# Properties of Dirac delta function in Integral

I was reading commutation relation of canonical momentum in KG Field from Lectures of Quantum Field Theory by Ashok Das. In page 179, He has used Integration to derive the result where he expressed integration as follow

I was unable to understand how the exponential value changed from $$k$$ to $$k^0$$.

• $k\cdot x = k^{0}x^{0} - \mathbf{k}\cdot \mathbf{x}\equiv \omega t - \mathbf{k}\cdot \mathbf{x}$ in "east-coast" Minkowski metric. – Frederic Thomas May 7 '19 at 10:11
As already pointed out in the comments $$(k-k^\prime)\cdot x$$ is the four vector product with $$k = \begin{pmatrix}k^0 & k^1 & k^2 & k^3\end{pmatrix}^T$$. But the integral is only done in three dimensions. Therefore
$$\int \frac{\mathrm{d}^3 x}{(2\pi)^3} e^{i(k-k^\prime)\cdot x}=\int \frac{\mathrm{d}^3 x}{(2\pi)^3} e^{i\left[(k^0-k^{\prime 0})x^0-(\vec{k}-\vec{k^\prime})\cdot\vec{x}\right]}= e^{i\left[(k^0-k^{\prime 0})x^0\right]}\int\frac{\mathrm{d}^3 x}{(2\pi)^3} e^{-i\left[(\vec{k}-\vec{k^\prime})\cdot\vec{x}\right]}$$