I want to show that $\frac{\partial{E}}{\partial{t}}-\Delta E=\delta(t,x) $.
So it suffices to show that $\langle \frac{\partial{E}}{\partial{t}}-\Delta E, \phi \rangle=\phi(0,0) $.
So far I have shown that $ \langle \frac{\partial{E}}{\partial{t}}-\Delta E, \phi \rangle=\lim_{\epsilon \to 0} \int_{\mathbb{R}^n} \frac{1}{(2 \sqrt{\pi \epsilon})^n} e^{-\frac{|x|^2}{4 \epsilon}} \phi(\epsilon,x) dx$
$E$ is the heat kernel, and it's a fundamental solution of the heat equation in the sense of distributions
How could we continue?
Hint: proof that
1) $(\partial_t - \Delta)K_t(x)=0$ $\forall x \in \mathbb{R}^n$, $\forall t >0$
2) $\displaystyle \int_{\mathbb{R}^n} K_t(x)dx)=1$ $\forall x \in \mathbb{R}^n$, $\forall t >0$
3) $\forall \delta > 0$ we have $\displaystyle \lim_{t \rightarrow 0+} \int_{\mathbb{R}^n \setminus B(0,\delta)} K_t(x) dx = 0$
where $K_t(x):=1/(4\pi t)^{n/2} e^{-|x|^2/4t}$ if $(x,t) \in \mathbb{R}^n \times (0,\infty)$ and zero otherwise.
You can conclude that $K_t(x) \in L^1_{loc}(\mathbb{R}^{n+1})$ is a distribution, and the distributional derivative can be written as
(i) $\displaystyle (\partial_t K_t)(\varphi):=-\lim_{\epsilon \rightarrow 0^+} \int_{\epsilon}^{\infty} \int_{\mathbb{R}^n} \partial_t \varphi(x,t) K_t(x) dx dt$ , $\forall \varphi \in \mathcal{D}(\mathbb{R}^{n+1})$
integrates parts respect $t$ the inner integral function, place $x=y\sqrt{4 \epsilon}$, and use Lebesgue theorem to show that
$\displaystyle \int_{\mathbb{R}^n} \varphi(x,\epsilon)K_\epsilon(x) dx \rightarrow \varphi(0) \cdot 1/\pi^{n/2} \int_{\mathbb{R}^n} e^{-|y|^2} dy =\varphi(0)$
in the previous step using a well-known integral. Now, uses the property 1) the heat kernel, and with an integration by parts, you can conclude that
$\displaystyle \int_{t > \epsilon} \int_{\mathbb{R}^n} \varphi(x,t) \partial_t K_t(x) dx dt = \int _{t > \epsilon} \int_{\mathbb{R}^n} \Delta \varphi(x,t) K_t(x) dx dt$
back to (i) and you can conclude.