The semiclassical limit you're describing says that the amplitude for a particle to get from here to there in a set time is equal to the exponential of the classical action for the corresponding classical trajectory. In symbols this reads
$$\langle x_b|U(T)|x_a\rangle=\int \mathcal{D}\varphi e^{iS[\phi]/\hbar} \approx
e^{{i}S[\varphi_\textrm{cl}(x_a,x_b,T)]/\hbar}.$$
In a general quantum state, however, particles are not "here" and don't end up "there": they have an initial probability amplitude $\langle x|\psi(0)\rangle$ for being at each position $x$ at time $t=0$ and will have a final probability amplitude $\langle x|\psi(T)\rangle$ for being at position $x$ at time $T$. To apply the approximation, you pull out the propagator and insert a resolution of the identity:
$$\langle x|\psi(T)\rangle=\int dy\langle x|U(T)|y\rangle\langle y|\psi(0)\rangle=\int dye^{{i}S[\varphi_\textrm{cl}(y,x,T)]/\hbar}\langle y|\psi(0)\rangle.$$

To get a full semiclassical limit, you also need a semiclassical initial state (since otherwise you've obviously got no hope!). You take, then, a state with (relatively) sharply defined position and momentum (of course, the state will occupy some finite region of phase space but you usually can assume, in these circumstances, that it is small enough), and this will make the amplitudes for points outside the classical trajectory interfere destructively and vanish.

EDIT

So how does this happen? For one, $y$ must be close to the initial position, $y_0$ in order to contribute to the integral. For small displacements of the endpoints, then, the action along the classical trajectory varies as $$\delta S=p_{\varphi,x}\delta x-p_{\varphi,y}\delta y$$ (cf. Lanczos, The Variational Principles of Mechanics, 4th edition, Dover, eqs 53.3 and 68.1, or simply do the standard integration by parts and set $\int\delta L dx=0$ along the classical trajectory). The main contribution of the initial state to the phase is of the form $e^{ip_\textrm{cl}y}$, which means that the integral has more or less the form, up to a phase,
$$\langle x|\psi(T)\rangle\approx \int_{y_0-\Delta x/2}^{y_0+\Delta x/2}e^{i(p_{\varphi,y}-p_\textrm{cl})y/\hbar}dy.$$

Here the momentum $p_{\varphi,y}$ is determined by $x$ and (to leading order) $y_0$, since there is a unique classical path that connects them. This momentum must match (to precision $\Delta p\approx \hbar/\Delta x$, which we assume negligible in this semiclassical limit) the classical momentum of the initial state, $p_\textrm{cl}$, and therefore only those $x$'s on the trajectory determined by the initial state will have nonzero amplitudes.

Further to Jonathan's answer, it seems to me that the integrals you're worried about are not actually infinite:
$$\int_{-\infty}^\infty da_1\cdots \int_{-\infty}^\infty da_N e^{\frac{im}{2\hbar}\sum_{n=1}^N\frac{(n\pi)^2}{2T}a_n^2}
=\prod_{n=1}^N\int_{-\infty}^\infty da_n e^{i\frac{m \pi^2}{4\hbar T}n^2 a_n^2}$$
and each of the individual integrals is a Fresnel integral with a finite result, including a nontrivial phase. However, the $a_n$ are lengths and therefore carry dimensional information, so that your final result (proportional to $(\hbar T/m)^{N/2}$ from dimensional analysis) is wrong by some $N$-dependent constant that comes from the measure normalization. Fixing that should let you get on with the fun.

## Best Answer

Yes, the Schaden and Spruch interpretation is correct. The interpretation is not used much because it's not as well connected into how experiments are run.

In the usual text books, the Fourier transform is taken over position and time $(\vec{x},t)$ to get energy and momentum $(E,\vec{p})$. This is done by four integrations, one each getting rid of one of the four variables $x_1, x_2, x_3, t$, and replacing with the corresponding one of the four Fourier transformed variables $p_1, p_2, p_3, E$. Mathematically, nothing is wrong with only doing one of the four transforms, and the interpretation is clear.