As an exercise, I'd like to write up something that I agonized over for 48 hours -- the propagator for a free particle in Quantum mechanics.
We begin with the following function, which I will call $U$.
\begin{eqnarray}
U &=& \langle x_f, t_f \vert x_i, t_i \rangle \\
&=& \int dx_1 dx_2 \dots dx_{n-1}dx_n \langle x_1, t_1 \vert e^{\frac{-i\hat{H}\tau}{\hbar}}\vert x_2, t_2 \rangle \langle x_2, t_2 \vert e^{\frac{-i\hat{H}\tau}{\hbar}}\vert x_3, t_3 \rangle \cdots \langle x_{n-1}, t_{n-1} \vert e^{\frac{-i\hat{H}\tau}{\hbar}}\vert x_n, t_n \rangle
\end{eqnarray}
$U$ has a peculiar property, because if we take the inner product of our final state with any wavefunction $\vert \psi \rangle$, we find
\begin{eqnarray}
\langle x_f,t_f \vert \psi \rangle &=& \int \langle x_f, t_f \vert x_i, t_i \rangle \langle x_i, t_i \vert \psi \rangle dx_i \\
&=& \int U \langle x_i, t_i \vert \psi \rangle dx_i \\
\psi_f &=& \int U \psi_i dx_i
\end{eqnarray}
Our function $U$ acts like a Green's function on our initial state, $\psi_i$. In the integral above, we have partitioned the transition between states into $n$ steps, propagating forward in time with the operator $e^{\frac{-i \hat{H}\tau}{\hbar}}$
If we examine a single transition between the $x_{j+1}$ state and the $x_j$, we find that we can represent our wave functions in terms of their fourier transforms, or in terms of the momentum eigenbasis:
\begin{eqnarray}
\vert x \rangle &=& \int \vert p \rangle \langle p \vert x \rangle \\
&=& \int \frac{1}{\sqrt{2\pi \hbar}}e^{ipx/\hbar} \phi(p) dp \\
\Rightarrow && \frac{1}{\sqrt{2\pi \hbar}}e^{ipx/\hbar} = \langle p \vert x \rangle \\
&& \frac{1}{\sqrt{-2\pi \hbar}}e^{ipx/\hbar} = \langle x \vert p \rangle
\end{eqnarray}
So rewriting our integral in terms of these bases,
\begin{eqnarray}
\langle x_{j+1} \vert e^{\frac{-i}{\hbar}\hat{H}\tau} \vert x_j \rangle &=& \int \langle x_{j+1} \vert q \rangle \langle q \vert e^{\frac{i}{\hbar}\hat{H}\tau} \vert p \rangle \langle p \vert x_j \rangle dp dq \\
&=& \int \frac{e^{\frac{-iqx_{j+1}}{\hbar}}}{\sqrt{2\pi \hbar}} \langle q \vert e^{\frac{-i}{\hbar}\hat{H}\tau} \vert p \rangle \langle p \vert x_j \rangle dp dq \\
&=& \int \frac{e^{\frac{-iqx_{j+1}}{\hbar}}}{\sqrt{2\pi \hbar}} \langle q \vert e^{\frac{-i}{\hbar}\hat{H}\tau} \vert p \rangle \frac{e^{\frac{ipx_{j}}{\hbar}}}{\sqrt{2\pi \hbar}} dp dq \\
&=& \int \frac{e^{\frac{-iqx_{j+1}}{\hbar}}}{\sqrt{2\pi \hbar}} \langle q \vert \left( 1 - \frac{i}{\hbar}\frac{p^2}{2m}\tau-iV(x)/ \hbar +\dots \right) \vert p \rangle \frac{e^{\frac{ipx_{j}}{\hbar}}}{\sqrt{2\pi \hbar}} dp dq \\
&=& \int \frac{e^{\frac{-iqx_{j+1}}{\hbar}}}{\sqrt{2\pi \hbar}} \left( 1 - \frac{i}{\hbar}\frac{p^2}{2m}\tau-iV(x)/ \hbar\tau + \dots \right) \frac{e^{\frac{ipx_{j}}{\hbar}}}{\sqrt{2\pi \hbar}}\delta(p-q) dp dq \\
&=& \int \frac{e^{\frac{-ip(x_{j+1}-x_j)}{\hbar}}}{2\pi \hbar} \left( e^{\frac{-i}{\hbar}\frac{p^2}{2m}\tau-iV(x)/ \hbar}\tau \right) dp \\
&=& \int \frac{e^{\frac{-ip(x_{j+1}-x_j)}{\hbar}}}{2\pi \hbar} e^{\frac{-i}{\hbar}\frac{p^2}{2m}\tau}dp e^{-iV(x)\tau/ \hbar}
\end{eqnarray}
We recognize the integral over $p$ to be a Gaussian integral, or the fourier transform of a Gaussian. The result is
\begin{eqnarray}
\langle x_{j+1} \vert e^{\frac{-i}{\hbar}\hat{H}\tau} \vert x_j \rangle &=& \sqrt{\frac{-im}{2 \pi \hbar \tau}} e^{\frac{i}{\hbar}\left(\frac{m}{2 \tau}(x_{j+1}-x_j)^2-V(x)\tau \right)}
\end{eqnarray}
Multiplying all of our transition amplitudes and integrating over the various possible $x_j$ variables, we find the Path integral formulation of Quantum Mechanics!
\begin{eqnarray}
U &=& \left(\frac{-im}{2 \pi \hbar \tau}\right)^{N/2} \int \cdots \int \left( \prod_{j=1}^N dx_j \right) e^{\frac{i}{\hbar}\sum_{j=1}^N \left(\frac{m}{2 \tau}(x_{j+1}-x_j)^2-V(x)\tau \right)}
\end{eqnarray}
If we take the limit as $N \to \infty$, then we get our action in the exponential argument, unlike the discrete quantity we currently have,
\begin{eqnarray}
\lim_{N\to \infty} U &=& \lim_{N\to \infty} \left(\frac{-im}{2 \pi \hbar \tau}\right)^{N/2} \int \cdots \int \left( \prod_{j=1}^N dx_j \right) e^{\frac{i}{\hbar} \int L(x,\dot{x},t) dt}
\end{eqnarray}
(I think what I have called $U$ is in fact the propagator. I also seem to be off by a negative sign in our normalization factor -- the factor $-im$ should be $im$.)
For a free particle, $V(x)=0$ and so we have for our inner product:
\begin{eqnarray}
U &=& \left(\frac{-im}{2 \pi \hbar \tau}\right)^{N/2} \int \cdots \int \left( \prod_{j=1}^N dx_j \right) e^{\frac{i}{\hbar}\sum_{j=1}^N \left(\frac{m}{2 \tau}(x_{j+1}-x_j)^2\right)}
\end{eqnarray}
Examining the first few terms of this nasty integral, we see that we have
\begin{eqnarray}
\int dx_1 e^{\frac{mi}{2\hbar \tau}\left(x_0^2 - 2x_0x_1+ x_1^2 -2x_1x_2 + 2x_2^2 -2x_2x_3+ \dots x_n^2\right)}
\end{eqnarray}
Integrating over each successive $x_{n-1}$ yields
\begin{eqnarray}
\sqrt{\frac{2\pi \hbar \tau}{-im(n-1)}}^{n-1}e^{\frac{-im}{2\hbar (n-1)\tau}\left(x_{n}-x_0\right)^2}
\end{eqnarray}
Notice that the factor next to our Gaussian like function is the perfect inverse of our normalization factor for the path integral earlier! (along with a factor of $n-1$, which will create our total time interval, $T=n\tau$ as you'll see in a minute.) So our final expression for $U$ becomes
\begin{eqnarray}
U = \langle x_f, t_f \vert x_i, t_i \rangle &=& \sqrt{\frac{2\pi \hbar }{-imT}}\mathrm{exp}\left[\frac{-im}{2\hbar T}\left(x_f-x_0\right)^2 \right]
\end{eqnarray}
So this expression gives the probability of transitioning from one state $\vert x_0, t_0 \rangle$ to another $\langle x_f, t_f \vert$. This expression is Gaussian, although I don't know what the complex phase means in the exponential argument. It is certainly normalized, since we are multiplying by $\sqrt{2\pi/\sigma^2}$ ...
We begin with the following function, which I will call $U$.
\begin{eqnarray}
U &=& \langle x_f, t_f \vert x_i, t_i \rangle \\
&=& \int dx_1 dx_2 \dots dx_{n-1}dx_n \langle x_1, t_1 \vert e^{\frac{-i\hat{H}\tau}{\hbar}}\vert x_2, t_2 \rangle \langle x_2, t_2 \vert e^{\frac{-i\hat{H}\tau}{\hbar}}\vert x_3, t_3 \rangle \cdots \langle x_{n-1}, t_{n-1} \vert e^{\frac{-i\hat{H}\tau}{\hbar}}\vert x_n, t_n \rangle
\end{eqnarray}
$U$ has a peculiar property, because if we take the inner product of our final state with any wavefunction $\vert \psi \rangle$, we find
\begin{eqnarray}
\langle x_f,t_f \vert \psi \rangle &=& \int \langle x_f, t_f \vert x_i, t_i \rangle \langle x_i, t_i \vert \psi \rangle dx_i \\
&=& \int U \langle x_i, t_i \vert \psi \rangle dx_i \\
\psi_f &=& \int U \psi_i dx_i
\end{eqnarray}
Our function $U$ acts like a Green's function on our initial state, $\psi_i$. In the integral above, we have partitioned the transition between states into $n$ steps, propagating forward in time with the operator $e^{\frac{-i \hat{H}\tau}{\hbar}}$
If we examine a single transition between the $x_{j+1}$ state and the $x_j$, we find that we can represent our wave functions in terms of their fourier transforms, or in terms of the momentum eigenbasis:
\begin{eqnarray}
\vert x \rangle &=& \int \vert p \rangle \langle p \vert x \rangle \\
&=& \int \frac{1}{\sqrt{2\pi \hbar}}e^{ipx/\hbar} \phi(p) dp \\
\Rightarrow && \frac{1}{\sqrt{2\pi \hbar}}e^{ipx/\hbar} = \langle p \vert x \rangle \\
&& \frac{1}{\sqrt{-2\pi \hbar}}e^{ipx/\hbar} = \langle x \vert p \rangle
\end{eqnarray}
So rewriting our integral in terms of these bases,
\begin{eqnarray}
\langle x_{j+1} \vert e^{\frac{-i}{\hbar}\hat{H}\tau} \vert x_j \rangle &=& \int \langle x_{j+1} \vert q \rangle \langle q \vert e^{\frac{i}{\hbar}\hat{H}\tau} \vert p \rangle \langle p \vert x_j \rangle dp dq \\
&=& \int \frac{e^{\frac{-iqx_{j+1}}{\hbar}}}{\sqrt{2\pi \hbar}} \langle q \vert e^{\frac{-i}{\hbar}\hat{H}\tau} \vert p \rangle \langle p \vert x_j \rangle dp dq \\
&=& \int \frac{e^{\frac{-iqx_{j+1}}{\hbar}}}{\sqrt{2\pi \hbar}} \langle q \vert e^{\frac{-i}{\hbar}\hat{H}\tau} \vert p \rangle \frac{e^{\frac{ipx_{j}}{\hbar}}}{\sqrt{2\pi \hbar}} dp dq \\
&=& \int \frac{e^{\frac{-iqx_{j+1}}{\hbar}}}{\sqrt{2\pi \hbar}} \langle q \vert \left( 1 - \frac{i}{\hbar}\frac{p^2}{2m}\tau-iV(x)/ \hbar +\dots \right) \vert p \rangle \frac{e^{\frac{ipx_{j}}{\hbar}}}{\sqrt{2\pi \hbar}} dp dq \\
&=& \int \frac{e^{\frac{-iqx_{j+1}}{\hbar}}}{\sqrt{2\pi \hbar}} \left( 1 - \frac{i}{\hbar}\frac{p^2}{2m}\tau-iV(x)/ \hbar\tau + \dots \right) \frac{e^{\frac{ipx_{j}}{\hbar}}}{\sqrt{2\pi \hbar}}\delta(p-q) dp dq \\
&=& \int \frac{e^{\frac{-ip(x_{j+1}-x_j)}{\hbar}}}{2\pi \hbar} \left( e^{\frac{-i}{\hbar}\frac{p^2}{2m}\tau-iV(x)/ \hbar}\tau \right) dp \\
&=& \int \frac{e^{\frac{-ip(x_{j+1}-x_j)}{\hbar}}}{2\pi \hbar} e^{\frac{-i}{\hbar}\frac{p^2}{2m}\tau}dp e^{-iV(x)\tau/ \hbar}
\end{eqnarray}
We recognize the integral over $p$ to be a Gaussian integral, or the fourier transform of a Gaussian. The result is
\begin{eqnarray}
\langle x_{j+1} \vert e^{\frac{-i}{\hbar}\hat{H}\tau} \vert x_j \rangle &=& \sqrt{\frac{-im}{2 \pi \hbar \tau}} e^{\frac{i}{\hbar}\left(\frac{m}{2 \tau}(x_{j+1}-x_j)^2-V(x)\tau \right)}
\end{eqnarray}
Multiplying all of our transition amplitudes and integrating over the various possible $x_j$ variables, we find the Path integral formulation of Quantum Mechanics!
\begin{eqnarray}
U &=& \left(\frac{-im}{2 \pi \hbar \tau}\right)^{N/2} \int \cdots \int \left( \prod_{j=1}^N dx_j \right) e^{\frac{i}{\hbar}\sum_{j=1}^N \left(\frac{m}{2 \tau}(x_{j+1}-x_j)^2-V(x)\tau \right)}
\end{eqnarray}
If we take the limit as $N \to \infty$, then we get our action in the exponential argument, unlike the discrete quantity we currently have,
\begin{eqnarray}
\lim_{N\to \infty} U &=& \lim_{N\to \infty} \left(\frac{-im}{2 \pi \hbar \tau}\right)^{N/2} \int \cdots \int \left( \prod_{j=1}^N dx_j \right) e^{\frac{i}{\hbar} \int L(x,\dot{x},t) dt}
\end{eqnarray}
(I think what I have called $U$ is in fact the propagator. I also seem to be off by a negative sign in our normalization factor -- the factor $-im$ should be $im$.)
For a free particle, $V(x)=0$ and so we have for our inner product:
\begin{eqnarray}
U &=& \left(\frac{-im}{2 \pi \hbar \tau}\right)^{N/2} \int \cdots \int \left( \prod_{j=1}^N dx_j \right) e^{\frac{i}{\hbar}\sum_{j=1}^N \left(\frac{m}{2 \tau}(x_{j+1}-x_j)^2\right)}
\end{eqnarray}
Examining the first few terms of this nasty integral, we see that we have
\begin{eqnarray}
\int dx_1 e^{\frac{mi}{2\hbar \tau}\left(x_0^2 - 2x_0x_1+ x_1^2 -2x_1x_2 + 2x_2^2 -2x_2x_3+ \dots x_n^2\right)}
\end{eqnarray}
Integrating over each successive $x_{n-1}$ yields
\begin{eqnarray}
\sqrt{\frac{2\pi \hbar \tau}{-im(n-1)}}^{n-1}e^{\frac{-im}{2\hbar (n-1)\tau}\left(x_{n}-x_0\right)^2}
\end{eqnarray}
Notice that the factor next to our Gaussian like function is the perfect inverse of our normalization factor for the path integral earlier! (along with a factor of $n-1$, which will create our total time interval, $T=n\tau$ as you'll see in a minute.) So our final expression for $U$ becomes
\begin{eqnarray}
U = \langle x_f, t_f \vert x_i, t_i \rangle &=& \sqrt{\frac{2\pi \hbar }{-imT}}\mathrm{exp}\left[\frac{-im}{2\hbar T}\left(x_f-x_0\right)^2 \right]
\end{eqnarray}
So this expression gives the probability of transitioning from one state $\vert x_0, t_0 \rangle$ to another $\langle x_f, t_f \vert$. This expression is Gaussian, although I don't know what the complex phase means in the exponential argument. It is certainly normalized, since we are multiplying by $\sqrt{2\pi/\sigma^2}$ ...
