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let be $ f(x)= x^{a} $ with $ -1<a<0 $ if i use Euler Maclaurin summation formula

$$ \int_{1}^{\infty}dxx^{a}= \frac{1}{2}+ \sum_{n=2}^{\infty}n^{a}+\sum_{r=1}^{\infty}\frac{B_{2r}}{2r!}D^{2r-1}f(1) $$

then my doubt is this

i can regularize the divernget series $$ \sum_{n=2}^{\infty}n^{a}= \zeta (-a,2) $$

so the divergent integral would be $ \int_{1}^{\infty}x^{a}dx =\zeta (-a,2)+constant $

but if i use Hadamard finite part so $ F.p \int_{1}^{\infty}x^{a}dx = \frac{-1}{a+1} $

would i get the zeta regularized value for the divergent series ?? then what regularization is CORRECT does the Euler maclaurin summation formula works in both senses for series and for itnegrals even they are divergent.

Jose Garcia
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  • I think it is best to truncate the sum or integral and manipulate the finite sums and integrals before removing the truncation with limits. – robjohn Jan 18 '14 at 14:13
  • I think you have a problem in the formula $$\int_1^\infty x^a,\mathrm{d}x=\zeta(-a,2)+\text{constant}$$ because the sum $$\sum_{r=1}^\infty\frac{B_{2r}}{(2r)!}D^{(2r-1)}f(1)$$ is not a constant in $a$. – robjohn Jan 18 '14 at 14:20

1 Answers1

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In this answer, I say

The Euler-Maclaurin Sum formula says that, as a function of $n$, $$ \sum_{k=1}^n\frac{1}{k^z}=\zeta^\ast(z)+\frac{1}{1-z}n^{1-z}+\frac12n^{-z}+O\left(zn^{-1-z}\right)\tag{1} $$ for some $\zeta^\ast(z)$. Note that for $\mathrm{Re}(z)>1$, $\zeta^\ast(z)=\zeta(z)$.

For all $z\in\mathbb{C}\setminus\{1\}$, define $$ \zeta_n(z)=\sum_{k=1}^n\frac{1}{k^z}-\frac{1}{1-z}n^{1-z}\tag{2} $$ Note that each $\zeta_n$ is analytic and equation $(1)$ says that $$ \zeta_n(z)=\zeta^\ast(z)+\frac12n^{-z}+O\left(zn^{-1-z}\right)\tag{3} $$ which says that for $\mathrm{Re}(z)>0$, $$ \lim_{n\to\infty}\zeta_n(z)=\zeta^\ast(z)\tag{4} $$ and the convergence is uniform on compact subsets of $\mathbb{C}\setminus\{1\}$. Thus, the $\zeta^\ast(z)$ defined in $(4)$ is analytic and agrees with $\zeta(z)$ for $\mathrm{Re}(z)>1$. Thus, $\zeta^\ast(z)$ is the analytic continuation of $\zeta(z)$ for $\mathrm{Re}(z)>0$.

In this manner, the Euler-Maclaurin Sum Formula allows us to analytically continue $\zeta(z)$ as far left of $\mathrm{Re}(z)=1$ as we wish.

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robjohn
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  • thanks :) however i would more interested in regularizing divegent integrals $ \int_{0}^{\infty}x^{a}dx $ from zeta regularization $ \zeta (-a) $ – Jose Garcia Jan 31 '14 at 16:41
  • I am not sure precisely what you mean. However, if we try to split up $$ \begin{align} \int_0^\infty x^a,\mathrm{d}x &=\overbrace{\int_0^1x^a,\mathrm{d}x}^{a\gt-1} +\overbrace{\int_1^\infty x^a,\mathrm{d}x}^{a\lt-1}\ &=\frac1{a+1}-\frac1{a+1} \end{align} $$ My guess is that you won't find a useful regularization of this integral. – robjohn Jan 31 '14 at 19:10