In evaluating the path integral for a relativistic point particle in the einbein formulation, it is common to see the following
$$Z = N\int_{x(t_1) = x_1}^{x(t_2) = x_2}\mathcal{D}e\mathcal{D}x\,e^{-\frac{i}{2}\int_0^1dt\,\Big(\frac{1}{e}\dot x^2 - em^2\Big)} = \int_0^{\infty}dL\int_{x(t_1) = x_1}^{x(t_2) = x_2}\mathcal{D}xe^{-\frac{i}{2}\int_0^1dt\,\Big(\frac{1}{L}\dot x^2 - Lm^2\Big)}\,.$$
This is done by gauge fixing the einbein $e$ to a constant $L$. Intuitively this formula makes perfect sense. We only want to integrate over physically distinct profiles of $e$, which are given by different values of the parameter $L$.
For my own understanding, I am trying to arrive at the same result via the Faddeev Popov trick, but unsuccessfully. My attempt is as follows. First, define the Faddeev Popov determinant via the equation
$$1 = \Delta(e)\int \mathcal{D}\xi\, \delta(e - L[\xi]),$$
where $L[\xi]$ is the value of a constant einbein $L$ after a (finite) gauge transformation has been applied. Substituting this into the expression for $Z$ gives
$$Z = N \int\mathcal{D}\xi \int_{x(t_1) = x_1}^{x(t_2) = x_2}\mathcal{D}e\mathcal{D}x\,e^{-\frac{i}{2}\int_0^1dt\,\Big(\frac{1}{e}\dot x^2 - em^2\Big)}\Delta(e)\delta(e - L[\xi])$$
Following the usual procedure, we may do the integral over $e$:
$$Z = N \int\mathcal{D}\xi \int_{x(t_1) = x_1}^{x(t_2) = x_2}\mathcal{D}x\,e^{-\frac{i}{2}\int_0^1\Big(\frac{1}{L[\xi]}\dot x^2 - L[\xi]m^2\Big)}\Delta(L[\xi]).$$
Finally, because the integrand is gauge invariant, we drop the $\xi$ and separate out the volume of the gauge group:
$$Z = N\text{ Vol} \int_{x(t_1) = x_1}^{x(t_2) = x_2}\mathcal{D}x\,e^{-\frac{i}{2}\int_0^1\Big(\frac{1}{L}\dot x^2 - Lm^2\Big)}\Delta(L).$$
I have tried to proceed by computing the determinant as follows:
$$1 = \Delta(e) \int \mathcal{D}\xi\,\delta(L - L - \dot L\xi - L\dot\xi) = \Delta(e) \int \mathcal{D}\xi\,\delta(- L\dot\xi) = \frac{\Delta}{L}\times \text{Const.}$$
In the first equality, I have used the infinitesimal form of the gauge transformation, which I think is allowed because the $\delta$ function will only have support for $\xi$ near zero. In the second equality, I have set $\dot L = 0$, and in the third equality, I have scaled $L$ out of the $\delta$ function. Substituting this back into our prior expression gives
$$Z = N \text{ Vol} \int_{x(t_1) = x_1}^{x(t_2) = x_2}\mathcal{D}x\,e^{-\frac{i}{2}\int_0^1dt\,\Big(\frac{1}{L}\dot x^2 - Lm^2\Big)}L\,,$$
where I have absorbed the constant from the evaluating the determinant into $N$. This final result is clearly incorrect, as the path integral evaluates to the propagator for the non-relativistic point particle.
One can try and salvage it as follows: the original $Z$ had no $L$ dependence, so we are free to integrate both sides of the above expression against a function $L$ without changing the answer. We can choose to integrate both sides against $1/L$, giving
$$Z = \int_0^{\infty}dL\int_{x(t_1) = x_1}^{x(t_2) = x_2}\mathcal{D}xe^{-\frac{i}{2}\int_0^1dt\,\Big(\frac{1}{L}\dot x^2 - Lm^2\Big)}\,.$$
This is the desired final result, but something else has gone wrong because we've clearly change the value of $Z$ in this final step.
What is the correct way to apply the Faddeev Popov trick in this example?
