My gut feeling is that

$\int_0^\infty (1-\frac1{x^2})dx=0$

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

$\int_0^\infty (x^2-\frac6{x^4})dx=0,$

etc, and in general,

$\int_0^\infty (x^k-(k+1)!x^{-(k+2)})dx=0,$

is the natural way to regularize these divergent integrals.

The approach is based on Laplace transform (it allows to transform a divergent integral of a function around a pole into a divergent integral of an unbounded function at infinity).

I wonder, whether this approach is known and used anywhere? Does it possess any properties that would justify its "naturalness"?

  • $\begingroup$ Could you add a precise definition, or at least a precise reference, for the meaning of "$\int_0^\infty f(x)dx$" in the case you are interested? $\endgroup$ Apr 9, 2021 at 9:24
  • 1
    $\begingroup$ @PietroMajer it seems this requires generalization. The operator $\mathcal{L}_t[t f(t)](x)$ preserves the area under the curve. So if we assume it still holds for divergent integrals, we can define classes of equivalence of them. Besides, this is supported by the theory of hyperfunctions, etc. $\endgroup$
    – Anixx
    Apr 9, 2021 at 9:27

2 Answers 2


Depends on how you choose to define it :)

A natural way is to define $\int_0^{\infty} = \lim_{a\rightarrow0} \int_a^{1/a}$, in which case: \begin{eqnarray*} \int_0^{\infty}\left(1-\frac1{x^2}\right)dx &=& \lim_{a\rightarrow0} \int_a^{1/a} \left(1-\frac1{x^2}\right)dx \\ &=& \lim_{a\rightarrow0} \left(x+\frac1x\right|_a^{1/a} \\ &=& \lim_{a\rightarrow0} 0 \\ &=& 0 \end{eqnarray*}

If you instead for some reason wanted $\int_0^{\infty} = \lim_{a\rightarrow0} \int_a^{1+1/a}$, this is perfectly valid for convergent integrals and will get the same answer, but for your integral you would now get the value of 1. Maybe this definition does not look as natural as the original one, but one can easily do natural operations and manipulations on the original expression and end up with a different answer like that.

This is very different from the study of divergent series (see Hardy's book) where you can often show that there's a canonical value you can give to a divergent series (i.e. the now famous $1+1/1+1/3+1/14+\cdots=-1/12$)

  • $\begingroup$ Well, my question was exactly because I feel this IS the canonical value. For instance, due to Laplace transform. $\endgroup$
    – Anixx
    Mar 7, 2021 at 19:17
  • 2
    $\begingroup$ Did you mean $1+2+3+4+\ldots = -1/12$? $\endgroup$
    – user76284
    Apr 8, 2021 at 5:52

Warning: The following contains formal manipulations that ignore convergence.


\begin{align} \mathrm{regularized} \int_0^\infty \mathrm{d}x^s = 0 \end{align} for all $s$ such that $\Re(s) \neq 0$.

“Proof” 1:

\begin{align} \int_0^\infty \mathrm{d}x^s &= \int_0^1 \mathrm{d}x^s + \int_1^\infty \mathrm{d}x^s \\ &= 1 - 1 \\ &= 0 \end{align}

where we simultaneously assumed $\Re(s) > 0$ and $\Re(s) < 0$ for the first and second integrals, respectively.

“Proof” 2:

Let \begin{align} f_\pm(\eta) = \mathrm{e}^{-\eta} \left( 1 \pm \frac{\eta}s \right) \end{align}

Case $\Re(s) > 0$: \begin{align} \int_0^\infty x^{s-1} \mathrm{d}x &= \int_0^\infty \lim_{\varepsilon \downarrow 0} x^{s-1} f_-(\varepsilon x) \mathrm{d}x \\ &\stackrel{!}{=} \lim_{\varepsilon \downarrow 0} \int_0^\infty x^{s-1} f_-(\varepsilon x) \mathrm{d}x \\ &= \lim_{\varepsilon \downarrow 0} 0 & \Re(s) > 0, \Re(\varepsilon) > 0 \\ &= 0 \end{align}

Case $\Re(s) < 0$: \begin{align} \int_0^\infty x^{s-1} \mathrm{d}x &= \int_0^\infty \lim_{\varepsilon \downarrow 0} x^{s-1} f_+(\varepsilon x^{-1}) \mathrm{d}x \\ &\stackrel{!}{=} \lim_{\varepsilon \downarrow 0} \int_0^\infty x^{s-1} f_+(\varepsilon x^{-1}) \mathrm{d}x \\ &= \lim_{\varepsilon \downarrow 0} 0 & \Re(s) < 0, \Re(\varepsilon) > 0 \\ &= 0 \end{align}

See here for more context about this regulator-based approach.

Applying linearity of integration yields your equalities.

  • $\begingroup$ "I'm inclined to believe that" - I agree with that. But the question was not exactly about regularization, but rather the idea that the infinite parts cancel each other, so that the values of the integrals are finite. $\endgroup$
    – Anixx
    Apr 8, 2021 at 7:40
  • $\begingroup$ By the way, this is interesting result. Why for any $s$ except from that line the regularization is zero, while for $s=-1$ the regularization is $\gamma$? What about other powers from that line? $\endgroup$
    – Anixx
    Apr 8, 2021 at 15:02
  • $\begingroup$ @Anixx I streamlined the answer somewhat. I'm not sure what the answer is for points on the excluded line. Maybe another regularization can reveal the answer. $\endgroup$
    – user76284
    Apr 9, 2021 at 8:31

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