A rough proof for infinitesimals?

I discovered the following relation for arbitrary $d_r$:

$$\lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \! \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \int_0^\infty f(x) \, dx$$

Where $f(x)$ is a smooth continuous function whose integral $\int_0^\infty f(x) d x$ is absolutely convergent. $d_r$ is the $r$'th number of a sequence.

What does this formula mean?

Consider the curve $f(x)$. Then splitting it to $k/n= h$ intervals then adding the first strip ($d_1$ times): $f(h) \cdot d_1$. Then the second strip ($d_2$ times) $f(2h) \cdot d_2$ times ... And so on . Hence. $d_r$ can be thought of as the number of infinitesimals at $f(rh)$!(?) (See proof as to why it makes sense to adopt this line of reasoning)

A sample example of $f(x)$ that should work is $f(x) = e^{-x}$

Heuristic Proof

Consider an integral such that $$\int_0^\infty f(x) \, dx = C,$$where, $f(x)$ is a smooth and continuous function and absolutely converges.

Now we raise both sides to the power s:

$$\left(\int_0^\infty f(x) \, dx\right)^s = C^s$$

We substitute $x$ with $rx$ to get:

$$\left(\int_0^\infty f(rx) \, dx\right)^s = (C/r)^s$$

Multiplying both sides by an arbitrary coefficient:

$$(b_r)\left(\int_0^\infty f(rx) \, dx\right)^s = (b_r)( C/r)^s$$

Taking their sum:

$$\sum_{r=1}^\infty b_r \left(\int_0^\infty f(rx) \, dx\right)^s = C^s \underbrace{\sum_{r=1}^\infty \frac{b_r}{r^s}}_{\text{dirichlet series}}$$

We write the integral as a limit of a Riemann sum:

$$\sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s = C^s \sum_{r=1}^\infty \frac{b_r}{r^s}$$

Using the mobius inversion formula:

$$\sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s = C^s \frac{1}{\zeta(s)}\sum_{r=1}^\infty \frac{d_r}{r^s}$$

We define $d_r = \sum_{e|r} b_e$

Note:

$$(\frac{b_1}{1^s} + \frac{b_2}{2^s} + \frac{b_3}{3^s} + \frac{b_4}{4^s} + \dots) \times (\frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \frac{1}{4^s} + \dots) = \frac{b_1}{1^s} + \frac{b_1 + b_2}{2^s} + \frac{b_1 + b_3}{3^s} + \frac{b_1 + b_2 + b_4}{4^s} + \dots$$

Now focusing on the L.H.S ($s \nearrow 1$):

$$\sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)^s = \sum_{r=1}^\infty \lim_{k \to \infty} \lim_{n \to \infty}\ b_r \left( \sum_{x=1}^n f(\frac{kx}{n}r)\frac{k}{n} \right)$$

Focusing on the L.H.S (and vertically summing):

$$\lim_{k \to \infty }\lim_{n \to \infty} b_1 ((f(\frac{k}{n}) + f(2 \frac{k}{n}) + f(3 \frac{k}{n}) +f(4 \frac{k}{n}) + \cdots)\frac{k}{n}$$ $$+$$ $$\lim_{n \to \infty} b_2 (0.f(\frac{k}{n}) + f(2 \frac{k}{n}) + 0.f(3 \frac{k}{n}) +f(4 \frac{k}{n}) +\cdots) \frac{k}{n}$$ $$+$$ $$\vdots$$

$$= \lim_{n \to \infty} (\underbrace{b_1}_{d_1} (f(\frac{k}{n}) + \underbrace{(b_1 + b_2)}_{d_2}f(2 \frac{k}{n}) + \underbrace{(b_1 + b_3)}_{d_3}f(3 \frac{k}{n}) +\underbrace{(b_1 + b_2 + b_4)}_{d_4}f(4 \frac{k}{n}) + \cdots)\frac{k}{n}$$

Note: this resummation trick can be only done for special functions ($f$ must absolutely converge)

Hence, for special $d_r$ the L.H.S converges:

$$\lim_{k \to \infty} \lim_{n \to \infty}\ \sum_{r=1}^n d_r \left( f(\frac{k}{n}r)\frac{k}{n} \right) = \lim_{s \to 1} \underbrace{\frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s}}_{\text{removable singularity}} \times \int_0^\infty f(x) \, dx$$

Example:

Let $f(x) = e^{-x}$

$$d_{2r} = 1$$ $$d_{2r+1} = 0$$

Hence,

Let us compute the R.H.S

$$\lim_{s \to 1} \frac{1}{\zeta(s)} (\frac{2^{(-s)}}{\zeta(s)}) . 1 = \frac{1}{2}$$

Looking at the L.H.S:

We are essentially adding all the even strips! This can be computed also by doing:

$$\int_{0}^\infty e^{-x} dx = 1$$ $$\implies \int_{0}^\infty e^{-2x} d(2x) = 1$$ $$\implies \int_{0}^\infty e^{-2x} d(x) = 1/2$$

Hence both answers match!

Questions

Geometry

Can $d_r$ be thought as of the number of infinitesimals (localised at a point $x$)? Can this argument be made rigorous? Is it possible to geometrically make sense of the formula without the idea of infinitesimals? Is it possible to obtain the same result with different definition of integration (Not Riemann–Stieltjes integral)?

Ananlysis & Asymotics

What is the $f(x)$ allow the re summation to happen? Can one find an asymptotic formula for the limiting cases?

• I know in the end the information is redundant, but it'd still be nice if you'd state all implicit prerequisites for your statement to hold in the beginning – Sudix Aug 20 '18 at 16:54
• Done ... There may be further requirements never been too good a rigour (Sorry I'm a physicist in training) – More Anonymous Aug 20 '18 at 17:01
• Your question is little bit unclear to me. What's $d_r$ actually? I think you have to state that.! – Empty Aug 20 '18 at 17:18
• @Empty Stated ... also if u look at the (long) proof one can see you get them by the specific sum of particular $b_r$ coefficients – More Anonymous Aug 20 '18 at 17:21

1 Answer

(I have no answer but a comment which is too long for the comment section:)

The r.h.s in your first equation $$\lim_{s \to 1} \! \frac{1}{\zeta(s)} \sum_{r=1}^\infty \frac{d_r}{r^s} \int_0^\infty f(x) \, dx$$

can be identified as a term containing the limit of the quotient of a Dirichlet series ($F(s)$) and the Riemann zeta function when approaching 1:

$$\lim_{s \to 1} \! \frac{F(s)}{\zeta(s)} \int_0^\infty f(x) \, dx$$

(here $F(s)=\sum_{n=1}^\infty d_n/n^s$). You maybe can think about the convergence and evaluation of that using known convergence conditions on $F(s)$. E.g in case F(1) would be finite the whole thing evaluates to $0$. The special case $d_n = 1$ which as you point out should break down to the Riemann integral unfortunately is just about the limit case $\sigma_0=1$ where we cannot say anything about convergence or divergence from that criteria alone.