The following question is a simple case of a type of problem that occurs in combinatorial enumeration problems.

Define $$F(x_1,\ldots,x_n) = \frac{1}{(2\pi)^{n/2}}\exp\biggl( -\frac12\sum_{j=1}^n x_j^2 + i\,n^{-1/4} \sum_{j=1}^n x_j^3 \biggr), $$ where $i$ is the imaginary unit. Let $\epsilon\gt 0$ be small enough. Let $$I_0(n) = \int_{-n^\epsilon}^{n^\epsilon}\cdots \int_{-n^\epsilon}^{n^\epsilon} F(x_1,\ldots,x_n) ~ dx_1\cdots dx_n. $$ We can estimate $I_0(n)$ as $n\to\infty$ by factoring $$F(x_1,\ldots,x_n) = \prod_{j=1}^n \frac{1}{\sqrt{2\pi}} \exp\bigl(-x_j^2/2+in^{-1/4}x_j^3\bigr),$$ which separates the integral into a product of $n$ 1-dimensional integrals. This very easily gives $$I_0(n) = \exp\bigl( -n^{1/2}/2 + O(n^{-3})\bigr).$$ So far so good. But now let $A=A(n)$ be a symmetric positive definite matrix. We can assume it is pretty nice, say all eigenvalues bounded between two positive constants independent of $n$. Now define $$G(x_1,\ldots,x_n) = \frac{1}{(2\pi)^{n/2}}\exp\biggl( -\frac12 \mathbf{x}^TA\mathbf{x} + i\,n^{-1/4} \sum_{j=1}^n x_j^3 \biggr). $$ How do we estimate $$I_1(n) = \int_{-n^\epsilon}^{n^\epsilon}\cdots \int_{-n^\epsilon}^{n^\epsilon} G(x_1,\ldots,x_n) ~ dx_1\cdots dx_n? $$ The source of the difficulty is that $\int G$ is exponentially smaller than $\int |G|$, so as soon as you approximate the integrand the answer goes away. What to do?

Note that in this case (and very commonly in practice) symmetry implies that $I_1(n)$ is real. So we can discard the imaginary part of the integrand and integrate only the real part, which is $$ \mathfrak{R}G(x_1,\ldots,x_n) = \frac{1}{(2\pi)^{n/2}}\exp\biggl( -\frac12 \mathbf{x}^TA\mathbf{x}\biggr)\cos\biggl(n^{-1/4} \sum_{j=1}^n x_j^3 \biggr). $$ Not sure that helps.

ADDED 2 Jan 2018. Mikhail Isaev and I now have a method that can precisely estimate such integrals in many general cases. It isn't completely written up yet.

  • $\begingroup$ What is the limit here? Is $n$, the number of dimensions, going to $\infty$? $\endgroup$ Commented Nov 25, 2013 at 12:50
  • $\begingroup$ @Igor : Yes, I forgot to mention that. $\endgroup$ Commented Nov 25, 2013 at 13:00
  • 1
    $\begingroup$ Is $i$ the imaginary unit? Also, if since $x_j^3$ is an odd function, doesn't its contribution disappear when you write the G(...) as a product? (typo in G, $x_i$ should be $x_j$) $\endgroup$
    – Suvrit
    Commented Nov 25, 2013 at 15:10
  • $\begingroup$ @Suvrit : Yes, $i$ is the imaginary unit. $x_j^3$ is an odd function but $\exp(x_j^3)$ isn't. I fixed the typos, thanks. $\endgroup$ Commented Nov 25, 2013 at 22:21
  • $\begingroup$ diagonal $A$ also seems to be ok. But after that, it seems to require a new idea....I wonder if numerically one can come up with a good guess about the asymptotics---and you mean that bounding using $\lambda_\min(A)x^Tx \le x^TAx \le \lambda_\max(A)x^Tx$ is too loose... $\endgroup$
    – Suvrit
    Commented Nov 25, 2013 at 23:19

1 Answer 1


I'll consider the integrals extended from $-\infty$ to $\infty$ and show that the integral of $G$ is exponentially small compared to the integral of $|G|$ -- precisely, the oscillating integral is smaller by a factor of $C_0\exp(-C \sqrt{n})$ for some positive constants $C_0$ and $C$ that depend only on the eigenvalues of the quadratic form $A$. One can get the same estimate for the truncated integrals. It may be possible to refine this to get asymptotics but I have not thought about that.

The problem asks for an estimate for $$ I = \frac{1}{(2\pi)^{n/2}} \int_{-\infty}^{\infty} \cdots \int_{-\infty}^{\infty} \exp\Big( -\frac{1}{2} A(x_1,\ldots,x_n) +\frac{i}{n^{\frac 14}} \sum_j x_j^3\Big) dx_1\ldots dx_n. $$ where $A$ is a positive definite quadratic form. We wish to show that $I$ is small compared to the same integral without the oscillating term; this is $$ \frac{1}{(2\pi)^{n/2} } \int_{-\infty}^{\infty}\cdots \int_{-\infty}^{\infty} \exp\Big(- \frac 12 A(x_1,\ldots,x_n )\Big) dx_1 \ldots dx_n = (\text{det} A)^{-\frac 12}. $$

Think of the integrals in $I$ as contour integrals being integrated on the real axis. The idea is to replace the integrals on the real axis by integrals along the line from $-\infty + i\alpha$ to $\infty +i\alpha$ for some real number $\alpha$ to be chosen carefully. Thus $$ I =\frac{1}{(2\pi)^{n/2}} \int_{-\infty+i\alpha}^{\infty+i\alpha} \cdots \int_{-\infty+i\alpha}^{\infty+i\alpha}\exp\Big(-\frac12 A(z_1,\ldots,z_n) + \frac{i}{n^{\frac 14}} \sum z_j^3 \Big) dz_1 \ldots dz_n, $$ and writing now $z_j =x_j +i\alpha$ this equals $$ \frac{1}{(2\pi)^{n/2}} \int_{-\infty}^{\infty}\cdots \int_{-\infty}^{\infty} \exp\Big( -\frac 12 A(x_1+i\alpha,\ldots, x_n+i\alpha) + \frac{i}{n^{\frac 14}} \sum_j (x_j+i\alpha)^3\Big) dx_1\ldots dx_n. $$

We now estimate the integral above by just taking the absolute value of the integrand (and choosing $\alpha$ carefully). The integrand is in modulus $$ \exp \Big( -\frac 12 A(x_1,\ldots, x_n) + \frac{\alpha^2 }{2} A(1,\ldots, 1) +\alpha^3 n^{\frac 34} -\frac{3\alpha}{n^{\frac 14} } \sum_j x_j^2\Big). $$ We will take $\alpha =\beta/n^{\frac 14}$ for a suitably small positive constant $\beta$. Since the eigenvalues of $A$ are bounded between two positive constants independent of $n$, we have that $A(1,\ldots,1) \le 2C_1 n$ for some positive constant $C_1$, and that $3\sum_j x_j^2 \ge \frac{C_2}{2} A(x_1,\ldots, x_n)$ for some positive constant $C_2$.
Thus the quantity above is $$ \le \exp\Big(C_1 \beta^2 \sqrt{n} + \beta^3 - \frac{1}{2} A(x_1,\ldots,x_n) \Big(1+\frac{C_2\beta}{\sqrt{n}}\Big) \Big). $$ Integrating this, we find that $$ I \le \frac{\exp(C_1 \beta^2\sqrt{n} +\beta^3)}{(2\pi)^{n/2}} \int_{\infty}^{\infty}\cdots\int_{-\infty}^{\infty} \exp\Big(-\frac 12 A(x_1,\ldots,x_n) \Big(1+\frac{\beta C_2}{\sqrt{n}}\Big) \Big) dx_1 \ldots dx_n $$ which is readily seen to be $$ \exp(C_1 \beta^2\sqrt{n} +\beta^3) (\text{det} A)^{-\frac 12} \Big( 1 +\frac{\beta C_2}{\sqrt{n}}\Big)^{-n/2}. $$

In other words we have shown that $I$ is smaller than the trivial bound by a factor of $$ \exp(C_1 \beta^2\sqrt{n} +\beta^3) \Big( 1 +\frac{\beta C_2}{\sqrt{n}}\Big)^{-n/2} \le \exp(C_1 \beta^2 \sqrt{n} +\beta^3 - C_3 \beta\sqrt{n} \Big) $$ for a suitable positive constant $C_3$. By choosing $\beta$ small enough, we find that this is $\le C_0 \exp(-C\sqrt{n})$ for some positive constants $C_0$ and $C$. This finishes the proof.

  • $\begingroup$ @BrendanMcKay: Is this not really what you wanted? $\endgroup$
    – Lucia
    Commented Dec 5, 2013 at 14:02
  • $\begingroup$ I'm looking for techniques that can be used to do the asymptotics for problems of this nature. In general the integrand is a lot more messy than the example I gave. It is interesting that a shift of the contours can give good bounds - probably I can use that but I think it won't be enough to get precise asymptotics in general. My feeling is that tweaking the contours individually will not be enough but the whole contour surface needs tweaking. Please see mathoverflow.net/questions/37779/… for a related question. $\endgroup$ Commented Dec 5, 2013 at 22:50

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