How to calculate the vacuum expectation value of energy momentum tensor of a scalar field? I have to prove that the vacuum expectation value of energy-momentum tensor of a scalar field (in flat spacetime) can be written as, $$\langle 0|T_{\mu\nu}|0\rangle=-\langle \rho\rangle\eta_{\mu\nu}.$$
where, $\langle\rho\rangle$ is the vacuum energy density. I have tried the following 

I am unable to see how the quantity in the parenthesis of the last line will convert to $-\omega^2_{\vec{k}}\eta_{\mu\nu}$. I guess the second and third term in the parenthesis cancel each other and we are left with $k^\mu k^\nu$ and I don't see how I can get equation $\langle 0|T_{\mu\nu}|0\rangle=-\langle \rho\rangle\eta_{\mu\nu}$ out of this.
Also in this paper by Weinberg, he writes the equation $\langle 0|T_{\mu\nu}|0\rangle=-\langle \rho\rangle g_{\mu\nu}$ (a generalization in curved spacetime), can be motivated from the principle of Lorentz invariance. How does this logic work?
Therefore, questions,
(1) How is $k^\mu k^\nu$ or the quantity in the last parenthesis is equal to $-\omega^2_{\vec{k}}\eta_{\mu\nu}$?
(2) How to derive the equation $\langle 0|T_{\mu\nu}|0\rangle=-\langle \rho\rangle g_{\mu\nu}$ from Lorentz invariance or in any other way?
EDIT: I again tried to solve it and got this . Am I correct here? What am I doing wrong? And the question (2) remains.
 A: The updated calculation contains a mistake. Here, $i$ is not a dummy index but a fixed index, thus sumover is not implied in the fourth line of the calculation for the vacuum expectation of $ii$-th component of energy-momentum tensor. \begin{array}
e \langle\hat{T}^{ii}\rangle_0 &=& \int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}k^ik^i\\
&=& \int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}\eta^{ij}k_jk^i\\
&=& -\int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}\delta^{i}_{j}k_jk^i \\
&=& -\int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}k_ik^i \qquad\quad\text{not summed over},\\
&=& -\frac{1}{3}\int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}|\vec{k}|^2 \qquad\text{due to the isotropy or rotational symmetry of the $k$-space}.
\end{array}
And $\int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}|\vec{k}|^2\neq\int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}(|\vec{k}|^2+m^2),$ because adding $\int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}m^2$ amounts to adding a divergent quantity. Also only at high $k$ limit your argument is true that $|\vec{k}|^2$ is arbitrarily larger than $m^2$, however, this is not true here because $|\vec{k}|$ goes from $0$ to $\infty$ and not for all these values,  $|\vec{k}|$ is necessarily larger than $m$. Therefore, the $ii$-th and $00$-th components are not the same. $$\langle\hat{T}^{ii}\rangle_0= -\frac{1}{3}\int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32\omega_{\vec{k}}}|\vec{k}|^2 \qquad \langle\hat{T}^{00}\rangle_0= \int\frac{\mathrm{d}^3\vec{k}}{(2\pi)^32}\sqrt{|\vec{k}|^2+m^2}. $$ But they're all divergent and $00$-th component has a different sign than the spatial components. This justifies the relation that the vacuum expectation value of energy momentum tensor is proportional to the metric (but the values are all divergent!), $$\langle T^{\mu\nu}\rangle_0\approx-\langle\rho\rangle\eta^{\mu\nu}\qquad \text{for signature $(-+++)$}.$$
This is only a weak equality because for spatial components this is only true in the sense that the integral is divergent. 
A: When you compute the integral, it is not difficult to verify that
$$
  \int\frac{d^3k}{(2\pi)^3\omega_k}k^\mu k^\nu=0\qquad \mu\ne\nu
$$
This is due to the property that
$$
\int\frac{d^3k}{(2\pi)^3\omega_k}k_i=0 \qquad i=1,2,3.
$$
