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Often when asking about a regularized value of an integral or series, I encounter a negative reaction of the sorts that "regularization is what you define it".

But in practice if we consider some widely used regularization methods: analytic continuation, Cesaro, Abel, Borel, Ramanujan etc we always get the same results where more than one method is applicable.

So why we cannot talk about THE regularization rather than consider it not-well-defined thing?

For instance, consider the following integral:

$$\operatorname{reg} \int_{0}^{1}\frac{e^{1/x}}{x^2}\,dx=-e$$

I bet this regularized value is the natural regularization of this integral, regardless what (mutually consistent) methods one choses to apply (in this case one can come to the value via analytic continuation of $\int_0^1 \frac{e^{a/x}}{x^2} \, dx$ over $a$).

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Preliminary comment: I second the idea that, for certain integrals, there should be a "well-behaved" class of regularization methods which all give the same value - of course, the important thing then is to give a good description of such a class of regularizations.

Answer to your question: Yet, your claim that we can always agree on THE natural regularization is certainly too bold. I can't resist the sweet temptation to beat you with your own example ;-):

  • The regularization which you suggest for the integral $\int_0^1 \frac{e^{1/x}}{x^2} \,dx$ yields the value $-e$.

  • Now consider the integral $\int_0^1 e^{x^a}x^{a -1} \,dx$. For, say, $a > 0$ this yields $\frac{e-1}{a}$, which has a (unique) analytic continuation with respect to $a$ in $\mathbb{C} \setminus \{0\}$. If we choose $a = -1$, we obtain the value $1-e$, though.

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  • $\begingroup$ Brilliant example, but I somehow insist that the natural regularization of this integral is $-e$. This is based on the general rule that $\int_0^1 f(x) dx=\int_1^\infty \frac1{x^2}f(\frac1x)dx$, which in my view has precedence and well verified. $\endgroup$
    – Anixx
    Commented Jun 6, 2018 at 18:39
  • $\begingroup$ Using Borel we can see that $\operatorname{reg}\int_1^\infty e^x dx=-e$ This was actually the method I used to derive this regularization, not the analytic continuation method from the question. $\endgroup$
    – Anixx
    Commented Jun 6, 2018 at 18:52
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    $\begingroup$ @Anixx: Well, but we can also compute $\int_1^\infty e^x \, dx$ by analytic continuation of $\int_1^\infty \exp(x^z) x^{z-1} \, dx$ which yields $\frac{1-e}{z}$ for $z < 0$ (and which is actually the method from my answer, plugged into the formula $\int_0^1 f(x) \,dx = \int_1^\infty \frac{1}{x^2} f(\frac{1}{x^2}) \, dx$, after substituting $z = -a$). I'm not sure that I understand why you prefer one value or regularization method to the other one. $\endgroup$
    – Jochen Glueck
    Commented Jun 6, 2018 at 19:01
  • $\begingroup$ Well, so it seems, analytic continuation needs caution, because it can give results contradicting other methods (in this case, Borel). $\endgroup$
    – Anixx
    Commented Jun 6, 2018 at 19:04
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    $\begingroup$ @Anixx: I agree, and after all your comment seconds the "preliminary comment" in my answer: there is certainly a class of regularization methods which all yield the result $-e$ for your integral. The question is: is there a good description/characterization of such a class? $\endgroup$
    – Jochen Glueck
    Commented Jun 6, 2018 at 19:35

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