Timeline for How is this expression for the regularization of integrals of monomials, given in a paper, justified? How strong is argument in favor?
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| Jul 21, 2021 at 16:33 | comment | added | Anixx | @STaf the only rule besides linearity you can use to transform divergent integrals are these area-preserving Laplace-based transforms: $\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_t[t f(t)](x)dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x)dx$. They keep the integrals equivalent, including the regularized values. | |
| Jul 21, 2021 at 16:30 | comment | added | Anixx | @STaf also, regarding your result $\operatorname{reg}\int_0^\infty x^2\sqrt{1+x^2}\,dx=-\frac{1}{10}$, change of variables does not work with divergent integrals with improper bounds. It does not keep the regularized value. Consider, for instance, $\int_1^\infty \frac1x dx$, which regularizes to $0$ and $\int_2^\infty \frac1t dt$ which has regularized value of $-\ln 2$. The later can be obtained from the former via $t=x/2$. | |
| Jun 6, 2021 at 19:56 | comment | added | Anixx | @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: $$\sum_{k=0}^\infty f(k)=\int_{-1/2}^\infty\sum_{k=0}^\infty\operatorname{rect}(x+k)f(k)dx$$ | |
| Jun 6, 2021 at 19:39 | comment | added | Anixx | @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 | |
| Jun 6, 2021 at 19:06 | comment | added | Anixx | @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). | |
| Jun 6, 2021 at 19:01 | comment | added | Anixx | @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). | |
| Jun 6, 2021 at 17:19 | review | Close votes | |||
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| Jun 6, 2021 at 16:46 | comment | added | S Taf | 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. | |
| Jun 6, 2021 at 13:57 | history | edited | Anixx | CC BY-SA 4.0 |
added 107 characters in body; edited title
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| Jun 6, 2021 at 13:57 | history | edited | LSpice | CC BY-SA 4.0 |
Name of "this paper"
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| Jun 6, 2021 at 11:58 | history | edited | YCor |
edited tags
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| Jun 6, 2021 at 11:39 | history | asked | Anixx | CC BY-SA 4.0 |