Greiner's Green's function for diffusion I am reading Greiner's "Quantum Electrodynamics". In example 1.5 he derives the Green's function for diffusion. I am stuck on a step in the derivation.
He has the defining differential equation as
$$
{\bf \nabla}^\prime G-a^2{\partial G \over \partial t^\prime}=-4\pi\delta^3({\bf x}^\prime-{\bf x})\delta(t^\prime-t)
\label{eq4}
\tag{1}
$$
where $a$ is a constant. He then defines
\begin{align}
\tau&=t^\prime-t\\
{\bf R}&={\bf x}^\prime-{\bf x}.
\tag{2}
\label{eq5}
\end{align}
The Fourier transform of the Green's function is given as
$$
G({\bf x}^\prime,t^\prime,x,t)={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot({\bf x}^\prime-{\bf x})} g({\bf p},\tau)\, .
\tag{3}
\label{eq6}
$$
We insert Eq. (\ref{eq6}) into the left hand side of Eq. (\ref{eq4}) to get
$$
{\bf \nabla}^\prime G-a^2{\partial G \over \partial \tau}={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot{\bf R}} \left(-p^2 g-a^2{\partial g \over \partial \tau}\right)\, .
\tag{4}
\label{eq8}
$$
Using,
$$
\delta^3({\bf R})={1\over (2\pi)^3}\int {\rm d}^3 p e^{i{\bf p}\cdot{\bf R}}
\tag{5}
\label{eq7}
$$
this results in a differential equation for $g$:
$$
a^2{\partial g\over \partial \tau}+p^2 g=4\pi\delta(\tau)\,.
\label{eq9}
\tag{6}
$$
Greiner then claims that this equation has the following solution:
$$
g={4\pi \over a^2}e^{-(p^2\tau/a^2)}\Theta(\tau)
\label{eq10}
\tag{7}
$$
To prove this he substitutes Eq. (\ref{eq10}) into Eq. (\ref{eq9}) and uses the relation
$$
{ {\rm d}\Theta \over {\rm d} \tau}=\delta(\tau)
\label{eq11}
\tag{8}
$$
to get 
$$
4\pi e^{-p^2\tau/a^2}{{\rm d} \Theta \over {\rm d} \tau}=4\pi\delta(\tau) \,.
\label{eq12}
\tag{9}
$$
As far as I can see 
 Eqs. (\ref{eq12}) and (\ref{eq11}) imply that $e^{-p^2\tau/a^2}=1$. But I don't think this is correct. My question is what am I missing or is this an error in the textbook?
 A: Let a real function $\;f(x)\;$ of the real variable $\;x\in\mathbb{R}\;$ for which
\begin{align}
f(x)\boldsymbol{=}0 \quad & \text{for any} \quad x\boldsymbol{\ne} x_{0} \quad \textbf{and}
\tag{01a}\label{01a}\\
\int\limits_{\boldsymbol{x_{0}-\varepsilon}}^{\boldsymbol{x_{0}+\varepsilon}}\!\!\!f(x)\mathrm dx\boldsymbol{=}1\quad &  \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0
\tag{01b}\label{01b} 
\end{align}
Under these conditions it seems that this function is not well-defined at  $\;x_{0}$, may be because of a singularity at this point. But we have good reasons to $^{\prime\prime}$believe$^{\prime\prime}$ that
\begin{equation}
f(x)\boldsymbol{\equiv}\delta\left(x\boldsymbol{-}x_{0}\right)
\tag{02}\label{02} 
\end{equation}
since equations \eqref{01a},\eqref{01b} remind us the defining properties of Dirac delta function on the real axis $\mathbb{R}$.
Now let the function
\begin{equation}
f(\tau)\boldsymbol{=}e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\dfrac{\mathrm d \Theta}{\mathrm d \tau}\,, \quad b\in\mathbb{R}
\tag{03}\label{03} 
\end{equation}
We have
\begin{align}
f(\tau)\boldsymbol{=}e^{\boldsymbol{-}b^2 \boldsymbol{\tau}}\dfrac{\mathrm d \Theta}{\mathrm d \tau}\boldsymbol{=}0 \quad & \text{for any} \quad \tau\boldsymbol{\ne} 0 \quad \textbf{and}
\tag{04a}\label{04a}\\
\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}f(\tau)\mathrm d\tau\boldsymbol{=}\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}e^{\boldsymbol{-}b^2 \boldsymbol{\tau}}\dfrac{\mathrm d \Theta}{\mathrm d \tau}\mathrm d\tau\boldsymbol{=}1\quad &  \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0
\tag{04b}\label{04b} 
\end{align}
Equation \eqref{04b} is proved by integrating by parts
\begin{align}
\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\mathrm d \Theta & \boldsymbol{=}\left[e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\Theta\right]_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}\boldsymbol{-}\int\limits_{\boldsymbol{-\varepsilon}}^{\boldsymbol{+\varepsilon}}\Theta\,\mathrm d \left(e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\right)
\nonumber\\
& \boldsymbol{=}e^{\boldsymbol{-}b^2\boldsymbol{\varepsilon}}\boldsymbol{-}\int\limits_{0}^{\boldsymbol{+\varepsilon}}\mathrm d \left(e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\right) \boldsymbol{=}e^{\boldsymbol{-}b^2\boldsymbol{\varepsilon}}\boldsymbol{-}\left[e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\right]_{0}^{\boldsymbol{\varepsilon}} \boldsymbol{=}1
\tag{05}\label{05} 
\end{align}
So
\begin{equation}
\boxed{\:\:
e^{\boldsymbol{-}b^2\boldsymbol{\tau}}\dfrac{\mathrm d \Theta}{\mathrm d \tau}\boldsymbol{=}\delta\left(\tau\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \quad b\in\mathbb{R}
\tag{06}\label{06} 
\end{equation}
$\boldsymbol{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}$

$\textbf{NOTE :}$


We treat the 3-dimensional case on the same reasoning :


So, let a real function $\;F(\mathbf{r})\;$ of the vector variable $\;\mathbf{r}\in\mathbb{R}^{\bf 3}\;$ for which
\begin{align}
F(\mathbf{r})\boldsymbol{=}0 \quad & \text{for any} \quad \mathbf{r}\boldsymbol{\ne} \mathbf{r}_{0} \quad \textbf{and}
\tag{n-01a}\label{n-01a}\\
\iiint\limits_{\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)}F(\mathbf{r})\mathrm d^{\bf 3}\mathbf{r}\boldsymbol{=}1\quad &  \text{for any} \quad \boldsymbol{\varepsilon} \boldsymbol{>}0
\tag{n-01b}\label{n-01b} 
\end{align}
where $\;\mathcal B\left(\mathbf{r}_{0},\boldsymbol{\varepsilon}\right)\;$ a ball with center at $\;\mathbf{r}_{0}\;$ and radius $\;\boldsymbol{\varepsilon}$.


Under these conditions it seems that this function is not well-defined at  $\;\mathbf{r}_{0}$, may be because of a singularity at this point. But we have good reasons to $^{\prime\prime}$believe$^{\prime\prime}$ that
\begin{equation}
F(\mathbf{r})\boldsymbol{\equiv}\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right)
\tag{n-02}\label{n-02} 
\end{equation}
since equations \eqref{n-01a},\eqref{n-01b} remind us the defining properties of Dirac delta function in the real space $\;\mathbb{R}^{\bf 3}$.


An example is the representation of the Laplacian of  $\;1/\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert\;$ by Dirac delta function 
\begin{equation}
\boxed{\:\:
\nabla^{\bf 2}\left(\!\dfrac{1}{\Vert\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\Vert}\right)\boldsymbol{=}\boldsymbol{-}4\pi\delta\left(\mathbf{r}\boldsymbol{-}\mathbf{r}_{0}\right)\vphantom{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:}
\tag{n-03}\label{n-03}
\end{equation}
useful in Electrostatics.


For a proof of \eqref{n-03} see my answer therein :Related : Divergence of  r/r2  , what is the 'paradox'?.

