The relativistic particle action is (in its more natural form)
$$A=McS=Mc\int_{\lambda_a}^{\lambda_b}d\lambda\sqrt{x'^2(\lambda)}.\tag{19.12}$$
That action doesn't lend itself easily to a calculation of the corresponding path integral, so (as done in many papers and textbooks) we can use the alternative form that is more suitable since it is quadratic:
$$\bar{A}=\int_{\lambda_a}^{\lambda_b}d\lambda \left[\frac{Mc}{2h(\lambda)}x'^2(\lambda)+h(\lambda)\frac{Mc}{2}\right].\tag{19.10}$$
Although the $\bar{A}$ describes the same classical physics as the original action (19.12), I wonder that it may lead to a different quantum physics. The textbooks give some arguments for equivalence but they don't actually "calculate" the integral, which seems imposible due to the square root. Kleinert's book1 "Path Integrals in Quantum Mechanics, Statistics, Polymer Physics, and ..." is the only one I have found (if you know another reference dealing with the equivalence issue please tell me), page 1424, Appendix 19A. It seems that he got
$$(x_b|x_a)=N''\int\frac{d^D k}{(2\pi)^D}\frac{1}{k^2+M_Rc^2/\hbar^2}e^{ik}(x_b-x_a)\tag{19A.15b}$$
with
$$z \equiv \nu\log\nu , \qquad \nu\equiv \frac{D+1}{\bar{\epsilon}}\lambda_C, \qquad \bar{\epsilon} = \frac{S}{N+1},\tag{19A.13}$$ $$M_R=M(1+z)^{1/2} .\tag{19A.16}$$ So, in principle, they are different, similar but different.
So, my questions are:
What is happening here? I mean with the mass, it looks like regularization or renormalization.
Is it necessary to calculate both path integrals in order to show they are equivalent?
Has the path integral been calculated correctly (by Kleinert) without cheating nor using any hocus-pocus process? Or what has he done?
1 If you don't have he book, please read this draft http://users.physik.fu-berlin.de/~kleinert/b5/psfiles/pthic19.pdf, page 1436.