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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?


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

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 $\ln\zeta(s)$:


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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If that is a multivariable integral and $y$ is a vector, then what does $e^y$ mean? – anon Jul 25 '12 at 19:45
I wonder what kind of math was used to arrive at this problem... – user31373 Jul 28 '12 at 4:15
I don't understand the question – Ethan May 25 '13 at 4:55

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