Determine $C$ from $\theta(x,t)= \frac{\theta_0 C}{\sqrt{kt}} e^{ \frac{-x^2}{4kt}}$ where $\theta(x,0)=\theta_0 \delta(x)$ I have solved a differential equation leading to a solution for arbitrary $C$: 
$$\theta(x,t)= \frac{\theta_0 C}{\sqrt{kt}} e^{ \frac{-x^2}{4kt}}$$
Now I need to determine the constant $C$ using the initial condition $\theta(x,0)=\theta_0 \delta(x)$ where $\delta(x)$ is the Dirac delta function. 
My attempt to solve:
$$\int\limits_{-r}^{r} \theta_0 \delta(x) dx =\theta_0=\int\limits_{-r}^{r} \theta(x,0)dx$$
However, when I try to evaluate $\theta(x,0)$ I run into trouble because I would be dividing by zero twice. Now I really don't know how to proceed, maybe I should try a different approach. I also tried 
$$\int\limits_{-r}^{r} \theta(x,t)dx=2\sqrt{\pi}\theta_0 C\int\limits_{-r}^{r}\frac{e^{ \frac{-x^2}{4kt}}}{\sqrt{\pi 4kt}}dx  =2\sqrt{\pi}\theta_0 C\int\limits_{-r}^{r}\delta\left(\frac{x}{\sqrt{4k}}\right)dx $$
as $t\to 0$. However I am not yet comfortable working with $\delta(x)$ so I am not sure if this is right and how to proceed from here. If anyone could help me out I would be very grateful.
Thanks in advance
 A: We know that $\theta(x,t) = \frac{\theta_0}{\sqrt{\kappa t}}q_2 =  \frac{\theta_0}{\sqrt{\kappa t}}C e^{ \frac{-x^2}{4\kappa t}} $. We consider $\lim\limits_{t\to 0}\theta(x,t)$:
$$\lim\limits_{t\to 0} \theta(x,t)=\lim\limits_{t\to 0} \frac{\theta_0}{\sqrt{\kappa t}}C e^{ \frac{-x^2}{4\kappa t}}=\frac{\theta_0 C}{\sqrt{\kappa }}\lim\limits_{t\to 0} \frac{1}{\sqrt{t}} e^{ \frac{-x^2}{4\kappa t}}$$ 
Let $x = \frac{\bar{x}}{\sqrt{4\kappa }}$ and $\sqrt{t} = a$, so $a \to 0$ as $t \to 0$:
Then 
$$\frac{\theta_0 C}{\sqrt{\kappa }}\lim\limits_{t\to 0} \frac{1}{\sqrt{t}} e^{ \frac{-x^2}{4\kappa t}} = \frac{\theta_0 C \sqrt{\pi}}{\sqrt{\kappa }}\lim\limits_{a\to 0} \frac{1}{a\sqrt{\pi}} e^{ \frac{-\bar{x}^2}{a^2}}=  \frac{\theta_0 C \sqrt{\pi}}{\sqrt{\kappa }} \delta\left(\frac{x}{\sqrt{4\kappa }}\right) $$
Now $\delta\left(\frac{x}{\sqrt{4\kappa }}\right)=\sqrt{4\kappa }\delta(x)$, and thus:
$$\lim\limits_{t\to 0} \theta(x,t) = \theta_0 C \sqrt{4\pi} \delta\left(x\right) $$
Now we want $\lim\limits_{t\to 0} \theta(x,t) $ to equal $\theta(x,0)= \theta_0 \delta(x)$ due to continuity so we choose $C$ such that:
$$\theta_0 C \sqrt{4\pi} \delta\left(x\right) = \theta_0 \delta(x)$$
Giving $C = \frac{1}{\sqrt{4\pi}}$. So since $\theta(x,t) =  \frac{\theta_0}{\sqrt{\kappa t}}C e^{ \frac{-x^2}{4\kappa t}} $ we conclude that $\theta(x,t) = \frac{\theta_0}{\sqrt{4\pi \kappa t}} e^{ \frac{-x^2}{4\kappa t}} $
A: $\displaystyle{\large%
\theta\left(x, t\right)= {\theta_{0}C \over \sqrt{kt\,}}\,
{\rm e}^{-x^{2}/4kt}}$
$$
\theta\left(x, t\right)
=
2\,\sqrt{\pi\,}\,\theta_{0} C\;
{{\rm e}^{-x^{2}/4kt}
 \over
 \sqrt{\vphantom{\large A}2\pi\,}\, \left(2kt\right)^{1/2}}\,,
\quad
\lim_{t \to 0^{+}}\theta\left(x,t\right)
=
2\,\sqrt{\pi\,}\,C\;\left[\theta_{0}\delta\left(x\right)\right]
$$
$$
\begin{array}{|c|}\hline\\
\color{#ff0000}{\large\quad%
 C \color{#000000}{\ =\ } {2\,\sqrt{\pi\,} \over \pi}\quad}
\\ \\ \hline
\end{array}
$$
