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Graf's book on hyperfunction theory says (page $36$) that

$$\frac1{(x-i0)^n}=\frac{(-1)^{n-1}\pi i}{(n-1)!}\delta^{(n-1)}(x)+\operatorname{fp}\frac1{x^n},$$

while the table of Fourier transforms says that the Fourier transform of $x^n$ is $2\pi i^n\delta^{(n)}(x)$ (here the Fourier transform is interpreted as ${\hat {f}}(\nu )=\int _{-\infty }^{\infty }f(x)e^{-i\nu x}\,dx$).

This gives us formally (at $x=0$) the following relation:

$$\int_{0^+}^\infty \frac1{t^{n+2}} dt=\frac1{(n+1)!}\int_0^\infty t^{n} dt$$

I wonder whether this relation is justified in any theory of divergent integrals, integral transforms, hyperfunctions or something else?

Is it justified philosophically?

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  • $\begingroup$ Your link is behind paywall. Is "$i0$" a typo? $\endgroup$
    – YCor
    Commented Oct 14, 2019 at 13:37
  • $\begingroup$ @YCor direct link: mobt3ath.com/uplode/book/book-35173.pdf This is not a typo, I interpret it as one-sided "limit" $\endgroup$
    – Anixx
    Commented Oct 14, 2019 at 13:41
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    $\begingroup$ @YCor I believe $\frac{1}{x -i0}$ is one notation for a particular element of the 1-dimensional family of distributions one can consider to represent $\frac{1}{x}$. The mapping $\phi \mapsto \int_{\mathbb{R}} \frac{1}{x}\phi \mathrm{d}x$ defines a linear functional on the space of smooth compactly supported functions that also vanish at $0$. This space is a closed hyperplane in $\mathcal{D}$, the space of compactly supported smooth functions, so the set of extensions to a distribution is 1-dimensional. $\endgroup$
    – Robert Furber
    Commented Oct 15, 2019 at 1:47
  • $\begingroup$ One can be obtained by taking the limit of the map $\phi \mapsto \int_{\mathbb{R}} \frac{1}{x-i\epsilon}\phi \mathrm{d}x$ as $\epsilon \to 0$, which I think is what is meant by $\frac{1}{x-0i}$, and another, $\frac{1}{x+0i}$ is obtained using $\frac{1}{x+i\epsilon}$. The average between the two is the Cauchy principal value $\mathrm{pv}\frac{1}{x}$, which can also be obtained through a two-sided limit in the usual way. $\endgroup$
    – Robert Furber
    Commented Oct 15, 2019 at 1:50
  • $\begingroup$ This can be generalized to $\frac{1}{x^n}$. You can also vary the space of test functions, to get compactly supported distributions, tempered distributions, or (compactly supported) hyperfunctions. $\endgroup$
    – Robert Furber
    Commented Oct 15, 2019 at 1:54

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I don't see where the $\frac{1}{(n+1)!}$ factor comes from.

If you take $II = \int_{a}^{b}\frac{1}{t^{n+2}}dt$ and perform the change of variables $t \to 1/x$, you get $$-\int_{1/a}^{1/b}\left(\frac{1}{x}\right)^{-n}dx$$ Taking $a>0$, $b>0$ lets one flip things around. As $x>0$, $\left(\frac{1}{x}\right)^{-n}$ is equal to $x^n$. Letting $a\to 0^{+}$ and $b\to\infty$, you get $$II = \int_{0}^{\infty} x^{n} dx$$ Purely formally, of course. But it is hard to see how that factor would 'sneak in'.

The way it can sneak in is by the process you use to derive things. If you go via a larger space (like $\mathbb{C}$ say, or a larger space of function, or an integral transform), it is then entirely possible to have that process 'see' things that are invisible from the purely extensional question (i.e. the values of the functions on exactly and only the positive reals).

Roughly, it boils down to:

  1. for convergent integrals, we have all sorts of theorems that say that the answer is unique, i.e. no matter what computation process is used, the answer will always be the same.
  2. for divergent integrals, it is necessarily the case that the answer is process-dependent. There are still uniqueness theorems, but they all say "given class of process X, then the answer is unique".
  3. there is no guarantee that two classes of processes will agree on divergent integrals.

The very important part about point 3 above is: you can't mix processes, else you will get inconsistencies. All interpretations of all identities must be within the same framework, and results between frameworks can't trivially be ported.

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  • $\begingroup$ No, I simply think the kind of variable change that you used is not applicable to divergent integrals. Rather the results produced by distribution theory and hpyperfunction theory are correct. $\endgroup$
    – Anixx
    Commented Nov 12, 2019 at 6:37
  • $\begingroup$ For instance if you apply variable change to $\int_0^\infty dx$ you can get arbitrary results. $\endgroup$
    – Anixx
    Commented Nov 12, 2019 at 6:54
  • $\begingroup$ When speaking about divergent integrals one should interpret $x$ not as an arbitrary variable but as a special symbol meaning the identity function $\operatorname{id}$ $\endgroup$
    – Anixx
    Commented Nov 12, 2019 at 8:56

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