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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?)

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

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    $\begingroup$ Interesting, thank you. However, this makes Peskin and Schroder very confusing. On page 286, chapter 9, they write "to justify the gaussian integration when the exponent is purely imaginary, recall that time is slightl imaginary", so why would they do this to make the exponent slightly real if the integral converges anyway? $\endgroup$ – Jonathan Lindgren Oct 2 '15 at 8:58
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    $\begingroup$ a complement to Javier's answer is that the integral does not converge in the sense of Lebesgue integration. It converges in the sense that the limit A->infinity of integral up to A exists and one safe way of obtaining this limit is to put the i epsilon and then take the epsilon->0 limit. For QFT putting a cut-off A on the size of the field is not very convenient. $\endgroup$ – Abdelmalek Abdesselam Nov 8 '16 at 23:13

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