Let $(M,g)$ be a simply connected 2-dimensional Riemannian manifold without boundary, and let $K$ be the Gausssian curvature defined on $M$. If $M$ is compact, then by Gauss-Bonnet Theorem, we have
$$\int\_M K dA = 4\pi,$$where $dA$ is the area element of $M$ under the metric $g$.


If $M$ is not compact, then the above equality is no longer true. For example, let $M=\mathbb{R}^2$ and we define the conformal metric $g=e^{2u}\delta$, where $u=\ln(sech x)$ and $\delta$ is the Euclidean metric, then one can verify that $K \equiv 1$ on $M$, but the total area of $M$ is $\infty$, not $4\pi$.


Motivated by this example, my question is that, if we assume $(M,g)$ is a noncompact simply connected surface without boundary, and the total area of $M$ is finite, then is it true that  

$$\int\_M K dA =4\pi?$$
Or is it true at least for the conformal case?



Gauss-Bonnet Theorem on noncompact surface without boundary
Consider the metric on $\mathbb R^2$ that is rotationally symmetric metric outside a compact set, namely, it is $dr^2+m(r)^2 d\phi^2$ for $r>R>0$. Here $m$ is a positive function on $[R,\infty)$.


The area form at points with $r>R$ is $dA=m(r)drd\phi$, so the surface has finite area if and only if $m$ is integrable on $[R,\infty)$. The total curvature of the rotationally symmetric end is $$\int\_R^\infty -\frac{m^{\prime\prime}}{\!\!\! m} dA=-2\pi\int\_R^\infty m^{\prime\prime} dr=2\pi\left(m'(R)-\lim\_{r\to\infty} m'(r)\right).$$
It is easy to find examples where the limit on the right hand side does not exist but $m$ is integrable. If the limit exists and $m$ is integrable, the limit is zero, and the total curvature of the end is $2\pi m^\prime(R)$.


The metric is arbitrary if $r<R$, and the total curvature of this region can be computed with the usual Gauss-Bonnet for surfaces with boundary. The geodesic curvature of the boundary is easy to compute from the $r\ge R$ side (I don't remember the answer).


EDIT: As Willie Wong points out by Gauss-Bonnet every smooth metric on the region $\{r<R\}$ will have the same total curvature. So just extend $m$ to a smooth function on $[0,R]$ so that $m(r)=r$ near $0$ and consider the metric $dr^2+m(r)^2d\phi^2$ for all $r>0$. Its metric completion is smooth at the origin. (More generally, the metric is smooth at the origin if and only if $m^\prime(0)=1$ and $m$ extends to an odd smooth function on $\mathbb R$). Now the above computation gives total curvature as $2\pi(m^\prime(0)-m^\prime(\infty))=2\pi(1-m^\prime(\infty))$ and if $m^\prime(\infty)$ exists and the area is finite, the total curvature is $2\pi$.


