# Gaussian path integrals and convergence

The Hamiltonian path integral in quantum mechanics, for a particle with coordinate $q$ and momentum $p$ and Hamiltonian $H=p^2/2m+V(q)$, is

$\int \mathcal{D}q(t)\mathcal{D}p(t)e^{i\int_0^T(p\dot{q}-\frac{p^2}{2m}-V(q))}$

Now, to go to the Lagrangian formulation, it seems like the standard procedure is to complete the square for $p$ and then evaluate the gaussian integral to "integrate out" $p$. My question here is, why can we do that? The exponent is purely imaginary, and the gaussian integral should only be well defined if the real part of coefficient of $p^2$ is negative, right?

(In Peskin and Schroder (Chapter 9, Functional methods), when they evaulate the full path integral for a free field, they comment on this and say that convergence is guaranteed because the time $T$ is slightly imaginary. However, when they earlier did the above operation to integrate out $p$, they did not comment on this issue at all. Are these two cases different or are they both solved by slightly imaginary $T$ in some way?)

• Remember the $i\epsilon$ prescription? This is needed to make this integral well-defined. – Prahar Oct 20 '16 at 14:42

The Gaussian integral with a purely imaginary exponent actually converges because of the increasingly fast oscillations. This Math.SE question has a bunch of (fully rigorous) proofs that $\int \sin(x^2)\, dx$ converges, which is of course the imaginary part of $\int \exp(ix^2)\, dx$. The real part can be similarly be proved to converge.