In a sense this is a follow up question to The mathematical theory of Feynman integrals although by all rights it should precede that question.

Let $S$ be a polynomial with real coefficients in $n$ variables. Is there a criterion which would say when the integral $$\int_{\mathbf{R}^n}e^{iS(x)}dx$$ converges? Here $dx$ is the Lebesgue measure on $\mathbf{R}^n$ and the integral is understood as the limit of the integrals over the balls of radius $r$ centered at the origin (with respect to the standard metric) as $r\to\infty$.

Some obvious remarks:

  1. If $\deg S=2$, the integral converges if and only if the quadratic part of $S$ is nondegenerate.

  2. If $n=1$, the integral converges if and only if $\deg S>1$.

  3. The answer to the above question is probably classical (but it is unknown to me).

upd: Conjecture (inspired by Jeff's answer below). A sufficient condition for the integral to converge is as follows: let $S_i$ be the degree $i$ part of $S$ and let $V_i,i=1,\ldots,d=\deg S$ be the subvariety of the real projective space $\mathbf{P}^{n-1}(\mathbf{R})$ given by $S_i=0$. The integral converges if $V_2\cap\cdots\cap V_d={\emptyset}$. Here is how one can try to prove this: the above condition is equivalent to saying that the integral along any line converges, so one can try to first integrate along all half-lines emanating from the origin, get a continuous function on the sphere (hopefully) and then integrate it along the sphere. As remark 1. above shows, this condition may be sufficient but it is not necessary.

  • $\begingroup$ See en.wikipedia.org/wiki/Stationary_phase_approximation and en.wikipedia.org/wiki/… $\endgroup$ Commented Jul 15, 2010 at 12:39
  • $\begingroup$ Steve -- thanks, but the pages you mention don't seem to address the convegence issues for general (non-quadratic) polynomials. $\endgroup$
    – algori
    Commented Jul 15, 2010 at 15:53
  • 1
    $\begingroup$ Great question! It would be nice to have a necessary and sufficient criterion, and also a precise connection between the different definitions (as in J.S. answer or via analytic continuation). Did you check Arnolʹd, V. I., Guseĭn-Zade, S. M., Varchenko, A. N., "Singularities of differentiable maps. Vol. II. Monodromy and asymptotics of integrals". Monographs in Mathematics, 83. Birkhäuser Boston, Inc., Boston, MA, 1988. viii+492 pp. ISBN: 0-8176-3185-2. It's the canonical reference for stationary phase with higher degree polynomials S. $\endgroup$ Commented Jul 15, 2010 at 18:19

3 Answers 3


Let $d=\deg S$. If $d\ge 2$ and $S$ is non-degenerate near infinity, in the sense that $|\nabla S(x)|^2 \ge c|x|^{2d-2}$ for $|x|\ge R$ for some positive $R$ and $c$ then the integral converges in a sense close to what you propose. Namely, $$\lim_{r\rightarrow \infty} \int_{\mathbb R^n} \phi(|x|/r) e^{iS(x)} dx$$ exists where $\phi$ is any sufficiently smooth function with $\phi(t)\equiv 1$ for $0\le t\le 1$ and $\phi (t)\equiv 0$ for $t\ge 2$.

To prove this use integration by parts. Fix $r < r'$ and let $$I_{r,r'} = \int_{\mathbb R^n} \phi_{r,r'}(x) e^{i S(x)}dx,$$where $\phi_{r,r'}(x)=\phi(|x|/r)-\phi(|x|/r')$. Assume $r >R$. Multiply and divide by $|\nabla S(x)|^2$, noting that $$|\nabla S(x)|^2e^{i S(x)}= -i\nabla S(x)\cdot \nabla e^{i S(x)}.$$
So we have, after IBP, $$I_{r,r'} = -i \int_{\mathbb R^n} \left [ \nabla \cdot \left ( \frac{\phi_{r,r'}(x)}{|\nabla S(x)|^2} \nabla S(x) \right ) \right ] e^{i S(x)} d x.$$ The integrand is supported in the region $r \le |x| \le 2 r'$, and you can check that it is bounded in magnitude by some constant times $|x|^{1-d}$. (It is useful to note that an $m$-th derivative of $S$ is bounded from above by $|x|^{d-m}$ since it is a polynomial of degree $d-2$.) If $d$ is large enough this may be enough to control the integral. If not repeat the procedure as many times as necessary to produce a factor that is integrable and you can use to show that $I_{r,r'}$ is small for $r,r'$ sufficiently large. Basically, each time you integrate by parts after multiplying and dividing by $|\nabla S(x)|^2$ you produce a an extra factor of size $|x|^{1-d}$.

All of the above can be extended to integrals with $S$ not necessarily a polynomial, but sufficiently smooth in a neighborhood of infinity and with derivatives that satisfy suitable estimates. The real difficulty with such integrals is not proving that they exist, but estimating their size. Here stationary phase is useful, when applicable, but I don't know of much else.

  • $\begingroup$ Jeff -- thanks. This is a nice sufficient condition. $\endgroup$
    – algori
    Commented Jul 15, 2010 at 17:49
  • $\begingroup$ I've added a conjecture which generalizes all this a bit. $\endgroup$
    – algori
    Commented Jul 15, 2010 at 18:09

That integral doesn't converge absolutely, so you'll have to choose some way of making sense of the sum. The one you've chosen above -- taking a limit of integrals on compact sets -- is not so easy to control. What physicists typically do is "analytic continuation": They find a family $[0,\infty) \ni t \mapsto S_t$ of actions for which (a) $S_0 = S$, and (b) $Z_t = \int e^{iS_t(x)}dx$ is absolutely convergent for $t \neq 0$. ($t$ is a real parameter.) Then they define the integral of $e^{iS}$ as the boundary value $\lim_{t \to 0} Z_t$.

For some reason, the physicists speak about this process as if they were analytically continuing from $Z_0$ rather than to $Z_0$.

  • $\begingroup$ A.J. -- thanks. Yes, we can make the integral converge by say adding a term $tq$ to $S$ where $t$ is a small parameter with a positive real part and $q$ is negative definite quadratic. But it is not clear to me how this makes things easier, since the same question arises: for which $S$ can this be analytically continued say to a punctured open neighborhood of 0? $\endgroup$
    – algori
    Commented Jul 15, 2010 at 16:27

I would just like to add a note about this notion of "analytic continuation" that is often employed to make sense of such integrals. There are two dual ways of thinking about it,

  • Often a "convergence factor" of $i\epsilon$ is added to the action and then the "path integration" done. At the end the $\epsilon \rightarrow 0$ limit is taken. Now there are lots of issues about why or when this limit will exist. I guess most often this limit does not exist and various methods are used to make sense of it like "Borel Resummation". This basically does kind of an asymptotic approximation but at low values of the "perturbation parameter". The series expansions of these integrals are generically divergent but a finite number of the terms sum to good approximations for low values of the perturbation parameter.

  • The other way to look at analytic continuation is what is also called the "Feynman Prescription" by way of which one makes a complex analytic continuation of a real function by choosing a contour on which things are well-behaved. A part of this contour will include the region on the real axis where the actual interest is. Most often these functions have singularities on the real line and one has to infinitesimally deform the contour around them and take limits using what is called the Sokhotsky-Plemlj equation.

I haven't seen a rigorous proof of equivalence between these two points of view but on a case by case basis they work out equivalently.

The simplest example where things work out easiest with no glitches is when the integration variables are on finite-dimensional vector spaces and the action is quadratic with no zero eigen values. One basically gets a determinant of the quadratic form. Now this is extrapolated to give "functional determinants" (determinants of differential operators) when the integration variables are Grassmann valued functions. Here open up a new possibility of trouble, i.e there being 0 modes. Some classical theorems assure that on compact manifolds this will not happen but Physics is more often than not done on non-compact manifolds. This invites the whole theory of "regularization". There is yet another beast that can make things go wild, what in physicists' parlance is called "short distance singularity". I would love to know what is the mathematical way of seeing their source and whether they are somehow related to there being 0 modes on non-compact manifolds. I guess these are related.

  • $\begingroup$ It's easy to see that the two approaches you list are equivalent. Just calculate the Feynman propagator for the deformed action S+ie. Also I'm not sure why you think that there are no zero modes on compact spaces. That depends on the action and is often false. Take eg super-symmetric quantum mechanics on any compact manifold. The space of zero modes is then isomorphic to the harmonic forms. Also there is no connection between zero modes and short-distance singularities. Zero modes are a feature of the global structure of spacetime but short distance singularities are the same on any manifold. $\endgroup$ Commented Jul 15, 2010 at 18:48
  • $\begingroup$ I am aware of this idea of yours where by taking the Feynman Propagator of the $i\epsilon$ deformed action one gets the contour integral prescription. (but this too on a case by case basis, I am not aware of a general proof) But I am not so confident of this because this entirely hinges on faith that whatever we did to get the propagator from the path-integral made perfect sense. But having this level of surety is tantamount to saying that I understand what "exactly" Feynman Path Integrals are and hence I am scared of making that claim. $\endgroup$
    – Anirbit
    Commented Jul 16, 2010 at 8:12
  • $\begingroup$ I think there is a large class of differential operators for which it can be shown that they don't have 0 modes on a compact manifold. Many of the common QFT's yield differential operators in this class, for example Klein-Gordon scalar field or the Dirac Operator. I am not sure what is the precise classification of such operators. It would be great if you can give the precise statement. I was only guessing that the existence of 0-modes might have something to do with existence of short distance singularities. Seems my guess was wrong. $\endgroup$
    – Anirbit
    Commented Jul 16, 2010 at 8:33
  • $\begingroup$ A slightly weaker statement that does hold for sure is that on compact connected manifold the laplacian can have a 0 eigenvalue with multiplcity only one. I suppose having a "mass" term should be able to remove this too. $\endgroup$
    – Anirbit
    Commented Jul 16, 2010 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.