I am currently working on a similar system. But your properties (2) and (3) would not work and need change.

Instead, the following properties would work much better:

$$\int_a^c f(x) dx=\int_a^b f(x)dx+\int_b^c f(x)dx\tag{1}$$

$$\int_a^b (f(x)+g(x)) dx=\int_a^b f(x)dx+\int_a^b g(x)dx\tag{2}$$

$$\int_a^b c f(x) dx =c \int_a^b f(x) dx\tag{3}$$

$$\int_{-\infty}^0 f(x) dx=\int_0^\infty f(-x) dx\tag{4}$$

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 Cesaro integrable are taken to be equal to their Cesaro sum:

$$\int_0^\infty f(x)\,dx=\lim_{\lambda \to \infty }\int _{0}^{\lambda }\left(1-{\frac {x}{\lambda }}\right)^{\alpha }f(x)\,dx\tag{6}$$

if there is such $\alpha\ge0$ that the above integral converges.

Two integrals $\int_0^\infty f(x) dx$ and $\int_0^\infty g(x) dx$ are thus equal if for some $\alpha\ge0$

$$\lim_{\lambda \to \infty }\int _{0}^{\lambda }\left(1-{\frac {x}{\lambda }}\right)^{\alpha }(f(x)-g(x))\,dx=0$$

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 (mentioned above), Abel, Ramanujan, Borel, Zeta 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:

$$\operatorname{reg} \sum _{n=0}^{\infty} f(n)= -\sum_{n=1}^{\infty} \frac{f^{(n-1)} (0)}{n!} B_n \tag{7}$$

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 interpreted as $\tau=\pi\delta(0)$ due to Fourier transform).

By interpreting formula (7) as a Taylor series, we come to a formula which allows to generalize the analytic functions to extended numbers (at least in the sense of determining the regular part of the result):
$$\operatorname{reg} f'(\omega_-+z)= \Delta f(z)\tag{8}$$ and in particular, to the generalizations of powers of our key series:
$$\operatorname{reg}\omega_-^n=B_n\tag{9}$$
$$\operatorname{reg}\omega_+^n=B^*_n\tag{9a}$$
where $B^*$ are the second Bernoulli numbers (those which have $B^*_1=1/2$ ).
A more general formula reveals the role of the Hurwitz zeta function:
$$\operatorname{reg}(\omega_-+x)^a= B_a(x)=-a\zeta(1-a,x)$$

Based on formula (7) we even can derive an **expression for a derivative of an analytic function which does not use limits**:

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

which works for any regular $x$.

Also, since many series expansions of trigonometric functions use Bernoulli numbers, we can interpret them as regular parts of similar series but involving extended numbers. This way, and using formula (8) we can obtain the following relations:

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

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

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

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

Particularly, $$\operatorname{reg}\sin \omega_-=-1/2;$$ $$\operatorname{reg}\sin \omega_+=1/2;$$ $$\operatorname{reg}\ln \omega_+=-\gamma;$$ $$\operatorname{reg} e^{\omega_-}=\frac{1}{e-1};$$ $$\operatorname{reg} e^{\omega_+}=\frac{e}{e-1}.$$

Another notable thing is the **possibility to express trigonometric functions via inverse trigonometric or logarithms**:

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

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

$\operatorname{reg}\frac{x}{\pi} \operatorname{arccoth}\left(\frac{\pi \omega _+}{x}\right)=x \coth (x)-1$

This, combined with the delta function form of $\tau=\pi\delta(0)$ can explain why logarithm has its imaginary part represented as a step function (which is integral of delta function).

Following from Faulhaber's formula,

$$\int_0^\infty x^n dx=i^n\pi\delta^{(n)}(0)=\frac{\left(\tau +\frac{1}{2}\right)^{n+2}-\left(\tau -\frac{1}{2}\right)^{n+2}}{(n+1)(n+2)}=\frac{\omega _+^{n+2}-\omega _-^{n+2}}{(n+1)(n+2)}$$

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

We introduce generalized limits in the following way:

$$
\operatorname{gen}\lim_{x\to u^+}f(x)=f(a)+\int_u^a f'(x)dx
$$

where $a>u$

and

$$\operatorname{gen}\lim_{x\to u^-}f(x)=f(a)+\int_a^u f'(x)dx$$
where $a<u$

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

Particularly, for $n\ge0$

$$\operatorname{gen}\lim_{x\to\infty}x^n=0^n+i^{n-1}\pi n\delta^{(n-1)}(0)=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{n+1}$$

Since (following Urs Graf, p.36)

$$\operatorname{gen}\lim_{x\to 0^+} \frac1{x^n}=0^n+\frac{i^{n-1}\pi\delta^{(n-1)}(0)}{(n-1)!}=\frac{\omega _+^{n+1}-\omega _-^{n+1}}{(n+1)!}$$

The last formula gives us

$$\int_{0^+}^\infty \frac1{x^n} dx=\frac1{(n-1)!}\int_0^\infty x^{n-2} dx=\frac{\omega _+^{n}-\omega _-^{n}}{(n-1)n!}$$

Particularly,

$$\int_{0^+}^\infty \frac1{x^2}=\tau$$

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

$$\operatorname{gen}\lim_{x\to0^\pm}\Gamma(0)=-\gamma\pm\tau$$

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

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

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

$$\operatorname{gen}\lim_{x\to1^\pm}\zeta(x)=\gamma\pm\tau$$

For Bernoulli polynomials we have these noticeable formulas:

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

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

And more generally,

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

The system forms an integral domain which can be extended to form a field.
Still, there are many white areas, particularly, there is no straightforward way to transform a divergent integral into an expression in terms of $\omega_\pm$ and vice versa. There is no formula to construct an integral representing a product of two integrals, etc.