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On a Riemannian Manifold $(M,g)$, for any $p \in M$, there exists a normal coordinate system at p, ${U, x_1, ..., x_n}$. I want to integrate over a normal coordinate system with U being a geodesic ball $B_\epsilon(p)$, where $\epsilon>0$ is small enough.

$$\int_{B_\epsilon(p)} \rho(x,p)^{n-\delta} |det(g)|^{1/2} dx_1 ... dx_n,$$

where $\rho{x,p}$ is the geodesic distance between $x$ and $p$. I know that $det(g) = 1 + O(r^2)$ since we are using normal coordinates. However, I would like to know if one can perform some kind of change of variables via polar coordinates to get something like that

$$ \int_{B_\epsilon(p)} \rho(x,p)^{-n+\delta} det(g)^{1/2} dx_1 ... dx_n \leq C\int_0^\epsilon r^{n-\delta} r^{n-1} (1+O(r^2))dr.$$

Basically, I do not understand what happens to $\rho(x,p)^{-n+\delta}$. I understand that normal coordinates allows us to transform the volume form into the usual Euclidean case + O(r^2), but I do not know what happens to the $\rho(x,p)^{-n+\delta}$.

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    $\begingroup$ In orthogonal normal coordinates, we have $\rho(x,p)^2 = (x_1)^2+\cdots+ (x_n)^2$ on the nose, at least within the cut locus of $p$, so there's not much mystery about it. $\endgroup$ Commented Mar 26, 2013 at 18:15
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    $\begingroup$ Basically, the geodesic distance is the usual Euclidean distance, at least on such normal coordinate system. That's what I thought. Thank you. $\endgroup$
    – Henry
    Commented Mar 26, 2013 at 18:18

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