Main Page

Basic idea
For a long time I was looking for a numerical system that would allow to compare infinite sets. In contrast to Cantor's approach that empathizes the possibility of on-to-one correspondence between sets, I want that adding an element to even an infinite set should increase its "quantity" and removing an element should decrease. For instance, 1,2,3,4,5... should have greater quantity than 1,2,4,5... Similarly I want more dense sets having greater quantity, that is, 1,2,3,4,5... having greater quantity than 1,3,5,7... and slmaller than 1,3/2,2,5/2,3,...

This way I came to the idea further generalization and considering divergent series and integrals. Let's suppose that any divergent integral represents an extended number. The following equivalence classes look natural in this case:

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

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

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

$$\int_{-\infty}^0 f(x) dx=\int_0^\infty f(-x) dx$$

where a,b,c,f(x) and g(x) take values from ℝ ∪ {−∞, +∞}.

Thus, any integral $$T=\int_a^b f(x) dx$$ represents an "extended" number. The integrals which are regularizable by a stable method like Cesaro or Abel (as opposed to Ramanujan or Dirichlet) are taken to be equal to their regularized sum: $$\int_0^\infty f(x)\,dx=\lim_{\epsilon\to 0}\int_0^\infty e^{-\epsilon x}f(x) \, dx$$ We also take integrals, regularizable with Cauchy principal value to be equal to that value.

Two integrals $$\int_0^\infty f(x) dx$$ and $$\int_0^\infty g(x) dx$$ are thus equal if

$$\lim_{\epsilon\to 0}\int_0^\infty e^{-\epsilon x}(f(x)-g(x)) \, dx=0$$

If $$\lim_{x\to\infty}f(x)=0$$, then $$\int_a^\infty f(x)dx=\int_{a+b}^\infty f(x-b)dx$$
 * Shift invariance for functions that do not diverge to infinity:


 * Area-preserving Laplace transform:

$$\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_t[t f(t)](x)dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x)dx$$

We also can equate some divergent series to the integrals:

$$\sum_{k=0}^\infty f(k)=\int_{-1/2}^\infty\sum_{k=0}^\infty\operatorname{rect}(x+k)f(k)dx$$

In our notation we will consider by definition $$\sum_{k=n}^\infty f(k)=\sum_{k=0}^\infty f(k)-\sum_{k=0}^{n-1}f(k)$$

Now, we postulate that the regularized value or the integral or corresponding series represents the regular part of an extended number, while the rest is the irregular part. Among the suitable regularization methods are Cesaro, Abel, Ramanujan, Borel, Dirichlet regulaization and some others (they agree with each other when applicable). We will denote the regularized value of an extended number $$w$$ as $$\operatorname{reg}\, w$$ ''

Particularly, very useful would be the Faulhaber's formula for Ramanujan's summation of analytic functions:

$$\operatorname{reg} \sum _{n=1}^{\infty} f(n)= -\sum_{n=0}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n,$$

where $$f^{(-1)}(x)$$ is natural integral.

We will use the following symbols for the three most key integrals and series:

$$\omega_+=\sum_{k=0}^\infty 1$$

$$\omega_-=\sum_{k=1}^\infty 1=\omega_+-1$$

$$\tau=\int_0^\infty dx=\omega_+-1/2=\omega_-+1/2$$ (this can also be formally denoted as τ = πδ(0) due to Fourier transform).

Exponentiation of non-standard numbers
Examining the Faulhaber's formula for Ramanujan's summation one can come to the following striking insight on the exponentiation of non-standard numbers.


 * $$\operatorname{reg}\,\omega_-^n=B_n$$
 * $$\operatorname{reg}\,\omega_+^{n}=B'_n$$

Where Bn are the first Bernoulli numbers and $$B^{'}_n$$ are the second Bernoulli numbers.

Indeed, we can see that $$\operatorname{reg}\,\omega_-=-1/2, \operatorname{reg}\,\omega_+=1/2, \operatorname{reg}\,\omega_-^2=1/6, \operatorname{reg}\,\omega_-^3=0$$ etc.

Given that Bernoulli numbers can be expressed through Hurwitz Zeta function, we can generalize:


 * $$\operatorname{reg}\,\omega_-^x=-x\zeta(1-x,0)

$$
 * $$\operatorname{reg}\,\omega_+^x=-x\zeta(1-x,1)=-x\zeta(1-x)$$

This allows to represent zeta function in exponential form:


 * $$(x-1)\zeta(x)= \operatorname{reg}\,\omega_-^{1-x}$$

or, more generally,



\operatorname{reg}(\omega_-+z)^n= B_n(z)$$


 * $$\operatorname{reg}(\tau+y)^x=-x\zeta(1-x,1/2+y)$$

Moreover, now any series containing Bernoulli numbers can be represented as power series over non-standard numbers.

There are the following relations:


 * $$\operatorname{reg}\,\tau^x = \operatorname{reg}\,\omega_+^x (2^{1 - x} - 1)$$


 * $$\operatorname{reg}\,(2\tau)^x = \operatorname{reg}\,\omega_+^x (2 - 2^x)$$

For x &gt; 1 ,


 * $$\operatorname{reg}\,\omega_-^x = \operatorname{reg}\,\omega_+^x $$

From the Riemann functional equation it follows:


 * $$\operatorname{reg}\,\omega_+^{-x}=\operatorname{reg}\frac{\omega_+^{x+1} 2^x\pi^{x+1}}{\sin(\pi x/2)\Gamma(x)(x+1)}$$

Expression for derivative using regular part
If $$f(x)$$ is analytic, the following holds:


 * $$f'(x)=\operatorname{reg}(f(\omega_++x)-f(\omega_-+x))=\operatorname{reg} \Delta f(\omega_-+x)$$


 * $$f'(x)=\operatorname{reg}(f(\omega_++x)-f(-\omega_++x))$$


 * $$f'(x)=\operatorname{reg}(f(-\omega_-+x)-f(\omega_-+x))$$

If $$f(x)$$ is odd,


 * $$\operatorname{reg}f(\omega_+)=-\operatorname{reg}f(\omega_-)= \frac12 f'(0)$$

Another consequence of Faulhaber's formula connects integral with the sum:


 * $$\operatorname{reg}\sum_{k=0}^\infty f(k)=-\operatorname{reg}\int_0^{\omega_-} f(x) dx$$

Standard part of improper integrals
Using the definition of integral as a limit of the sum, we can derive the formula for standard part of an improper integral.

$$\operatorname{reg}\int_0^\infty f(x)\,dx=\lim_{s\to0} \operatorname{reg} s\sum_{k=1}^\infty f(sk)$$

This can be implemented in Mathematica using the following code:

Particularly,

$$\operatorname{reg}\int_0^\infty e^x dx=-1$$

Modulus
One can define the modulus of non-standard numbers, by analogy with complex numbers:


 * $$\|w\|=\exp(\Re(\operatorname{reg}\ln w))$$

If so, the following holds:


 * $$\|\omega_+\|=e^{-\gamma}$$

This modulus is not a norm, and does not satisfy the triangle inequality.It is not continuous either.

Distribution form
Integral form of Dirac Delta function is


 * $$\delta\left(z\right)=\frac1{2\pi}\int_{-\infty}^{+\infty}e^{-ixz}dx

$$

It becomes $$\tau/\pi$$ at $$x=0$$.

Thus we can interpret extended numbers as derivatives of step-functions at the point of discontinuity, with step size determining the non-standard part and limits of derivatives from right and left determine the standard part.

Particularly,

$$\delta(0)=\tau/\pi$$

$$\operatorname{sign}'(0)=2\tau/\pi$$

Standard parts of some expressions
Given the above definitions, we have a lot of relations between trigonometric functions, for instance,


 * $$\operatorname{reg} \cos (z\omega_-)=\operatorname{reg} \cos (z\omega_+)=\frac z2 \cot \left(\frac z2\right)$$


 * $$\operatorname{reg} \cosh (z\omega_-)=\operatorname{reg} \cosh (z\omega_+)=\frac z2 \coth \left(\frac z2\right)$$


 * $$\operatorname{reg} \cos (z\tau)=\frac z2 \csc \left(\frac z2\right)$$


 * $$\operatorname{reg} \cosh (z\tau)=\frac{z}{2} \operatorname{csch}\left(\frac{z}{2}\right)$$


 * $$\operatorname{reg} e^{z\omega_-}=\frac{z}{e^{z}-1}$$


 * $$\operatorname{reg} \left(\frac{1}{\pi^2 \tau+\pi x}+\frac{1}{\pi^2  \tau-\pi x}\right)=(\sec x)^2$$


 * $$\operatorname{reg}\ln (\omega_-+z)=\psi(z)$$

Particularly,


 * $$\operatorname{reg}\ln \omega_+=-\gamma$$


 * $$\operatorname{reg}\frac1{\pi }\ln \left(\frac{\omega _+-\frac{z}{\pi }}{\omega _-+\frac{z}{\pi }}\right)=\cot z$$


 * $$\operatorname{reg} \frac1\pi\ln \left(\frac{\tau +\frac{z}{\pi }}{\tau -\frac{z}{\pi }}\right)=\tan z$$


 * $$\operatorname{reg}\sin (\omega_-+x) = \frac{1}{2} \cot \left(\frac{1}{2}\right) \sin x -\frac{1}{2} \cos x $$



Particularly,


 * $$\operatorname{reg}\sin \omega_-=-1/2$$,
 * $$\operatorname{reg}\sin \omega_+=1/2$$,
 * $$\operatorname{reg}\sin \tau=0$$


 * $$\operatorname{reg}\cos (\omega_-+x) = \frac{1}{2} \csc \left(\frac{1}{2}\right) \cos \left(\frac{1}{2}- x \right)$$




 * $$\operatorname{reg}\sin (a\omega_-+x) = \frac{a}{2} \cot \left(\frac{a}{2}\right) \sin x -\frac{a}{2} \cos x $$


 * $$\operatorname{reg}\cos (a\omega_-+x) = \frac{a}{2} \csc \left(\frac{a}{2}\right) \cos \left(\frac{a}{2}- x\right)$$


 * $$\operatorname{reg}\cos (\pi\tau+x)=-\frac\pi{2}\cos x$$


 * $$\operatorname{reg}\sin (\pi\tau+x)=-\frac\pi{2}\sin x$$

Relations between standard parts of trigonometric and inverse trigonometric functions

 * $$\operatorname{reg}\left(\cosh \left(2 x \omega _\pm\right)-1\right)=\operatorname{reg}\frac{x}{\pi} \operatorname{artanh}\left(\frac{x}{\pi\omega _\pm}\right)=\operatorname{reg}\frac{x}{\pi} \operatorname{arcoth}\left(\frac{\pi \omega _\pm}{x}\right)=x \coth (x)-1$$


 * $$\operatorname{reg}\frac{z}{2\pi }\ln \left(\frac{\omega _+-\frac{z}{2\pi }}{\omega _-+\frac{z}{2\pi }}\right)=\operatorname{reg} \cos (z\omega_-)=\operatorname{reg} \cos (z\omega_+)=\frac z2 \cot \left(\frac z2\right)$$

For regular z the following holds:


 * $$\tan z=\frac2\pi\operatorname{artanh} \frac{z}{\tau\pi}$$

Improper integrals of monomials
Following from Faulhaber's formula (for $$n\ge0$$),


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

Interpreting Fourier transform formally, for even $$n$$ we also have


 * $$\int_0^\infty x^n dx=i^n\pi\delta^{(n)}(0)$$

For $$n>1$$
 * $$\int_0^\infty \frac1{x^n} dx=\frac1{(n-1)!}\int_0^\infty x^{n-2} dx=\frac{\omega _+^{n}-\omega _-^{n}}{(n-1)n!}$$

Particularly,
 * $$\int_0^\infty 1 dx =\tau$$
 * $$\int_0^\infty x dx=\frac{\tau^2}2+\frac1{24}$$
 * $$\int_0^\infty x^2 dx=\frac{\tau^3}3+\frac{\tau}{12}$$

Using generalized limits (see below) one can write down the general formula for powers of $$\omega_\pm$$, for $$n>0$$:


 * $$\omega_-^n=\operatorname{gen}\lim_{x\to\infty}B_n(x)$$


 * $$\omega_+^n=\operatorname{gen}\lim_{x\to\infty}B_n(x+1)$$

Or in integral form:


 * $$\omega_-^n=B_n+n\int_0^\infty B_{n-1}(x)dx$$


 * $$\omega_+^n=B^*_n+n\int_0^\infty B_{n-1}(x+1)dx$$

They give us the general formulas:


 * $$(\omega_-+a)^n=\operatorname{gen}\lim_{x\to\infty}B_n(x+a)$$


 * $$(\omega_-+a)^n=B_n(a)+n\int_0^\infty B_{n-1}(x+a)dx$$

This in turn gives the formula for the exponential function:


 * $$e^{a+\omega _-}=\int_{-\infty}^{\infty } \frac{e^{a+t}}{e-1} \, dt+\frac{e^a}{e-1}$$

Generalized limits (germs)
We introduce generalized limits (germs) in the following way:

$$ \underset{x\to u^+}{\operatorname{germ}}f(x)=f(a)-\int_u^a f'(x)dx $$

where $$a>u$$

and

$$\underset{x\to u^-}{\operatorname{germ}}f(x)=f(a)+\int_a^u f'(x)dx$$

where $$a<u$$

This value can serve as a measure of the growth rate of a function.

Particularly, for $$n\ge0$$

$$\underset{x\to\infty}{\operatorname{germ}}x^n=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{n+1}$$

For odd $$n$$,

$$\underset{x\to\infty}{\operatorname{germ}}x^n=i^{n-1}\pi n\delta^{(n-1)}(0)$$

And

$$\underset{x\to 0^+}{\operatorname{germ}} \frac1{x^n}=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{(n+1)!}$$

Following Urs Graf, p.36, for odd $$n$$,

$$\underset{x\to 0^+}{\operatorname{germ}} \frac1{x^n}=\frac{i^{n-1}\pi\delta^{(n-1)}(0)}{(n-1)!}$$

The Cauchy principal value of an analytic function at a pole corresponds to the regular part of its generalized limits at the pole. Thus,

$$\underset{x\to0^\pm}{\operatorname{germ}}\Gamma(x)=-\gamma\pm\tau$$

$$\underset{x\to{-1}^\pm}{\operatorname{germ}}\Gamma(x)=\gamma-1\mp\tau$$

$$\underset{x\to{-2}^\pm}{\operatorname{germ}}\Gamma(x)=\frac{3}{4}-\frac{\gamma }{2}\pm\frac\tau 2$$

$$\underset{x\to{-3}^\pm}{\operatorname{germ}}\Gamma(x)=\frac{\gamma }{6}-\frac{11}{36}\mp\frac\tau 6$$

Conversion formula
Generalizing the above formulas we can write down the general formula for conversion of divergent integral representation to omega-tau representation:


 * $$\int_0^{\infty } f(x) \, dx=\operatorname{reg}\int_0^{\infty } f(x) \, dx+\int _{\omega_-}^{\omega_+} \int _0^x f(t) dtdx$$

Here the second term is always purely irregular. And, for a function f(x) such that it is equal to its Newton series expansion, the following formula holds:


 * $$\sum_{k=0}^{\infty } f(k) =\operatorname{reg}\sum_{k=0}^{\infty } f(k)+\int _{\omega_-}^{\omega_+} \int _0^x f(t) dtdx$$

where the regular part can be found as


 * $$\operatorname{reg} \sum _{k=0}^{\infty} f(k)=\frac{f(0)}{2}+i \int_0^{\infty } \frac{f(i t)-f(-i t)}{e^{2 \pi t}-1} \, dt$$

Some extended numbers
$$ \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/Modulus}\\ \pi \delta (0) & \tau  & 0 & \int_0^{\infty } \, dx;\int_0^{\infty } \frac{1}{x^2} \, dx & \underset{x\to\infty}{\operatorname{germ}} x;\underset{x\to0^+}{\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; \int_{1/2}^\infty dx; \int_0^\infty \frac{e^{-\frac{x}{2}} (x+2)}{2 x^2}dx;\int_0^2\frac1{x^2} dx& \underset{x\to\infty}{\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; \int_{-1/2}^\infty dx; \int_0^\infty \frac{4-e^{-\frac{x}{2}} (x+2)}{2 x^2} dx & \underset{x\to\infty}{\operatorname{germ}} (x+1/2) & e^{-\gamma}  \\ 2 \pi \delta (i) & e^{\omega_+}-e^{\omega_-}-1 & 0 & \int_{-\infty }^{\infty } e^x \, dx & \underset{x\to\infty}{\operatorname{germ}} e^x &(1)\\ & \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{x\to\infty}{\operatorname{germ}}\frac{x^2}2;\underset{x\to0^+}{\operatorname{germ}} \frac1{x^2}&(2)&\\ & \frac{\tau ^2}{2}-\frac{1}{24} & -\frac1{12} & \sum _{k=0}^{\infty } k & \underset{x\to\infty}{\operatorname{germ}} \left(\frac{x^2}2-\frac1{12}\right)&(3) \\ -\pi \delta''(0) &\frac {\tau^3}3 +\frac\tau{12};\frac{\omega_+^4-\omega_-^4}{12};& 0 & \int_0^\infty x^2dx;\int_0^\infty\frac6{x^4}dx&\underset{x\to\infty}{\operatorname{germ}}\frac{x^3}3;\underset{x\to0^+}{\operatorname{germ}} \frac2{x^3}&\\ \pi^2\delta(0)^2-\pi\delta(0)+1/4&\omega_-^2&\frac16&2 \int_0^{\infty } \left(x-\frac{1}{2}\right) \, dx+\frac{1}{6}&\underset{x\to\infty}{\operatorname{germ}}B_2(x)&e^{-2\gamma}\\ \pi^2\delta(0)^2+\pi\delta(0)+1/4&\omega_+^2&\frac16&2 \int_0^{\infty } \left(x+\frac{1}{2}\right) \, dx+\frac{1}{6}&\underset{x\to\infty}{\operatorname{germ}}B_2(x+1)&e^{-2\gamma}\\ \pi^2\delta(0)^2&\tau^2&-\frac1{12}&\int_{-\infty}^{\infty } |x| \, dx-\frac{1}{12}&\underset{x\to\infty}{\operatorname{germ}}B_2(x+1/2)&\frac{e^{-2\gamma}}{16} \\ &\left(\omega _++1\right) \ln \left(\omega _++1\right)-\omega _+ \ln \left(\omega _+\right)-1&0&\int_1^\infty \frac{dx}x;\sum_{k=1}^\infty \frac1x -\gamma;\int_0^\infty\frac{e^{-x}}{x}dx;\int_0^\infty\frac{x-\ln x-1}{(x-1)^2}dx&\underset{x\to\infty}{\operatorname{germ}}\ln x &\\ &\left(\omega _++1\right) \ln \left(\omega _++1\right)-\omega _+ \ln \left(\omega _+\right)+\gamma-1&\gamma&{\int_0^1 \frac{dx}x;\int_0^\infty\frac{1-e^{-x}}{x}dx;\sum_{k=1}^\infty \frac1x;\int_0^\infty\frac{dx}{x^2+x};\int_0^\infty\frac{x\ln x+1-x}{(x-1)^2 x}dx}&\underset{x\to\infty}{\operatorname{germ}}\ln (x e^\gamma);-\underset{x\to0^+}{\operatorname{germ}}\ln x &\\ -3\pi\delta''(0)-\frac14 \pi\delta(0);\pi^3\delta(0)^3&\tau^3&0&\int_0^\infty \left(3x^2-\frac1{4}\right)dx&\underset{x\to\infty}{\operatorname{germ}}B_3(x+1/2)&\frac{e^{-3\gamma}}{64} \\ \frac{2\pi\delta(i)+1}{e-1}&e^{\omega_-}&\frac1{e-1}&\frac1{e-1}+\frac1{e-1}\int_{-\infty}^\infty e^x dx&\underset{x\to\infty}{\operatorname{germ}} \frac{e^x+1}{e-1}&\frac1{\sqrt{e}}\\ \frac{2\pi\delta(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{x\to\infty}{\operatorname{germ}} \frac{e^x+1}{1-e^{-1}}&\sqrt{e}\\ &(-1)^\tau&\frac\pi{2}&&&1\\ \end{array} $$

Missing from the table above:

$$(1)=e^{\psi _e(\ln (e-1))}$$

$$(2)=e^{\psi\left(\frac12+\frac{i}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{i}{2\sqrt{3}}\right)}$$

$$(3)=e^{\psi\left(\frac12+\frac{1}{2\sqrt{3}}\right)+\psi\left(\frac12-\frac{1}{2\sqrt{3}}\right)}$$