Volkenborn integral

In mathematics, in the field of p-adic analysis, the Volkenborn integral is a method of integration for p-adic functions.

Definition

Let

${\displaystyle f:\mathbb {Z} _{p}\rightarrow {\mathbb {\mathbb {C} }}_{p}}$

be a function from the p-adic integers taking values in the p-adic numbers. The Volkenborn integral is defined by the limit, if it exists:

${\displaystyle \int _{{\mathbb {Z}}_{p}}f(x)\,{\rm {d}}x=\lim _{n\to \infty }{\frac {1}{p^{n}}}\sum _{x=0}^{p^{n}-1}f(x).}$

More generally, if

${\displaystyle R_{n}=\left\{x=\sum _{i=r}^{n-1}b_{i}x^{i}|b_{i}=0,\ldots ,p-1{\text{ for }}r

then

${\displaystyle \int _{K}f(x)\,{\rm {d}}x=\lim _{n\to \infty }{\frac {1}{p^{n}}}\sum _{x\in R_{n}\cap K}f(x).}$

This integral was defined by Arnt Volkenborn.

Examples

${\displaystyle \int _{{\mathbb {Z}}_{p}}1\,{\rm {d}}x=1}$
${\displaystyle \int _{{\mathbb {Z}}_{p}}x\,{\rm {d}}x=-{\frac {1}{2}}}$
${\displaystyle \int _{{\mathbb {Z}}_{p}}x^{2}\,{\rm {d}}x={\frac {1}{6}}}$
${\displaystyle \int _{{\mathbb {Z}}_{p}}x^{k}\,{\rm {d}}x=B_{k}}$ , the k-th Bernoulli number

The above four examples can be easily checked by direct use of the definition and Faulhaber's formula.

${\displaystyle \int _{{\mathbb {Z}}_{p}}{x \choose k}\,{\rm {d}}x={\frac {(-1)^{k}}{k+1}}}$
${\displaystyle \int _{{\mathbb {Z}}_{p}}(1+a)^{x}\,{\rm {d}}x={\frac {\log(1+a)}{a}}}$
${\displaystyle \int _{{\mathbb {Z}}_{p}}e^{ax}\,{\rm {d}}x={\frac {a}{e^{a}-1}}}$

The last two examples can be formally checked by expanding in the Taylor series and integrating term-wise.

${\displaystyle \int _{{\mathbb {Z}}_{p}}\log _{p}(x+u)\,{\rm {d}}u=\psi _{p}(x)}$

with ${\displaystyle \log _{p}}$ the p-adic logarithmic function and ${\displaystyle \psi _{p}}$ the p-adic digamma function

Properties

${\displaystyle \int _{{\mathbb {Z}}_{p}}f(x+m)\,{\rm {d}}x=\int _{{\mathbb {Z}}_{p}}f(x)\,{\rm {d}}x+\sum _{x=0}^{m-1}f'(x)}$

From this it follows that the Volkenborn-integral is not translation invariant.

If ${\displaystyle P^{t}=p^{t}{\mathbb {Z} }_{p}}$ then

${\displaystyle \int _{P^{t}}f(x)\,{\rm {d}}x={\frac {1}{p^{t}}}\int _{{\mathbb {Z}}_{p}}f(p^{t}x)\,{\rm {d}}x}$