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In this answer by Carlo Beenakker he cites the following regularization formula: $$\int_0^\infty x^p\,dx\mathrel{"="}\frac{(-1)^{p+1}}{(p+1)(p+2)},\;\;p=0,1,2,\dotsc,$$ citing Tafazoli - Calculation of the vacuum energy density using zeta function regularization.

The formula cries manifestly wrong to me. In my opinion, the correct regularization formula is $$\operatorname{reg} \int_0^\infty x^p dx=\frac{B_{p+2}(1)-B_{p+2}(0)}{(p+1)(p+2)},$$ which essentially gives $0$ for positive $p$. Indeed, the zero value for regularization of integrals of monomials from zero to infinity can be verified with Mathematica, using the following code:

f[x_] := x^p; Limit[s Sum[f[s x],{x,1,Infinity},Regularization->"Dirichlet"],s->0]

Out := ConditionalExpression[0, p > -1]

This code uses Dirichlet regularization, which is essentially the same as Zeta regularization, which the linked paper claims to use.

Moreover, applying Fourier and Laplace transforms hint at the same regularization value, $0$.

That said, I wonder whether I missed something (like assumptions) in the linked paper? Is this result considered controversial at all?

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  • $\begingroup$ The derivation of the formula is in the appendix of the paper (at the link above), with appropriate references. The equal sign is equality in the sense of analytical continuation. $\endgroup$
    – S Taf
    Commented Jun 6, 2021 at 16:46
  • $\begingroup$ @STaf I think the paper has some kind of a mistake or bad generalization/definition. I wonder what other sources may point to this result (I am actually sure the regularization of the integrals of monomials should be zero, from Faulhaber's formula, Ramanujan's formula, Mathematica verification and manual calculation of the difference between the integral and corresponding series, which is always opposite to the regularized sum of the corresponding series, so the integral should be zero). $\endgroup$
    – Anixx
    Commented Jun 6, 2021 at 19:01
  • $\begingroup$ @STaf for instance, it is well known that $\operatorname{reg}\sum_{k=1}^\infty 1=-1/2$. Since this sum differs from integral $\int_0^\infty dx$ by half of a unit square (which the sum lacks), the integral's regularized value is $0$. But from this linked formula it should be $-1/2$ (which coincides with the series, not the integral). $\endgroup$
    – Anixx
    Commented Jun 6, 2021 at 19:06
  • $\begingroup$ @STaf Just get it is you who are the author. You may be interested to look through my thoughts on this issue: exnumbers.miraheze.org/wiki/… The approach provides for complete expression of integrals of monomials in terms of other divergent integrals $$\int_0^\infty x^p dx=\frac{\omega _+^{p+2}-\omega _-^{p+2}}{(p+1)(p+2)},$$ and the expression for regularization arises naturally $\text{reg} \int_0^\infty x^p dx=\frac{B_{p+2}(1)-B_{p+2}(0)}{(p+1)(p+2)}$. Most of these formulas are based of Faulhaber and Euler-Maclaurin formulas $\endgroup$
    – Anixx
    Commented Jun 6, 2021 at 19:39
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    $\begingroup$ @STaf After looking through your paper, I think I see some misapproaches you took. When replacing the integral with the series, you cannot just make a sum of all elements of the integral. The elements of the integral should have limits centered symmetrically around the summation points. For instance, $\sum_0^\infty 1=\int_{-1/2}^\infty dx$. The first summation point is $0$, so the first integral "brick" should be from $-1/2$ to $1/2$, not from $0$. Something like this: $$\sum_{k=0}^\infty f(k)=\int_{-1/2}^\infty\sum_{k=0}^\infty\operatorname{rect}(x+k)f(k)dx$$ $\endgroup$
    – Anixx
    Commented Jun 6, 2021 at 19:56

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