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

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

and

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

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.

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

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

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

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 $M(N)=0$.

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 $M(N)=0$.

Formula (7) above is a result related to this answer I posted to another one of my questions on Math Overflow and this answer I posted to a related question on Math StackExchange.

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.

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

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

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

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.

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

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

and

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

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.

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

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

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

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 $M(N)=0$.

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 $M(N)=0$.

Formula (7) above is a result related to this answer I posted to another one of my questions on Math Overflow and this answer I posted to a related question on Math StackExchange.

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.

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

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

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

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.

Now consider the slightly simpler analytic formula

$$u'(x)=\delta(x+1)+\delta(x-1)=\underset{N,f\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N\frac{\mu(2 n-1)}{2 n-1}\left(\frac{1}{2}+\sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k \cos\left(\frac{\pi k x}{2 n-1}\right)\right)\right)=\underset{N,f\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N\frac{\mu(2 n-1)}{2 n-1} \sec\left(\frac{\pi x}{4 n-2}\right) \cos\left(\pi x \left(2 f+\frac{1}{4 n-2}\right)\right)\right)\tag{12}$$

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

I believe formula (12) above is exactly equivalent to formulas (7) and (8) above in that all three formulas have the same Maclaurin series.

The Maclaurin series terms for formula (12) above can be derived based on the relationship

$$\sum\limits_{n=1}^\infty\frac{\mu(2 n-1)}{(2 n-1)^s}=\frac{1}{\lambda(s)}\,,\quad\Re(s)\ge 1\tag{13}$$

where $\lambda(s)=\left(1-2^{-s}\right)\,\zeta(s)$ is the Dirichlet lambda function. I believe formula (13) above is valid for $\Re(s)>\frac{1}{2}$ assuming the Riemann hypothesis.

The relationship in formula (9) above and formula (12) for $u'(x)$ above leads to

$\delta(x)=\underset{N,f\to\infty}{\text{lim}}\left(\frac{1}{2}\sum\limits_{n=1}^N\frac{\mu(2 n-1)}{2 n-1}\left(\frac{3}{4}+\sum\limits_{k=1}^{2 f (2 n-1)} (-1)^k \left(\cos\left(\frac{\pi k (x+1)}{2 n-1}\right)+\cos\left(\frac{\pi k (x-1)}{2 n-1}\right)\right)\right.\right.$ $\left.\left.-\frac{1}{2}\sum\limits_{k=1}^{4 f (2 n-1)} (-1)^k \cos\left(\frac{\pi k x}{2 (2 n-1)}\right)\right)\right)$

$=\underset{N,f\to\infty}{\text{lim}}\left(\frac{1}{4}\sum\limits_{n=1}^N\frac{\mu(2 n-1)}{2 n-1} \left(\sec\left(\frac{\pi (x+1)}{4 n-2}\right) \cos\left(\pi (x+1) \left(2 f+\frac{1}{4 n-2}\right)\right)+\sec\left(\frac{\pi (x-1)}{4 n-2}\right) \cos\left(\pi (x-1) \left(2 f+\frac{1}{4 n-2}\right)\right)-\frac{1}{2} \sec\left(\frac{\pi x}{2 (4 n-2)}\right) \cos\left(\frac{1}{2} \pi x \left(4 f+\frac{1}{4 n-2}\right)\right)\right)\right)\tag{14}$

which I believe is exactly equivalent to formula (10) above and the integral representation in formula (11) above in that all three formulas have the same Maclaurin series.

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

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

and

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

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.

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

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

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

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 $M(N)=0$.

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 $M(N)=0$.

Formula (7) above is a result related to this answer I posted to another one of my questions on Math Overflow and this answer I posted to a related question on Math StackExchange.

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.

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

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

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

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.

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

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

and

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

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.

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

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

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

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 $M(N)=0$.

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 $M(N)=0$.

Formula (7) above is a result related to this answer I posted to another one of my questions on Math Overflow and this answer I posted to a related question on Math StackExchange.

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.

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

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

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

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.

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

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

and

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

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.

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

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

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

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 $M(N)=0$.

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 $M(N)=0$.

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

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.

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

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

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

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.

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

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

and

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

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.

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

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

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

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 $M(N)=0$.

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 $M(N)=0$.

Formula (7) above is a result related to this answer I posted to another one of my questions on Math Overflow and this answer I posted to a related question on Math StackExchange.

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.

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

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

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

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.