# Did anyone ever propose the distinction between "divergent to infinity" as opposed to "divergent but with finite average"?

There are different regularization methods that allow us to ascribe finite values to divergent integrals, series or sequences.

Still, in my view there is fundamental difference between divergent integrals or series that diverge to infinity, monotonously growing as opposed to those which have infinite parts of opposite signs that cancel each other or can be averaged so to arrive at finite values. I would call them strongly divergent and weakly divergent. If we apply an absolute value to a function under a weakly divergent integral, it becomes strongly divergent. The concept is similar to absolute and non-absolute convergence.

Examples of integrals that "diverge" but can be averaged (weakly divergent):

$$\int_0^\infty\sin x dx=1$$ (Cesaro average)

$$\int_{-1}^1 \frac1xdx=0$$ (Cauchy mean value)

$$\int_0^\infty \left(x-\frac2{x^3}\right)dx=0$$

(applying area-preserving operator $$\mathcal{L}_t[t f(t)](x)$$ to one term gives the other with opposite sign, so they cancel each other)

Examples of strongly-divergent integrals:

$$\int_{-1}^1 \frac1{x^2}dx=2\pi\delta(0)-2$$

Using Hadamard finite part can be regularized to $$-2$$.

$$\int_0^\infty 1 dx=\pi\delta(0)$$

Can be obtained from the previous one using the area-preserving operator, regularizes to $$0$$.

$$\int_0^1 \frac1x dx$$

Using Ramanujan regularization of its Riemann sum (and other methods), can be regularized to $$\gamma$$.

$$\sum_{k=0}^\infty k$$

(can be regularized to $$-\frac1{12}$$ using Ramanujan's, Dirichlet's or Zeta regularization)

Thus, all summation methods can be divided into two very distinct parts: those that somehow average the result, attempting to cancel the infinite parts and those that do not make such attempts, but extract the finite part, dropping the infinite part.

Was such classification of summation methods ever been proposed?

• I'm not convinced one can really make these distinctions. It seems to me one can justify $\sum_{k=0}^\infty k=17$ or $-41$ or $-1/12$ equally well, and the same for any divergent quantity with other weak definitions ad hoc. Apr 28 at 9:38
• @PietroMajer oh no, please, regularization techniques are numerous, well established and widely used. Apr 28 at 9:40
• A minor comment: it is better to use some other synonym of "hard", e. g. "strong", since "hardly" has the connotation of "rarely". Apr 28 at 9:41
• @auniket You are right. But maybe there is already some term for it. Apr 28 at 9:42
• @PietroMajer That’s not correct. There’s a good reason why $\sum_{k=1}^\infty k = -1/12$ is the unique answer. Apr 28 at 17:27