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Suppose that $(M,g)$ is a compact smooth Riemannian manifold with a smooth boundary and suppose that $f$ is a smooth function on $M$ with the property that $$ \int_I f(\gamma(t))\,dt=0,$$ for any inextendible unit speed geodesic $\gamma:I\to M$.

Does it follow that $$ \int_M f\,dV_g=0?$$

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    $\begingroup$ are you assuming that the boundary of $M$ is convex? you also need to assume that $M$ is orientable to have an integral. under those assumptions this indeed immediately follows from Santalo's formula. You just have to lift $f$ in the obvious way to the unit normal bundle. $\endgroup$
    – Vitali Kapovitch
    Commented Mar 30, 2022 at 14:52
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    $\begingroup$ I was curious about the optimal assumptions. Orientability is assumed but I was not sure if nontrapping assumption and convexity of the boundary is required. $\endgroup$
    – Ali
    Commented Mar 30, 2022 at 14:59
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    $\begingroup$ oh I see. I forgot about nontrapping, it is of course also necessary for the standard Sanatlo's formula. I am not sure how necessary those conditions are. I think convexity of the boundry can be dropped. you can extend the manifold by a collar to get the boundary of the extended manifold to be convex and extend $f$ by 0. $\endgroup$
    – Vitali Kapovitch
    Commented Mar 30, 2022 at 15:15

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