TITLE: Oscillatory integrals of algebraic functions
QUESTION [6 upvotes]: Consider an algebraic function $\phi$ on $R^{d}$.
By this I mean that there exists a polynomial $P$
with coefficients in $R[x_1,...,x_d]$ (coefficients are polynomials!)
such that $P(\phi) = 0$
Let $r$ be a real number >0, I would like to know
whether $1/r^{d}$ times $\int_{[-r,r]^d} e^{i \phi}$
tends to zero when $r$ tends to infinity.
Of course it would be enough to show that this integral is
bounded by $r^{d-\epsilon}$...
I am not at all familiar with the subject, but from the literature I looked at,
I understand that this is the case when $\phi$ itself is a polynomial. I was wondering whether
this holds true for any algebraic function.

REPLY [3 votes]: The answer is no. Consider $\phi(x)=1+1/(x^2 +1)$. Then as $r\rightarrow\infty$ your integral approaches $e^i$.
If you want instead your function $\phi$ to be integral over the polynomial ring, meaning it satisfies a monic polynomial equation, then the answer is yes as long as $\phi$ is not a constant function. To see this, note that you can by partial integration replace the integrand with 
$e^{i\phi(x)}\frac{\phi''(x)}{\phi'(x)^2}$.  Now if $\phi(x)$ grows like $x^a$ then $\frac{\phi''(x)}{\phi'(x)^2}$ grows like $x^{-a}$. The integrality assumption forces $a>0$, which completes the proof.
Note that I assume you want $\phi$ real valued, else all hell breaks loose.