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$$δ^3(q⃗ )=\frac{δ(q)δ(\theta)}{2\pi q^2\sin(\theta)}$$ is wrong. The delta function is spherically symmetric, and thus has no θ dependence. Simply use: $$d^3(q⃗ )=\frac{δ(q)}{2\pi q^2}$$ instead. Use the Jacobian when you switch coordinate systems (from Cartesian to spherical) ($r^2 \sin(\theta)$), and you should get the result.

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One may only talk about a discrete sum over $k^\mu$ vectors if all the spacetime directions are compact. In that case, $k^\mu$ is quantized. If the spacetime is a periodic box with periodicities $L_x,L_y,L_z,L_t$, then $V=L_x L_y L_z$ and $T=L_t$. The component $k^\mu$ in such a spacetime is a multiple of $2\pi \hbar / L_\mu$ (I added $\hbar$ to allow any ...

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