## simplifying expressions – Want to realize this operation (multiplication of divergent integrals of polynomials) in Mathematica

I am currently researching divergent integrals.

1. Definition. An extended number is an expression of the form $$int_a^b f(x)dx$$, where function $$f(x)$$ is defined almost everywhere at $$(a,b)$$. Generally (when Riemann or Lebesque sum converges or when the equivalence follows from the rules expressed below), an extended number can be equal to a real or complex number.

2. There are four simple equivalence rules based on linearity:

$$int_a^c f(x) dx=int_a^b f(x)dx+int_b^c f(x)dx$$

$$int_a^b (f(x)+g(x)) dx=int_a^b f(x)dx+int_a^b g(x)dx$$

$$int_a^b c f(x) dx =c int_a^b f(x) dx$$

$$int_{-infty}^{-a} f(x) dx=int_a^infty f(-x) dx$$

1. There is one complicated Laplace-transform based rule:

$$int_0^infty f(x)dx=int_0^inftymathcal{L}_t(t f(t))left(xright)dx=int_0^inftyfrac1xmathcal{L}^{-1}_t( f(t))left(xright)dx$$

1. There is a rule that allows to represent divergent integrals of polynomials via the most basic divergent integral $$tau=int_0^infty dx$$:

$$int_0^infty x^n dx=frac{left(tau +frac{1}{2}right)^{n+2}-left(tau -frac{1}{2}right)^{n+2}}{(n+1)(n+2)}$$

1. Following Laplace transform, there is a similar rule (for $$n>1$$):

$$int_0^infty frac1{x^n} dx=frac1{(n-1)!}int_0^infty x^{n-2} dx=frac{left(tau +frac{1}{2}right)^{n}-left(tau -frac{1}{2}right)^{n}}{(n-1)n!}$$

1. There is the opposite rule, converting in the opposite direction:

$$tau^n=B_n(1/2)+nint_0^infty B_{n-1}(x+1/2)dx$$

That said, one can use these rules to multiply divergent integrals of polynomials.

Example.

$$int_0^infty left(2x^3-3x^2+x-4right) dx cdot int_0^infty left(2x^2-3x+1right) dx=left(frac{2 tau ^3}{3}-frac{3 tau ^2}{2}+frac{7 tau }{6}-frac{1}{8}right) left(frac{tau ^4}{2}-tau ^3+frac{3 tau ^2}{4}-frac{17 tau }{4}+frac{23}{480}right)=frac{tau ^7}{3}-frac{17 tau ^6}{12}+frac{31 tau ^5}{12}-frac{83 tau ^4}{16}+frac{5333 tau ^3}{720}-frac{4919 tau ^2}{960}+frac{1691 tau }{2880}-frac{23}{3840}=int_0^{infty } left(frac{7 x^6}{3}-frac{17 x^5}{2}+10 x^4-frac{41 x^3}{3}+frac{1007 x^2}{60}-frac{63 x}{10}-frac{113}{120}right) dx+frac{127}{420}$$

I did the previous example by hand. Is it possible to optimally realize it in Mathematica?

I mean, 1. Enter the coefficients of two polynomials under the integrals 2. Obtain the coefficients of the resulting polynomial under integral plus the free term.

I suspect, this may be some kind of convolution.

## Tell if the series below are convergent or divergent

a) `sum_{n=2}^{infty }{{{log n}over{n}}}=sum_{n=2}^{infty }{{{ log n}over{n}}}`

enter image description here

b) `sum_{n=2}^{infty }{{{1}over{log n}}}=sum_{n=2}^{infty }{{{1 }over{log n}}}`
enter image description here

c) `sum_{n=2}^{infty }{{{1}over{left(log nright)^{k}}}}=sum_{n=2 }^{infty }{{{1}over{left(log nright)^{k}}}}`

enter image description here

Sorry, i don’t make in code latex

Can you help me?

## divergent integrals – Do the infinite derivatives make sense?

I have stumped upon a question about whether one can generalize the notion of derivatives onto a function such as $$x mapsto x^{frac 13}$$ at $$x=0$$.

Spicifically, can we compare infinite derivatives to each other?

I am currently working on the theory of infinite divergent integrals/series (divergent quantities), and in that theory I introduce the notion of “germs”, which is a generalization of limits. In finite case they give usual limits, but at a pole of a function they are infinite quantities, containing information of growth rate and the regularized value. They are defined via divergent integrals.

I only had considered germs of monomials and inverses of monomials of integer powers though.

Still, if to generalize the formula to a fractional case, we formally obtain the derivative of $$f(x)=x^{1/3}$$ at $$x=0$$ as

$$f'(0)=frac{omega _+^{5/3}-omega _-^{5/3}}{3 Gamma left(frac{8}{3}right)}$$

where $$omega_+=sum_{0}^infty 1=int_{-1/2}^infty dx$$ and $$omega_-=omega_+-1=sum_1^infty 1=int_{1/2}^infty dx$$ are two divergent integrals/series.

This is an infinite quantity, but comparable to other infinite quatities as well. Its regularized value is zero.

Thus, I wonder, whether it would make sense to speak about infinite derivatives?

## calculus – Show that \$frac{1}{2k+ksin{k}}\$ is divergent using direct comparison test

I’m not quite sure if I got how the direct comparison test works.

For a series

$$sum_{k=1}^{n}{frac{1}{2k+ksin{k}}}$$
Is the following correct:

$$frac{1}{2k+ksin{k}} sim frac{1}{3k} le frac{1}{2k+ksin{k}} implies sum_{k=1}^{infty}{frac{1}{2k+ksin{k}}} text{ is divergent. }$$

## convergence divergence – Why is this correct for a divergent series?

I recently read something and I can’t wrap my head around it:

For a given divergent sequence $$a_n$$, we have the following property:
$$lim limits_{n toinfty} |a_{n+p} – a_n| = 0$$
for a fixed value $$pinmathbb{Q}$$. To me it seems like this violates the cauchy sequence for convergent sequences, since we are considering a divergent series here. What is it that I am missing?

## regularization – A set of divergent integrals that I think, equal to \$-gamma\$

So, we take $$frac{text{sgn}(x-1)}{x}$$ and apply $$mathcal{L}_t(t f(t))(x)$$ four times. The transform is known to keep area under the curve. These integrals, I think, are equal to minus Euler-Mascheroni constant. Since they all have infinite parts that cancel each other, their values are finite. I have already applied Laplace transforms to regularize divergent integrals in a similar way.

$$int_0^infty frac{text{sgn}(x-1)}{x}dx=int_0^inftyfrac{2 e^{-x}-1}{x}dx=int_0^inftyfrac{x-1}{x (x+1)}dx=int_0^infty left(2 e^x text{Ei}(-x)+frac{1}{x}right) dx=$$
$$int_0^infty frac{x^2-2 x log (x)-1}{(x-1)^2 x} dx=-gamma$$

Yes?

Proof: take 2 of them and find average:

$$int frac12left(frac{x-1}{x (x+1)}+ left(2 e^x text{Ei}(-x)+frac{1}{x}right)right)dx=-gamma$$

Right?

## laplace transform – Fractional power of the operator \$mathcal{L}_t[t f(t)](x)\$ and equivalence of divergent integrals

I wonder whether an expression for fractional power of operator $$mathcal{L}_t(t f(t))(x)$$ that involves Laplace transform can be derived?

I am asking this because this operator preserves the area under the function:

$$int_0^infty f(x)dx=int_0^infty mathcal{L}_t(t f(t))(x) dx$$

But more importantly, this still works well even with divergent integrals, thus allowing us to define equivalent classes of divergent integrals. For instance, the following divergent integrals are obtained by consecutive applying this transform, and as such, can be postulated to be equal (the first one and the third one are similar up to a shift by the way):

$$int_0^inftyfrac{theta (x-1)}{x}dx=int_0^inftyfrac{e^{-x}}{x}dx=int_0^inftyfrac{dx}{x+1}=int_0^inftyfrac{e^x x text{Ei}(-x)+1}{x}dx=int_0^inftyfrac{x-log (x)-1}{(x-1)^2}dx$$

The following plot illustrates the comparison:

So, is there a way to express fractional powers of this transform?

## real analysis – Comparing divergent and convergent sums

Let $$(x_n)$$ be a monotonically decreasing sequence of positive real numbers that is also summable.

Let $$(y_n)$$ be a sequence of positive real numbers such that
$$sum_n x_n y_n$$ converges.

Let $$(z_n)$$ be a monotonically increasing sequence of positive real numbers such that $$sum_n x_n z_n =infty.$$

Assume that the sequences $$y_n$$ and $$z_n$$ are such that $$2^{-varepsilon y_n}$$ and $$2^{-varepsilon z_n}$$ are summable for every $$varepsilon>0.$$ Does it follow that there is some $$delta>0$$ such that

$$sum_n Big(2^{-varepsilon y_n}-2^{-varepsilon z_n}Big) ge 0 text{ for all } varepsilon in (0,delta)?$$

The motivation for this statement to be true is that $$z_n$$ should be larger most of the time than $$y_n$$ and we capture this most of the time by taking $$varepsilon$$ small.

## intuition – What intuitive meaning “determinant” of a divergency (divergent integral or series) can have?

I am working on the algebra of “divergencies”, that is, infinite integrals, series and germs.
So, I decided to construct something similar to determinant of a matrix of these entities.

$$det w=exp(operatorname{reg }ln w)$$
which is analogous to how determinant of a matrix can be expressed, except we take finite part (regularize) instead of taking trace.

Like determinant of a matrix, determinant of a divergency can be negative. It does not follow the requirements for a norm (Pythagorean theorem), and is not even continuous. Still, it has some usual properties, like $$1/det w=det 1/w$$.

Below is a table of some divergencies with their finite parts and determinants.

An interesting property is that the constant $$e^{-gamma}$$ often appears in the expressions for these determinants.

I wonder, what intuitively can indicate such determinant of a divirgent integral or series? Can it tell something about its properties?

$$begin{array}{cccccc} text{Delta form} & text{In terms of } tau, omega_+,omega_- & text{Finite part} & text{Integral or series form} & text{Germ form} &text{Determinant}\ pi delta (0) & tau & 0 & int_0^{infty } , dx;int_0^{infty } frac{1}{x^2} , dx & underset{xtoinfty}{operatorname{germ}} x;underset{xto0^+}{operatorname{germ}}frac1x&frac{e^{-gamma}}4 \ pi delta (0)-frac{1}{2} & omega _-;tau-frac{1}{2} & -frac{1}{2} & sum _{k=1}^{infty } 1 & underset{xtoinfty}{operatorname{germ}} (x-1/2) &e^{-gamma} \ pi delta (0)+frac{1}{2} & omega _+;tau+frac{1}{2} & frac{1}{2} &sum _{k=0}^{infty } 1 & underset{xtoinfty}{operatorname{germ}} (x+1/2) & e^{-gamma} \ 2 pi delta (i) & e^{omega_+}-e^{omega_-}-1 & 0 & int_{-infty }^{infty } e^x , dx & underset{xtoinfty}{operatorname{germ}} e^x \ & frac{tau ^2}{2}+frac{1}{24};frac{omega_+^3-omega_-^3}6 & 0 & int_0^{infty} x , dx;int_0^infty frac2{x^3}dx & underset{xtoinfty}{operatorname{germ}}frac{x^2}2;underset{xto0^+}{operatorname{germ}} frac1{x^2}\ & frac{tau ^2}{2}-frac{1}{24} & -frac1{12} & sum _{k=0}^{infty } k & underset{xtoinfty}{operatorname{germ}} left(frac{x^2}2-frac1{12}right) \ -pi delta”(0) &frac {tau^3}3 +fractau{12};frac{omega_+^4-omega_-^4}{12}& 0 & int_0^infty x^2dx;int_0^inftyfrac6{x^4}dx&underset{xtoinfty}{operatorname{germ}}frac{x^3}3;underset{xto0^+}{operatorname{germ}} frac2{x^3}\ pi^2delta(0)^2-pidelta(0)+1/4&omega_-^2&frac16&2 int_0^{infty } left(x-frac{1}{2}right) , dx+frac{1}{6}&underset{xtoinfty}{operatorname{germ}}B_2(x)&e^{-2gamma}\ pi^2delta(0)^2+pidelta(0)+1/4&omega_+^2&frac16&2 int_0^{infty } left(x+frac{1}{2}right) , dx+frac{1}{6}&underset{xtoinfty}{operatorname{germ}}B_2(x+1)&e^{-2gamma}\ pi^2delta(0)^2&tau^2&-frac1{12}&int_{-infty}^{infty } |x| , dx-frac{1}{12}&underset{xtoinfty}{operatorname{germ}}B_2(x+1/2)&frac{e^{-2gamma}}{16} \ &ln omega_++gamma&0&int_1^infty frac{dx}x;sum_{k=1}^infty frac1x -gamma&underset{xtoinfty}{operatorname{germ}}ln x\ -3pidelta”(0)-frac14 pidelta(0);pi^3delta(0)^3&tau^3&0&int_0^infty left(3x^2-frac1{4}right)dx&underset{xtoinfty}{operatorname{germ}}B_3(x+1/2)&frac{e^{-3gamma}}{64} \ frac{2pidelta(i)+1}{e-1}&e^{omega_-}&frac1{e-1}&frac1{e-1}+frac1{e-1}int_{-infty}^infty e^x dx&underset{xtoinfty}{operatorname{germ}} frac{e^x+1}{e-1}&frac1{sqrt{e}}\ frac{2pidelta(i)+1}{1-e^{-1}}&e^{omega_+}&frac1{1-e^{-1}}&frac1{1-e^{-1}}+frac1{1-e^{-1}}int_{-infty}^infty e^x dx&underset{xtoinfty}{operatorname{germ}} frac{e^x+1}{1-e^{-1}}&sqrt{e}\ &(-1)^tau&fracpi{2}&&&1\ end{array}$$

## the limit as n approaches infinity of a divergent series divided by n

If $$lim_{ ntoinfty} x_n = x < infty$$, prove that:
$$lim_{ntoinfty} frac{ x_1 + x_2 + ldots + x_n}{n} = x.$$

So far I deduced that since the sequence xn converges to x, the sum of the series will be divergent but I can’t think of anything else.