In Chapter I.5 Appendix 3 of Zee's text, the Voss-Weyl formula for the divergence of a vector field $W^\mu (x) $,
$$ D_\mu W^\mu = \frac{1}{\sqrt{\det g}}\partial_\mu \big(\sqrt{\det g} W^\mu \big)$$
is cleverly derived by invoking and manipulating the integral
$$ \mathrm{I} = \int d^Dx \sqrt{\det g} W^\mu (x) \partial_\mu \phi(x) $$
which is invariant under coordinate transformations. The manipulation involves solving the integral by parts. In the text it is said that doing this leads to
$$ \mathrm{I} = -\int d^Dx \partial_\mu(\sqrt{\det g} W^\mu) \phi $$
I do not understand how the first term from the integration by parts vanishes, leaving only the second term.
[To clarify, in response to Arctic Char's observation, by that I mean the first term and second term of the integration by parts result
$$ \mathrm{I} = \int d^Dxf(x)g'(x)=f(x)g(x) - \int d^Dxf'(x)g(x) $$
I do not understand why $ f(x)g(x) $ vanishes in the above case where $ f(x)=\sqrt{\det g} W^\mu (x) $ and $ g(x)=\phi(x) $.]
Can anyone help?
I got a straightforward answer from the author of physicspages.com. It can be found here: https://physicspagescomments.wordpress.com/2020/04/23/divergence-in-curved-space/comment-page-1/#comment-2946
"growescience says: 27 September 2020 at 11:28 If you mean the sentence just before Zee’s eqn (26), it’s standard practice in these types of integrals to assume that the integrated term goes to zero fast enough at infinity that we can just ignore it. Usually, textbooks will state this, but Zee has the habit of doing things without really explaining what he’s doing. I’ve given up on his books for now."