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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.

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1 Answer 1

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.

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