\right),\qquad M(N)=0$
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Formulas (4), (5), and (6) defined above are illustrated in the following three figures where the blue curves are the reference functions, the orange curves represent formulas (4), (5), and (6) above evaluated at $f=4$ and $N=39$, and the green curves represent formulas (4), (5), and (6) above evaluated at $f=4$ and $N=101$. The three figures below illustrate formulas (4), (5), and (6) above seem to converge to the corresponding reference function for $x\in\mathbb{R}$ as the evaluation limit $N$ is increased. Note formula (6) above for $\sin(y)\ e^{-y^2}$ illustrated in Figure (4) below seems to converge much faster than formulas (4) and (5) above perhaps because formula (6) represents an odd function whereas formulas (4) and (5) both represent even functions.
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[![Illustration of formula (4)][2]][2]
**Figure (2)**: Illustration of formula (4) for $e^{-\left|y\right|}$ evaluated at $N=39$ (orange curve) and $N=101$ (green curve) overlaid on the reference function in blue
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[![Illustration of formula (5)][3]][3]
**Figure (3)**: Illustration of formula (5) for $e^{-y^2}$ evaluated at $N=39$ (orange curve) and $N=101$ (green curve) overlaid on the reference function in blue
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[![Illustration of formula (6)][4]][4]
**Figure (4)**: Illustration of formula (6) for $\sin(y)\ e^{-y^2}$ evaluated at $N=39$ (orange curve) and $N=101$ (green curve) overlaid on the reference function in blue
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>**Question (1)**: Is it true formula (1) above is an example of a series representation of the Dirac delta function $\delta(x)$?
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>**Question (2)**: What is the class or space of functions $f(x)$ for which the integral $f(0)=\int\limits_{-\infty}^\infty\delta(x)\ f(x)\ dx$ and Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ are both valid when using formula (1) above for $\delta(x)$ to evaluate the integral and Fourier convolution?
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>**Question (3)**: Is formula (1) above for $\delta(x)$ an example of what is referred to as a tempered distribution, or is formula (1) for $\delta(x)$ more general than a tempered distribution?
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Formula (1) above for $\delta(x)$ is based on the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$. The conditional convergence requirement $M(N)=0$ for formula (1) is because the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ only evaluates to zero at $x=0$ when $M(N)=0$.
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Whereas the Fourier convolution $f(y)=\int\limits_{-\infty}^\infty\delta(x)\ f(y-x)\ dx$ evaluated with $\delta(x)$ defined in formula (1) above seems to converge for $x\in\mathbb{R}$, Mellin convolutions such as $f(y)=\int\limits_0^\infty\delta(x-1)\ f\left(\frac{y}{x}\right)\ \frac{dx}{x}$ and $f(y)=\int\limits_0^\infty\delta(x-1)\ f(y\ x)\ dx$ evaluated using the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ typically seem to converge on the half-plane (e.g. $\Re(s)>0$ or $\Re(s)<0$ depending on the function $f(x)$), but in some cases these Mellin convolutions are globally convergent for $s\in\mathbb{C}$. I'll note that in general formulas derived from Fourier convolutions evaluated using formula(1) above for $\delta(x)$ seem to be more complicated than formulas derived from Mellin convolutions using the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ which I suspect is at least partially related to the extra complexity of the piece-wise nature of formula (1) above.
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See [this answer][5] I posted to my own question [What is Relationship Between Distributional and Fourier Series Frameworks for Prime Counting Functions?][6] for more information on the nested Fourier series representation of $\delta(x+1)+\delta(x-1)$ and examples of formulas derived from Mellin convolutions using this representation. See my [Question related to nested Fourier series representation of $h(s)=\frac{i s}{s^2-1}$][7] for information on the more general topic of nested Fourier series representations of other non-periodic functions.
[1]: https://i.sstatic.net/FPcZy.jpg
[2]: https://i.sstatic.net/YThBd.jpg
[3]: https://i.sstatic.net/VhjK9.jpg
[4]: https://i.sstatic.net/gnJ22.jpg
[5]: https://math.stackexchange.com/q/2380164
[6]: https://math.stackexchange.com/q/1980231
[7]: https://math.stackexchange.com/q/3551922
[8]: https://math.stackexchange.com/q/3552092