# Integral of the exponential of a homogeneous quartic - reference request

For a calculation I am doing, I have to calculate an integral of the form

$$I = \int_{\mathbf{R}^n} \exp[-Q(\mathbf{x})] d^n\mathbf{x} \text,$$

where $$Q(\mathbf{x})$$ is a homogenous, degree-4 polynomial in $$n$$ real variables that is positive for nonzero $$\mathbf{x}$$. After a lot of searching in the usual places, I tracked down the following from Wikipedia:

For multivariate quartic Gaussian integrals is:

$$\int_{-\infty}^\infty \exp\left(- \sum_{i,j,k,l=1}^{n}A_{ijkl} x_i x_j x_k x_l \right) \, d^nx =\frac{(\frac{1}{2}\Gamma(\frac{1}{4}))^n}{\det(A)^{1/4}}$$

where det(A) is a hyperdeterminant of the 4-tensor A (which is simply the contraction of A with Levi Civita symbols as in the quadratic case).

If true, this is exactly what I need, but it is provided without citation. This was added to Wikipedia by an unregistered user on July 1, 2014. Can anyone provide a reference for this purported integration formula? Can anyone think where I might look for a reference that would have this?

UPDATE: The plot thickens. The stated identity appears to be false, but I found another source that seems to somewhat agree with it.

Let us limit ourselves to looking at the simplest nontrivial case, $$n=2$$.

In the preprint arXiv:hep-th/9706001, at the bottom of page 11, the author has

Let $$\mathrm{dim}\, V = 2$$ (case of binary forms, when our physical space consists only of 2 points), so $$S(v) = a_d x^d + a_{d−1} x^{d-1} y + \ldots a_0 y^d$$. In this case $$\deg S = 2(d − 1)$$. Then:

$$Z(S) := \int_V \exp(i(a_d x^d + a_{d−1} x^{d-1} y + \ldots a_0 y^d))dxdy$$ $$=\frac{\Lambda(2,d)}{|\mathrm{Dis}(S)|^{1/d(d-1)}}\exp(i\pi\mathrm{sgn}(S))$$ here $$\mathrm{Dis}(S)$$ is the usual discriminant of polynomial of degree $$d$$.

$$\mathrm{sgn}(S)$$ is explained to be a phase constant that depends on which component of the complement of the zero set of $$\mathrm{Dis}$$ contains $$S$$. This differs from my original question because of the imaginary unit in there, but it's close enough.

Now consider $$S_c= x^4+c x^2 y^2+ y^4$$. We have $$\mathrm{Dis}(S_0)=256$$ and $$\mathrm{Dis}(S_{1/2})=225$$, and $$\mathrm{Dis}>0$$ for all $$0, so we expect $$Z(S_{1/2})/Z(S_0) = (256/225)^{1/12}= 1.0108$$. I performed the integrals $$Z(S_0)$$ and $$S(S_{1/2})$$ both numerically and symbolically in Mathematica, and I get $$Z(S_{1/2})/Z(S_0) = 1.0531$$. No other rational power instead of $$1/12$$ fixes the discrepancy.

Moreover, $$S_2$$ has a vanishing discriminant but $$Z(S_2)$$ is finite.

• Have you tried it out on some examples? One counter example is all it takes... Jun 16, 2015 at 2:30
• @Zach466920, I have now tried some examples and the formula appears to be wrong. Jun 16, 2015 at 19:12
• You might want to go edit the Wikipedia page to help out future users... Jun 17, 2015 at 15:57
• @Zach466920 I removed the offending part of the Wikipedia article. I might put in something about integral discriminants later, when I have the time. Jun 17, 2015 at 16:01

I found a paper that derives what appears to be the correct solution for $I$ in the case $n=2$. The answer turns out to be significantly more complicated already for $n=2$ than the formula given in the question and it is given in terms of hypergeometric functions of the $SL(n)$ invariants of the form $Q$ (the hyperdeterminant is only one of several invariants).