My question above, [original answer][1], and this new answer are all based on analytic formulas for

$$u(x)=-1+\theta(x+1)+\theta(x-1)\tag{1}\,.$$

and

$$u'(x)=\delta(x+1)+\delta(x-1)\tag{2}\,.$$

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My question and original answer are both based on the analytic formula

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

where the evaluation frequency $f$ is assumed to be a positive integer and

$$M(N)=\sum\limits_{n=1}^N \mu(n)\tag{4}$$

is the Mertens function.

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Formula (3) above simplifies to

$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{\underset{M(N)=0}{N,f\to\infty}}{\text{lim}}\left(2\sum\limits_{n=1}^N\frac{\mu(n)}{n}\sum\limits_{k=1}^{f\ n}\cos\left(\frac{2 k \pi x}{n}\right)\right)\tag{5}$$

since

$$\sum\limits_{n=1}^\infty\frac{\mu(n)}{n}=\frac{1}{\zeta(1)}=0\,.\tag{6}$$

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I originally defined formulas (3) and (5) above in [this answer][2] I posted to one of my own questions on Math StackExchange.

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The formula in my question above wasn't quite right as it wasn't a smooth function at $x=0$ (there's a discontinuity in the first-order derivative corresponding to $\delta'(x)$). My original answer fixed this problem but still required the upper evaluation limit $N$ be selected such that $M(N)=0$.

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This new answer is based on the analytic formula

$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \mu(n) \left(-2 f \text{sinc}(2 \pi f x)+\frac{1}{n}\sum\limits_{k=1}^{f\,n} \left(\cos\left(\frac{2 \pi (k-1) x}{n}\right)+\cos\left(\frac{2 \pi k x}{n}\right)\right)\right)\right)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \mu(n) \left(-2 f\,\text{sinc}(2 \pi f x)+\frac{\sin(2 \pi f x) \cot\left(\frac{\pi x}{n}\right)}{n}\right)\right)\tag{7}$$

which no longer requires $N$ to be selected such that $M(N)=0$.

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Formula (7) above is a result related to [this answer][3] I posted to another one of my questions on Math Overflow.

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I believe formula (7) above is exactly equivalent to

$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{f\to\infty}{\text{lim}}\left(2 f\ \text{sinc}(2 \pi f (x+1))+2 f\ \text{sinc}(2 \pi f (x-1))\right)\tag{8}$$

in that formulas (7) and (8) above both have the same Maclaurin series.

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My original answer and this new answer are based on the relationship

$$\delta(x)=\frac{1}{2}\left(u'(x+1)+u'(x-1)-\frac{1}{2} u'\left(\frac{x}{2}\right)\right)\tag{9}$$

which using formula (7) above for $u'(x)$ leads to

$\delta(x)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\mu(n) \Bigg(f (-\text{sinc}(2 \pi f (x+1))-\text{sinc}(2 \pi f (x-1))+\text{sinc}(2 \pi f x))+\right.$ $\left.\frac{1}{2 n}\left(\sum\limits_{k=1}^{f\,n}\left(\cos\left(\frac{2 \pi (k-1) (x+1)}{n}\right)+\cos\left(\frac{2 \pi k (x+1)}{n}\right)+\cos\left(\frac{2 \pi (k-1) (x-1)}{n}\right)+\cos\left(\frac{2 \pi k (x-1)}{n}\right)\right)-\frac{1}{2} \sum\limits_{k=1}^{2 f\,n}\left(\cos\left(\frac{\pi (k-1) x}{n}\right)+\cos\left(\frac{\pi k x}{n}\right)\right)\right)\Bigg)\right)$

$$=\underset{N,f\to\infty}{\text{lim}}\left(\mu (n)\left(f (-\text{sinc}(2 \pi f (x+1))-\text{sinc}(2 \pi f (x-1))+\text{sinc}(2 \pi f x))+\frac{\sin(2 \pi f (x+1)) \cot\left(\frac{\pi (x+1)}{n}\right)+\sin(2 \pi f (x-1)) \cot\left(\frac{\pi (x-1)}{n}\right)-\frac{1}{2} \sin(2 \pi f x) \cot\left(\frac{\pi x}{2 n}\right)}{2 n}\right)\right)\tag{10}$$

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I believe the formula (10) above is exactly equivalent to the integral representation

$$\delta(x)=\underset{f\to\infty}{\text{lim}}\left(\int\limits_{-f}^f e^{2 i \pi t x}\,dt\right)=\underset{f\to\infty}{\text{lim}}\left(2 f\ \text{sinc}(2 \pi f x)\right)\tag{11}$$

in that formulas (10) and (11) above both have the same Maclaurin series.

  [1]: https://mathoverflow.net/q/380261
  [2]: https://math.stackexchange.com/q/2380164
  [3]: https://mathoverflow.net/q/395266