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We have the following 'transform' of a real valued, piecewise continuous function $f(x)$ :

$$T[f(x)]=1+\sum_{n=1}^{\infty}\int_{\mathbb{R}^{n}_{+}}f\left(\frac{x}{\Lambda _{n}} \right )\frac{1}{n!}\left(\prod_{i=1}^{n-1}f(u_{i}) \right )\Theta^{n-1}_{u_{n}}f(u_{n})\frac{d\Lambda_{n}}{\Lambda _{n}}$$

where :

$$\Lambda_{n} =\prod_{i=1}^{n}u_{i}$$ $$d\Lambda_{n}=\prod_{n=1}^{n}du_{i}$$ $$\Theta_{u_{k}}=u_{k}\frac{d}{du_{k}}$$

and we wish to recover $f(x)$ from $T[f(x)]$.

What kind of mathematics should be used to study this problem?

EDIT:

$$J(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\ln\zeta(s)\frac{x^{s}}{s}ds$$

where $J(x)$ is the Riemann prime counting function . The above relation is the well known Perron's formula for Dirichlet series. This induced me to express $\zeta(s)$ as an exponential series expansion in terms of $\ln\zeta(s)$:

$$\zeta(s)=\sum_{n=0}^{\infty}\frac{(\ln\zeta(s))^{n}}{n!}$$

Applying Perron's formula :

$$\left \lfloor x \right \rfloor=\frac{1}{2\pi i}\int_{\alpha-i\infty}^{\alpha+i\infty}\zeta(s)\frac{x^{s}}{s}ds=\frac{1}{2\pi i}\sum_{n=0}^{\infty}\frac{1}{n!}\int_{c-i\infty}^{c+i\infty}(\ln\zeta(s))^{n}\frac{x^{s}}{s}ds$$

The first two terms corresponding to $n=0,n=1$ are trivial. The other terms starting at $n=2$ could be done using Mellin convolution. Namely, if two functions, say $F(s)$ and $G(s)$ are given by:

$$F(s)=\int_{0}^{\infty}f(x)x^{-s-1}dx$$ $$G(s)=\int_{0}^{\infty}g(x)x^{-s-1}dx$$ then the following holds :

$$F(s)G(s)=\int_{0}^{\infty}f(x)\bigstar g(x) x^{-s-1}dx$$ Where the star stands for Mellin convolution, and is defined by : $$f(x)\bigstar g(x)=\int_{0}^{\infty}f\left( \frac{x}{u}\right)g(u)\frac{du}{u}$$

Another property of the Mellin transform we will need is that, if a function, say $h(x)$ is a Mellin inverse of of $H(s)$, then : $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}sH(s)x^{s}ds=x\frac{d}{dx}h(x)$$

Using these facts about the Mellin transform, we can evaluate the integrals :

$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}(\ln\zeta(s))^{n}\frac{x^{s}}{s}ds=J(x)\bigstar \left( x\frac{d}{dx}J(x)\right)^{\bigstar n-1}$$

Where the star and the power $n-1$ mean repeated convolution for $n-1$ times .

Furthermore, the Mellin convolution has the property :

$$\left(x\frac{d}{dx}\right)^{n}(f(x)\bigstar g(x))=f(x)\bigstar \left(x\frac{d}{dx}\right)^{n}g(x)=g(x) \bigstar\left(x\frac{d}{dx}\right)^{n}f(x) $$ Therefore :

$$\left \lfloor x \right \rfloor=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\zeta(s)\frac{x^{s}}{s}ds$$

Therefore :

$$\left \lfloor x \right \rfloor=1+ \sum_{n=0}^{\infty}\frac{J(x)^{\bigstar n}}{(n+1)!} \bigstar \left(x\frac{d}{dx}\right)^{n}J(x)$$

This relation could be given more explicitly by:

$$\left \lfloor x \right \rfloor=1+\sum_{n=1}^{\infty}\int_{\mathbb{R}^{n}_{+}}J\left(\frac{x}{\Lambda _{n}} \right )\frac{1}{n!}\left(\prod_{i=1}^{n-1}J(u_{i}) \right )\Theta^{n-1}_{u_{n}}J(u_{n})\frac{d\Lambda_{n}}{\Lambda _{n}}$$ where :

$$\Lambda_{n} =\prod_{i=1}^{n}u_{i}$$ $$d\Lambda_{n}=\prod_{n=1}^{n}du_{i}$$ $$\Theta_{u_{k}}=u_{k}\frac{d}{du_{k}}$$

We define the following operator acting on a distribution $f(x)$: $$T[f(x)]=1+\sum_{n=1}^{\infty}\int_{\mathbb{R}^{n}_{+}}f\left(\frac{x}{\Lambda _{n}} \right )\frac{1}{n!}\left(\prod_{i=1}^{n-1}f(u_{i}) \right )\Theta^{n-1}_{u_{n}}f(u_{n})\frac{d\Lambda_{n}}{\Lambda _{n}}$$

if $f(x)$ is given by :

$$f(x)=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\ln g(s)\frac{x^{s}}{s}ds$$

then, the following holds:

$$T[f(x)]=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty} g(s)\frac{x^{s}}{s}ds$$

The plan here is to investigate the geometric and algebraic properties of the mapping above. One thing that comes in mind is trying to find an inverse of the operator $T[f(x)]$, such that : $$T^{-1}T[f(x)]=f(x)$$

assuming such an operator exists, and applying the operator to $\left \lfloor x \right \rfloor$:

$$T^{-1}[\left \lfloor x \right \rfloor]=J(x)$$ and the prime counting function $\pi(x)$ could be given by:

$$\pi(x)=\sum_{n=1}^{\infty}\frac{\mu(n)}{n}T^{-1}\left[\left \lfloor x^{1/n} \right \rfloor\right]$$

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  • 8
    $\begingroup$ If that is a multivariable integral and $y$ is a vector, then what does $e^y$ mean? $\endgroup$ – anon Jul 25 '12 at 19:45
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    $\begingroup$ I wonder what kind of math was used to arrive at this problem... $\endgroup$ – user31373 Jul 28 '12 at 4:15
  • $\begingroup$ I don't understand the question $\endgroup$ – Ethan May 25 '13 at 4:55
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The fundamental prime-power counting function $\pi(x)$ and the Riemann prime-power counting function $J(x)$ (also referred to as $\Pi(x)$) defined in (1) and (2) below can be expressed in terms of the floor function $\lfloor x\rfloor$ as illustrated in (3) and (4) below where $b_{\pi}(n)$ and $b_J(n)$ are the Möbius transforms of $a_{\pi}(n)$ and $a_J(n)$ (see Möbius transformation - Wikipedia and Weisstein, Eric W. "Möbius Inversion Formula." From MathWorld--A Wolfram Web Resource).

(1) $\quad\pi(x)=\sum\limits_{n=1}^x a_{\pi}(n)\,,\quad a_{\pi}(n)=\begin{array}{cc} \{ & \begin{array}{cc} 1 & n\in \mathbb{P} \\ 0 & \text{True} \\ \end{array} \\ \end{array}\quad$(see https://oeis.org/A010051)

(2) $\quad J(x)=\sum_\limits{n=1}^x a_J(n)\,,\quad a_J(n)=\begin{array}{cc} \{ & \begin{array}{cc} 0 & n=1 \\ \frac{\Lambda (n)}{\log (n)} & \text{True} \\ \end{array} \\ \end{array}$

(3) $\quad\pi(x)=\sum\limits_{n=1}^x b_{\pi}(n)\,\lfloor \frac{x}{n}\rfloor\,,\quad b_{\pi}(n)=\sum\limits_{d|n}a_{\pi}(d)\,\mu(\frac{n}{d})=\sum\limits_{d|n}\mu(d)\,a_{\pi}(\frac{n}{d})\,,\\$ $\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$(see https://oeis.org/A143519)

(4) $\quad J(x)=\sum\limits_{n=1}^x b_J(n)\,\lfloor \frac{x}{n}\rfloor\,,\quad b_{J}(n)=\sum\limits_{d|n}a_{J}(d)\,\mu(\frac{n}{d})=\sum\limits_{d|n}\mu(d)\,a_{J}(\frac{n}{d})$

Formulas (3) and (4) above are the basis of the Fourier series representations of $\pi(x)$ and $J(x)$ (see Fourier Series for Prime Counting Functions).

I believe the Fourier series representation of $U'(x)=\delta(x+1)+\delta(x-1)$ is probably of more significant importance because it can be used to evaluate Mellin convolutions such as $g(y)=\int_0^\infty\delta(x-1)\,g\left(\frac{y}{x}\right)\,\frac{dx}{x}$ to derive new formulas for a variety of functions thereby providing new insights into functions and their relationships (see my answer to my own question What is Relationship Between Distributional and Fourier Series Frameworks for Prime Counting Functions?).

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