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Has anybody ever proposed a generalization of integration in which an integral of a function that has non-zero values only on a countable set produces non-zero (but maybe infinitesimal) result?

For instance, say, something like this (but not necessarily exactly like this):

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

(with corresponding generalization of the delta function)?

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    $\begingroup$ I guess you discount integration against an atomic measure …. $\endgroup$
    – LSpice
    Mar 31, 2017 at 13:46

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In some sense, that's what the theory of tempered distributions provides.

If $f$ is integrable, we have

$$\hat{f}(0) = \int_{\bf R} f(x) \, dx$$

If $f$ is a tempered distribution, the expression $\hat{f}$ still makes senses as a distribution and depends in a nice way on $f$. We may then quotient ${\mathscr S}'$ by the set of continuous functions that are zero at zero. This quotient contains the set ${\bf R}$ of standard values of integrals of integrable functions and gives a new target space for this extended integral.

So for example we get $$\int_{\bf R} \sin(x)\, dx = i\pi(\delta_1 -\delta_{-1})$$ $$\int_{\bf R} x^k\, dx = 2\pi i^k \delta_0^{(k)}$$ and so on.

EDIT: Thinking about it, maybe you are refering to integration as defined in nonstandard analysis? See the following link.

https://math.stackexchange.com/questions/464565/definition-of-the-integral-in-non-standard-calculus

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  • $\begingroup$ I am sorry, there was a typo in the question, that makes it totally wrong! Now fixed. The function under integral was wrong $\endgroup$
    – Anixx
    Mar 31, 2017 at 11:39
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    $\begingroup$ Indeed, this does not really answer your question. I was adressing the case when the function is too "large" (ie the integral of the absolute value is infinite) but you are interested by the case when the function is too "small", if I understand well. $\endgroup$
    – coudy
    Mar 31, 2017 at 11:40
  • $\begingroup$ Yes, see the fixed question. For f(x)=1 at x=0 but otherwise 0, the standard Fourier transform is zero. I wonder whether one considered a generalization where it is not zero. $\endgroup$
    – Anixx
    Mar 31, 2017 at 11:43
  • $\begingroup$ I am not aware of such theory. $\endgroup$
    – coudy
    Mar 31, 2017 at 11:45

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