I'm reading through a part of W. H. Schikof's Ultrametric Calculus and on page 170 and a specific case of an identity presented there is,
$$\int_{\mathbb{Z}_p} f(x) dx = \frac{1}{p} \sum_{n=0}^{p-1} \int_{\mathbb{Z}_p} f(n+px) dx$$
To me this seems intuitive, since this would run over all of the representatives of $\mathbb{Z}_p$, each from $(0+p \mathbb{Z}_p) \cup (1+p \mathbb{Z}_p)\cup (2+p \mathbb{Z}_p)\cup \cdots \cup (p-1+p \mathbb{Z}_p)$, so to give an explicit example of two calculations that I believe should give the same result,
This gives the correct result $$\int_{\mathbb{Z}_5} f(x)dx = \frac{1}{5} \sum_{n=0}^4 \int_{\mathbb{Z}_5} f(n+5x) dx$$
This gives the incorrect result $$\int_{\mathbb{Z}_5} f(x)dx = \frac{1}{5} \int_{\mathbb{Z}_5} f(0+5x) dx + \frac{1}{5} \sum_{n=1}^4 \int_{\mathbb{Z}_5} f(i^n+5x) dx$$
Despite it being that 0, 1, i, -1, and -i are all representatives corresponding to some $n+p \mathbb{Z}_p$. So what is my misconception here and/or how do I amend this in the spirit of my reasoning if possible?
EDIT:
To try to clear up my working of a specific case, let's pick the simple f(x)=x. It is already known that $$\int_{\mathbb{Z}_5} x dx = \frac{-1}{2}$$. I'll go ahead and expand the integral in the way I described,
$$\int_{\mathbb{Z}_5} x dx$$ $$=\frac{1}{5} \int_{\mathbb{Z}_5} (0+5x) dx + \frac{1}{5} \sum_{n=1}^4 \int_{\mathbb{Z}_5} (i^n+5x) dx$$ $$= \int_{\mathbb{Z}_5} x dx + \frac{1}{5} \sum_{n=1}^4 \left(i^n\int_{\mathbb{Z}_5} dx+5\int_{\mathbb{Z}_5} x dx\right)$$ $$= \int_{\mathbb{Z}_5} x dx + \frac{1}{5} \sum_{n=1}^4 i^n \left(\int_{\mathbb{Z}_5} dx\right)+ \sum_{n=1}^4 \left(\int_{\mathbb{Z}_5} x dx\right)$$ $$= \int_{\mathbb{Z}_5} x dx + 0\int_{\mathbb{Z}_5}dx+ 4\int_{\mathbb{Z}_5} x dx$$ $$= 5\int_{\mathbb{Z}_5} x dx $$
In other words, I've just shown that,
$$\int_{\mathbb{Z}_5} x dx = 5\int_{\mathbb{Z}_5} x dx$$ implying that, $$\int_{\mathbb{Z}_5} x dx = 0$$ Which contradicts with it being equal to -1/2. I'm looking very closely at each of my steps but I can't find anything objectionable.