I've been trying to calculate the electric field of an infinite sheet via the differential form, i.e. using
\begin{align}
\nabla\cdot\text{E}=\frac{\rho}{\epsilon\_0}.
\end{align}
I am aware the answer is
\begin{align}
\text{E}=\frac{\sigma}{2\epsilon\_0}\hat{n},
\end{align}
as well as how to use Gauss's law to arrive at it, but I am having trouble matching it without Gauss's law.

The charge distribution is
\begin{align}
\rho=\sigma\delta(z).
\end{align}
Since the sheet is infinite the electric field only points in the $z$ direction,
\begin{align}
\frac{\mathrm d\text{E}\_z}{\mathrm dz}=\frac{\sigma}{\epsilon\_0}\delta(z).
\end{align}
Integrating gives that
\begin{align}
\text{E}\_z=\frac{\sigma}{\epsilon\_0}.
\end{align}
Which is missing a factor of 1/2. Perhaps I am going wrong with my integration, as the delta function must be integrated around the point in order for it to output 1,
\begin{align}
\int\_{a-\varepsilon}^{a+\varepsilon}\delta(z-a)\mathrm dz=1,
\end{align}
which makes me suspicious of my indefinite integration.

Perhaps the correct methodology may lie in using Heaviside step function, as I believe H$(0)=1/2$ depending on who you ask,
\begin{align}
\rho&=\sigma\frac{\mathrm d}{\mathrm dz}\mathrm H(z)\\
\mathrm H(z)&=\int\_{-\infty}^z\delta(s)\mathrm ds,
\end{align}
but when I try that I get
\begin{align}
\mathrm E\_z=\frac{\sigma}{\epsilon\_0}\mathrm H(z),
\end{align}
and I can't justify evaluating both sides at 0.