Show $\frac{d^3\underline{k}}{2\omega_{k}(2\pi)^3}$ is Lorentz invariant.
Hint: try to evaluate $\int dk_0\delta(k_0^2 - M^2)\theta(k_0)$ where $M^2 = \underline{k} + m^2$
My attempt is as follows,
$$ I = \int d^4k \delta(k_0^2 - M^2)\theta(k_0) = \int dk_0 \int d^3\vec{k}\ \delta(k_0^2 - (|k|^2 + m^2))\theta(k_0)\\ = \int dk_0 \int d^3\vec{k} \delta(k_0^2 - (|k|^2 + m^2))\theta(k_0) $$
then using $M^2 = m^2 + |k|^2 = \omega_k^2$ and $\delta(a^2-b^2) = \frac{1}{2|\omega_k|}\delta(a-b) + \frac{1}{2|\omega_k|}\delta(a+b)$ we get
$$ I = \int dk_0 \int d^3\vec{k} \Big( \delta(k_0 - \omega_k) + \delta(k_0 + \omega_k) \Big) \frac{\theta(k_0)}{2\omega_k} $$ using the property of the theta function, we get, $$ I = \int_{0}^{\infty} dk_0 \int d^3\vec{k} \Big( \delta(k_0 - \omega_k) + \delta(k_0 + \omega_k) \Big) \frac{1}{2\omega_k} $$ integrating over $dk_0$ and using the fact that $\theta(\omega_k) + \theta(-\omega_k) = 1$ $$ I = \int \frac{d^3\vec{k}}{2\omega_k(2\pi)^3} \Big( \theta(\omega_k) + \theta(-\omega_k) \Big) = \int \frac{d^3\vec{k}}{2\omega_k(2\pi)^3} $$ however, I still have an integral for $\int d^3\vec{k}$ I don't know how to loose this integral without loosing the $d^3\vec{k}$!
