Are there Schwarzschild solutions to EFE with the Landau-Lifschitz Pseudotensor? I read that solving the Einstein Field Equations can sometimes lead to the problem of non-conservation of energy and that the Landau-Lifschitz Pseudotensor resolves this problem.
I can't however find literature on a Schwarzschild solution with a Energy-Stress tensor that also takes into account the Landau-Lifschitz Pseudotensor. Hasn't this been worked out yet? And what happens to the singularity? Is the curvature at the center of black hole still infinite? Wouldn't this imply a divergent energy?
 A: As stated in the comments, for the Schwarzschild vacuum spacetime there is no physical insight to be gained from examining the Landau–Lifshitz pseudotensor. It is of course a coordinate dependent object, and in the standard Schwarzschild coordinates it's nonzero components are
$$t^{11}_{LL}= \frac{2 (5r - 6r_s + r \cot^2\theta )}{r^2(r-r_s)} $$
$$ t^{22}_{LL}= -2 \frac{(r-r_s) (\cot^2 \theta -1)}{r^3}
$$
$$ t^{23}_{LL}= t^{32}_{LL}= 2 \frac{(r-r_s) \cot \theta }{r^4}  
$$
$$ t^{33}_{LL}= -\frac{2}{r^4}
 \quad \quad t^{44}_{LL}= -\frac{2 \csc^2 \theta}{r^4}
$$
Change to some other coordinates and you'll have different expressions. This doesn't represent the energy of the gravitational field, and at a point you can always find coordinates where $t_{LL}$ is zero. (I'm pretty sure in isotropic Cartesian coordinates at $R=0$ this object is zero, but you can check).
A: The result Eletie gave is correct if you plug in spherical Schwarzschild coordinates, but since the Landau Lifschitz formulation is designed to work in cartesian coordinates only:

 Fromholz, Poisson & Will wrote: "The LL formulation is defined to work in Cartesian-like coordinates, in which ηₐᵦ=diag(−1,1,1,1)"


 Landau & Lifschitz wrote: "On the other hand, we can get values of the tᶦᵏ different from zero in flat space, i.e. in the absence of a gravitational field, if we simply use curvilinear coordinates instead of cartesian."

the results make more sense if you transform your metric into the cartesian form before feeding it into the Pseudotensor, otherwise you get nonsensical $\theta$-dependencies all over the place, infinite energy density at the poles and weird crossterms even if the metric you plug in is spherically symmetric.
In cartesian Raindrop and Finkelstein coordinates the density of the Pseudoenergy is 0 everywhere, but in cartesian Droste coordinates we get
$$t_{LL}^{ \ tt} = -\frac{r_s \ M}{4  \ \pi  \ r^2 \ (r-r_s)^2} \ \ , \ \ \ \frac{t_{LL}^{ \ tt}}{g^{tt}} = -\frac{r_s \ M}{4  \ \pi  \ r^3 \ (r-r_s)}$$
so in the frame of the free falling raindrop there is no Pseudoenergy, while in the frame of a stationary observer there is. All the other components except the ${t}_{LL}^{tt}$ component are 0 in these coordinates, although it is the same metric Eletie used, just in the cartesian form.
As in the case of the electromagnetic field energy or the Newtonian gravitational field energy the Pseudoenergy density also falls off approximately proportional to $r^{-4}$.
References on the subject can be found at 42(4):261-264.2005 and arxiv:1308.0394, with more examples here.
