Lets address of the question if it is possible to make a sum over an uncontable set and discuss integration rules involving uncountably infinite constants.
I will provide introduction in very condensed form to get quicker to the essense.
==Conservative part (countable sets)==
First of all, let us introduce a concept of numerocity of a set, starting from the subsets of integers.
$N(S)=\sum_{k=-\infty}^\infty p_s(k)=p_s(0)+\sum_{k=1}^\infty p_s(k)+\sum_{k=1}^\infty p_s(-k)$
here, $p_s$ is the indicator function of $S$. As we can see, in general, numerocity is a divergent series, characterized by the rate of growth and regularized value.
The concept of numerocity is, unlike cardinality, additive. Numerocity of union of two non-intersecting sets is the sum of the numerocities of each one (Euclid's principle as opposed to Cantor-Hume principle implemented in cardinality).
Now, what if we want to generalize a numerocity to other countable sets, which are not necessarily subsets of integers? For instance, we need to count the numerocity of of the roots of a given function. It turns out, that the following expression of distributions often works: $\delta(f(x))|f'(x)|$. The following integral $\int_{(a,b)}\delta(f(x))|f'(x)|dx$ gives the number of roots of the function $f(x)$ on the interval $(a,b)$, so we take this integral (which also can be divergent if the numerocity of the roots is infinite) as the numerocity of the roots.
We will denote $\overline{\delta}(x)=\delta(f(x))|f'(x)|$ as the "[squarable delta function][1]", and generalize it to other cases via the definition that $\int_{S}\overline{\delta}(f(x))dx$ gives the number of roots of $f(x)$ on $S$.
Unlike the conventional Delta distribution, squarable delta function satisfies the property $\overline{\delta}(a x)=\overline{\delta}(x)$, so it allows piece-wise definition.
Via Fourier and Laplace transforms we can see that $\frac 1\pi\int_0^\infty dx=\overline{\delta}(0)$, so we can define it piecewise
$\overline{\delta}(x)={\begin{cases}\frac 1\pi\int_0^\infty dx,&{\text{if }}x=0\\ 0,&{\text{if }}x\ne 0.\end{cases}}$
or, more generally,
$\overline{\delta}(x)^p={\begin{cases}\frac p{\pi^p}\int_0^\infty x^{p-1}dx,&{\text{if }}x=0\\ 0,&{\text{if }}x\ne 0.\end{cases}}$
Here we introduce some more notations. First, we introduce the constant $\omega=\pi \overline{\delta}(0)=\int_0^\infty dx$. We can see this constant as the germ at infinity of the function $f(x)=x$. Since the constant is equal to $\frac12 \sum_{k=-\infty}^\infty1$, it is half of the numerocity of integers, equal to the numerocity of even or odd numbers: $\omega=1/2 N(\mathbb Z)$. This $\omega$ is the most simple infinite quantity, which can be encountered in the theory of formal power series or Hardy fields. Powers of $\omega$ are given as $\omega^p=p\int_0^\infty x^{p-1}dx$.
Using this constant, we can express numerocities in integral form: $N(S)=\frac1\pi\int_S p_s(x)\omega dx$.
Second notation to introduce is $u(x)$, a unit impulse function, or discrete delta function: $u(x)=0^{|x|}$, it is equal to $1$ at zero and otherwise, $0$. So, $\overline{\delta}(x)=u(x)\omega/\pi$.
Third note on notation is that we will understand $\int_a^b=\frac12 \int_{(a,b)}+\frac12\int_{[a,b]}$, in other words, the limits of an integral are half-included in the set over which we integrate.
Using these new notations we can write down the general rule for $\omega$:
$\int_{-1}^1 u(x)f(\omega ) \, dx=f'(\omega )$ for any analytic $f$. For functions, satisfying the requirements of Hardy filds (that is, which have germs at infinity), it is also equal to $f'(0)+\int_0^\infty f''(x)dx$.
An interesting consequence is that $\int_{-1}^1 u(x)e^\omega dx=e^\omega$.
==Advanced part (uncountable sets)==
A question: what is the numerocity of an uncountable set, say an interval $[0,1)$?
We can express this numerocity using an integral:
$N([0,1))=\int_0^1 \overline{\delta}(0) dx=\frac1\pi\int_0^1 \omega dx$
But this quantity cannot be represented as analytic function or power series of $\omega$ or germ of a real function at infinity because otherwise it could be equal to a numerocity of a countable set. In this sense, it is different from cardinal arithmetic, where $2^{\aleph_0}=\aleph_1$.
In this light, we introduce another constant, $\alpha$, as a numerocity of interval $[0,\pi)$:
$\alpha=N([0,\pi))=\pi N([0,1))=\int_0^\pi \overline{\delta}(0)dx=\int_0^1 \omega dx$
For some reason, chosing the interval $[0,\pi)$ instead of $[0,1)$ simplifies further expressions.
The regularized value (finite part) of numerocity of a set is equal to that set's Euler's characteristic, thus, the regularized value of $\alpha$ is $0$.
We can now express the numerocity of reals: $N(\mathbb{R})=N(\mathbb Z)\cdot N([0,1))=\frac{2\omega\alpha}\pi=\frac1\pi\int_{-\infty}^\infty \omega dx$.
Like $\omega$, $\alpha$ does not behave under integrals like a finite constant. We need to derive rules of dealing with this constant under integrals.
From Laplace transform we have:
$\int_{-\infty }^{\infty } \left(
\begin{cases}
f(\omega/\pi) & x=0 \\
0 & x\neq 0 \\
\end{cases}
\right) \, dx=f'(0)+\frac1\pi\int_0^{\infty } f''(x) \, dx=f'( \omega/\pi). \tag1$
This gives a rule for integration:
$\int_0^1 f(\omega ) \, dx=\alpha f'(\omega )+\operatorname{reg}f(\omega)\tag2$
where $\operatorname{reg}f(\omega)$ is the finite part of the germ of $f(x)$ at infinity.
More generally,
$\int_0^1 f(\omega ) g(x) \, dx=\left(\alpha f'(\omega )+\operatorname{reg}f(\omega)\right) \int_0^1 g(x) \, dx \tag3$
==Some examples==
Introducing yet another constant $\lambda=-\ln\omega-\gamma=-\int_0^1\frac1x dx$ (which due to the last expression we also can call "logarithm of zero", it is negatively infinite and has regularized value $-\gamma$), from the rule $(3)$ we can derive interesting equalities:
$\int_0^1 \frac{\omega }{x} \, dx=-\lambda\alpha$
$\int_0^{\infty } \ln \omega \, dx= \alpha$
$\int_0^{\infty } \lambda \, dx=-\alpha-\gamma \omega$
$\int_0^{\infty } | \tan x| \, dx=-\frac{\lambda\omega}\pi$
