In Schutz's Relativity Chapter 4, problem 23b) states:
Use the identity $T^{\mu\nu} _{~~~~~,\nu} = 0 $ to prove the following results for a bounded system (i.e. a system for which $T^{\mu\nu} = 0 $ outside a bounded region of space):
$$\frac{\partial^2}{\partial t^2}\int T^{00}x^i x^j d^3 x=2\int T^{ij}d^3x ~(\text{tensor virial theorem})$$
I have derived the following result:
\begin{align*} \left(T^{\alpha\beta}x^ix^j\right)_{,\alpha ,\beta}&=\left(\left(T^{\alpha\beta}x^ix^j\right)_{,\alpha }\right)_{,\beta}\\ &=\left(T^{\alpha\beta}_{~~~~~,\alpha}~x^ix^j + T^{\alpha\beta}(x^i_{~,\alpha}x^j+x^j_{~,\alpha}x^i)\right)_{,\beta}\\ &=\left(T^{\alpha\beta}x^i_{~,\alpha}x^j+T^{\alpha\beta}x^j_{~,\alpha}x^i\right)_{,\beta} ~~~\text{(using }T^{\mu\nu} _{~~~~~,\nu} = 0 \text{)}\\ &=\left(T^{i\beta}x^j+T^{j\beta}x^i\right)_{,\beta}\\ &=T^{i\beta}x^j_{~,\beta}+T^{j\beta}x^i_{~,\beta} ~~~\text{(using }T^{\mu\nu} _{~~~~~,\nu} = 0 \text{ again)}\\ &=T^{ij}+T^{ji}=2T^{ij} ~~~\text{(by symmetry of } T\text{)} \end{align*}
Expanding this expression spits out the integrand of the LHS of the theorem in the first term:
\begin{align*} \left(T^{\alpha\beta}x^ix^j\right)_{,\alpha ,\beta}&=\frac{\partial^2}{\partial t^2}{T^{00}}x^i x^j + ~... \end{align*}
I'm stuck here because I can't find a way to cancel out the other terms in the expansion. I'm not sure if it needs some version of the divergence theorem. I could also be going in a completely wrong direction.