Gaussian integral on a Riemannian manifold How do I estimate the Gaussian integral $\int d^nx \sqrt{g(x)}~e^{-x^2} $ on a Riemannian manifold $(M,g=det~g_{\mu\nu})$? I've tried to consider $\sqrt{g(x)}$ as an analytic function and expanded it. 
My questions are: 
Under what conditions can I assume that the determinant of the metric is a polynomial? How would you approach this integral?
Context: 
My real problem involves the free energy of a harmonic oscillator on a Riemannian manifold which leads to an integral similar to the one mentioned above. Indeed, potential energy of harmonic oscillator is $U=\frac{1}{2}kd(x,x_0)^2$ which $k$ is a constant, $d(x,x_0)$ is distance and distance needs metric.Therefore, the real integral is $\int d^nx \sqrt{g(x)}~exp(-\frac{1}{2}kd(x,x_0)^2)$. I think that considering this kind of potential is so rigorous. I've simply tried to estimate the potential by summing up all individual harmonic terms (i.e., using a Euclidean metric: $U=\frac{1}{2}k(x_1^2+x_2^2+...)$). 
Can anyone suggest a better approach to solve/estimate this integral?
 A: You can get a nice expression for the leading-order correction to the flat-manifold result via the use of Riemann normal coordinates.  Basically, imagine expanding the metric in a power series at the point $x_0$:
$$
g_{\mu \nu}(x) = g_{\mu \nu} (x_0) + \partial_\rho g_{\mu \nu} (x^\rho - x_0^\rho) + \frac{1}{2} \partial_\rho \partial_\sigma g_{\mu \nu} (x^\rho - x_0^\rho) (x^\sigma - x_0^\sigma) + \dots
$$
where the derivatives are understood to be evaluated at $x_0$.  Riemann local coordinates are constructed in such a way that the following three properties hold:


*

*The geodesic distance between an arbitrary point $x$ and the base point $x_0$ is just the Euclidean distance $\sqrt{(x^\rho - x_0^\rho) (x_\rho - (x_0)_\rho)}$.  Thus, the distance function in your exponential is just the sum of squares, which will make it easier to do the integrals.

*The metric at the point $x_0$ is the flat metric $\delta_{\mu \nu}$.

*The first derivatives of the metric with respect to the coordinates vanish at $x_0$, and the second-order derivatives are expressible as components of the Riemann tensor.  Specifically, in Riemann normal coordinates, we have
$$
g_{\mu \nu} = \delta_{\mu \nu} - \frac{1}{6} (R_{\mu \rho \nu \sigma} + R_{\mu \sigma \nu \rho} ) x^\rho x^\sigma + \dots
$$ 
where I've assumed that we've picked the coordinates such that $x_0^\mu = (0,0,\dots)$.
Finally, we have the very useful identity that for a "small" matrix $M_{\mu \nu}$, we have the approximation
$$
\det (\delta_{\mu \nu} + M_{\mu \nu}) \approx 1 + M^\mu_\mu.
$$
In our case, this implies that
$$
\sqrt{ \det g_{\mu \nu} } \approx 1 - \frac{1}{12} (R_{\mu \rho} {}^\mu {}_\sigma + R_{\mu \sigma} {}^\mu {}_\rho ) x^\rho x^\sigma = 1 - \frac{1}{6} R_{\rho \sigma} x^\rho x^\sigma
$$
and so your integral can be approximated as
$$
I = \int d^n x \sqrt{\det g} e^{-k d(x,x_0)^2/2} \approx \int d^n x \left(1 - \frac{1}{6} R_{\rho \sigma} x^\rho x^\sigma \right) e^{-k \vec{x}^2/2}.
$$
Finally, we can do a similarity transformation among the coordinates $x^\rho$ that diagonalizes the Ricci tensor;  this allows us to write
$$
I \approx \int d^n x \left(1 - \frac{1}{6} \sum_\rho R_{\rho \rho} (x^\rho)^2 \right) e^{-k \vec{x}^2/2}.
$$
If we do these integrals out, we get (I think)
$$
I \approx \frac{\pi^{n/2}}{k^{n/2}} \left( 1 - \frac{R}{12 k} \right)
$$
where $R$ is the Ricci scalar.  This implies that the second term will be negligible compared to the first term if $k$ is large compared to the components of the Ricci tensor (or, which is equivalent, the length scale of the manifold's curvature is much greater than the dispersion of the exponential.)  Seems reasonable.
In principle, one could carry this expansion to higher order, by expanding the determinant of $g$ to higher order in powers of $x$.  There would be two sources of new terms:  terms from the expansion the metric to greater order in the Taylor series, yielding terms that depend on the derivatives of the Riemann tensor at $x_0$;  and terms that arise from the expansion of the determinant in terms of the trace, yielding terms that would be quadratic (and higher-order) in the curvature (and its derivatives.)  
A: This question is subtler than one might think at a first glance. Three points may be worth  noting.


*

*It will make a difference whether the manifold $M$ is compact or not. The nice answer by Michael Seifert implicitly assumes the non-compact case. Otherwise the infinite Gaussian integrals (over $\mathbb{R}^n$) will not make sense. In the compact situation (say circle, sphere etc.) one expects curvature and finite size effects (mathematically speaking, "global" rather than just local issues) - and it will depend on parameter choices which of the two are going to dominate (if any).

*As the standard Gaussian integral is naturally defined on infinite Euclidean space only, the following question arises: What is the natural generalisation of a Gaussian integral on a (compact) Riemannian manifold? I am not sure whether there is a general answer to this (cf. this maths stackexchange entry), but one candidate is the heat kernel, which carries a lot of information about the underlying space. Physically, this may be viewed as the temperature distribution resulting, after time $t$, from an initial "hot spot" (delta function) placed on the manifold. In Euclidean space, the heat kernel is indeed a Gaussian (with $t$ dependent maximum and width). It will be narrow for small times and uniformly "cover" the manifold for asymptotic times. This concept should work on any Riemannian manifold. As far as I'm aware (I'm not an expert on this), heat kernels are known in closed form for a few types of manifolds only. For the two-sphere $S^2$, a rather mathematical discussion can be found here. For hyperbolic spaces some results can be found in this paper.

*In the OP, the original context is that of free energy or, I believe, the canonical partition function of the harmonic oscillator on a Riemannian manifold. But what is a harmonic oscillator on $M$? Normally, in Euclidean space, it results from expanding an arbitrary potential to second order around a local minimum. Again, on compact manifolds, global issues may arise. A simple example is the "planar rotor" aka the particle on a circle (parameterised, say, by angle $\theta$), where a natural potential is $V = \lambda \cos \theta$ with some coupling $\lambda$. Obviously, this reflects the periodicity of the circle - unlike its Taylor expansion around a minimum, which at fourth order, for instance, leads to a Hamiltonian unbounded from below (see the nice discussion in this PRD).
For the fun of it I tried to work out the classical partition function for this rather simple example of a Riemannian manifold. The Hamiltonian is 
$$ H = \frac{\ell^2}{2I} + \lambda (1 - \cos \theta) $$
with angular momentum $\ell$ conjugate to angle $\theta$, $-\pi < \theta < \pi$, and moment of inertia $I = mr^2$, $m$ and $r$ denoting particle mass and circle radius, respectively.
The partition function is then given by an integral over phase space (which must reflect the Riemannian structure),
$$ Z = \int\limits_{-\infty}^\infty d\ell \int\limits_{-\pi}^\pi d\theta \, e^{-\beta H} = (2\pi)^{3/2} \sqrt{I/\beta} \, e^{-\beta \lambda} \, I_0(\beta\lambda) \; , $$     
with $I_0$ a modified Bessel function and $\beta = 1/kT$. In the low temperature limit, $\beta \to \infty$, this becomes 
$$Z \to \frac{2\pi}{\beta} \sqrt{I/\lambda} \; . $$ 
The same asymptotic value is obtained upon expanding the cosine potential to second order (the "angular harmonic oscillator") and extending the angular integration over all $\mathbb{R}$ to have a standard Gaussian integral. This is only valid for narrow width, $\beta \lambda \gg 1$, such that the angular integral is insensitive to the finite size of the circle. Reassuringly, the narrow width approximation is basically the same as the low temperature limit, where small thermal energies do not allow the particle to "explore" the global features of the manifold.
