Feynman path integral normalisation from completeness condition The path integral for some potential can be evaluated explicitly by discretion the space and performing $N$ Gaussian integrals then taking the limit as $N \to \infty$ for the case of a free particle. However, there is a simpler method which takes advantage of the completeness condition for a transition amplitude.
After splitting the action into the classical path plus a variation, $x = x_{cl} + \eta$ on it we have:
$$ \langle x_b, t_b|x_a, t_a\rangle = \int^{x_b}_{x_a}Dx e^{\frac{i}{h}S[x]} = e^{\frac{i}{h}S[x_{cl}]} \int^0_0 D\eta e^{\frac{i}{h} S[\eta]}$$
Now we can we must evaluate the prefactor, $F(T)$, given by the $\eta$ integral which does not depend on the endpoints.
$$F(T) =  \int^0_0 D\eta e^{\frac{i}{h} S[\eta]} = \langle 0, t_b|0, t_a\rangle = \langle 0, T|0, 0\rangle $$
Inserting a complete set of states yields:
$$ \langle 0, T|0, 0\rangle = \int^\infty_{-\infty} dx \langle 0, T|x, t\rangle\langle x, t|0, 0\rangle $$
I have a few questions about the use of this identity to evaluate the normalisation for the free particle and harmonic oscillator.
1. Free Particle
The free particle action is $$S[x] = \frac{m}{2}\frac{(x_b-x_a)^2}{t_b-t_a},$$ plugging this in gives:
$$ F(T) = \int_{-\infty}^\infty dx~ F(T)\exp(\frac{imx^2}{2h(T-t)} )F(t-0)\exp(\frac{imx^2}{2ht}) = \sqrt{\frac{2\pi ih(T-t)t}{mT}} F(T-t)F(t)\tag{2.27}$$
I am okay with the calculation up to this point. Now in my course notes it states that taking the limit for large times ($T \gg  t$) gives:
$$F(t) = \sqrt{\frac{m}{2\pi iht}} $$
which is the correct free particle normalisation. Firstly, why do we evaluate it in this limit? The last step seems true but only through "hand-wavy" arguments can this process be made more clear or rigorous somehow?
2. Harmonic Oscillator
The classical action for the harmonic oscillator is $$S[x_{cl}] = \frac{m\omega}{2\sin\omega T}[(x_a^2 + x_b^2)\cos\omega T -2x_ax_b]$$ 
Following a similar process as for the free particle gives:
$$F(T) = \int_{-\infty}^\infty dx~F(T-t)\exp(\frac{im\omega x^2 \cos\omega (T-t)}{2h\sin\omega (T-t)}) F(t)\exp(\frac{im\omega x^2 \cos\omega t}{2h\sin\omega t}). $$
Performing the Gaussian integral yields:
$$ F(T) = F(T-t)F(t) \sqrt{\frac{2\pi i h \sin\omega (T-t)\sin\omega t}{m\omega \sin\omega T}} \tag{2.35}$$
And finally taking in the limit $T >> t$ we have:
$$F(t) = \sqrt{\frac{m\omega}{2\pi ih\sin\omega t}}$$
Again, for this case I cannot see why this final step is justified.
3. Explicit Calculation
I originally suspected that the previous calculations are only justifiable because they are in agreement with the explicit calculation by discretisation of the path (for a free particle):
$$ \langle x_b, t_b | x_a, t_a \rangle = \lim_{N\to\infty} A_N (\prod^N_{n=1} \int^\infty_{-\infty} dx_n) \exp(\frac{i\epsilon m}{2h}\sum^{N+1}_{n=1}(\frac{x_n-x_{n-1}}{\epsilon}^2))$$
but in that case the normalisation factor for each infinitesimal step is assumed to take the following form:
$$A_N = (\nu(\epsilon))^{N+1} , \nu(\epsilon) = \sqrt{\frac{m}{2\pi ih\epsilon}} $$
for which I can find no justification, other than agreement with the previous calculation.
Any attempt at an explanation will be greatly appreciated.
References:


*

*Lecture Notes, Quantum Theory, The University of Edinburgh,
School of Physics and Astronomy, September 2015. The course notes are available here: https://www2.ph.ed.ac.uk/~bjp/qt/qt.pdf ~p.13-15 for the relevant section.

 A: OP seems to have a point that the argument presented in the lecture notes is not watertight. Let us argue as follows:


*

*Divide the $F$-functions with their sought-for formulas, and call the quotient $f$. Then eqs. (2.27) & (2.35) become on the form
$$f(T)~=~f(T-t)f(t), \tag{A}$$
or equivalently, 
$$f(t+t^{\prime})~=~f(t)f(t^{\prime}). \tag{B}$$

*Let us additionally assume that $f$ is continuous, and not identically zero $f\not\equiv 0$.
Then eq. (B) implies that
$$ f(0)~=~1. \tag{C}$$

*Ignoring some mathematical technicalities, the functional eq. (B) implies that $f$ is an exponential function, i.e. there exists a constant $c$, so that
$$ f(t)~=~e^{ct},\tag{D} $$ 
see e.g. my Phys.SE answer here.

*For small $t\lesssim\tau $ much smaller than some characteristic timescale$^1$ $\tau$, we can evaluate the Hamiltonian path integral directly (with no Feynman fudge factors!). The result is
$$ F(t)~\simeq~  \sqrt{\frac{m}{2\pi ih t}} \quad\text{for} \quad t~\lesssim~ \tau, \tag{E}$$
see e.g. Section V of my Phys.SE answer here. Equivalently,
$$ f(t)~\simeq~  1 \quad\text{for} \quad t~\lesssim~ \tau, \tag{F}$$

*Comparing eqs. (D) & (F), we deduce that 
$$c ~\ll ~1/\tau .\tag{G}$$
Of course, the correct path integral normalization of the free particle and the harmonic oscillator can be directly calculated, and it is known that $c=0$ so that
$$ f(t)~=~1. \tag{H}$$
And hence that the $F$-functions are given by their sought-for formulas.
--
$^1$ E.g. for the harmonic oscillator $\tau=\omega^{-1}$.
