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I have questions about an oscillatory integral. Physics papers say the oscillations should "cancel each other out". By this logic, does this integral converge?

$$ \int_0^{\infty} e^{-i x^3} \, dx < \infty$$

Can we even replace it with a monic polynomial $p(x) = x^n + o(x^n)$ and still get convergence?

$$ \int_0^{\infty} e^{-i p(x)} \, dx < \infty$$

What is an upper bound for the constant this converges?

According to this thesis one can use Van der Corput trick to establish convergence, but I have only seen it to prove that $\{ p(n)\}_{n \in \mathbb{N}}$ is equidistributed $(\mod 1)$ is there a relation?

cactus314
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1 Answers1

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The integral $\int_1^\infty e^{-ix^r}\,dx$ is convergent if $r>1$. It is enough to study the integral between $1$ and $\infty$. The change of variables $x^r=t$ gives $$ \int_1^\infty e^{-ix^r}\,dx=\frac1r\int_1^\infty e^{-it}\,t^{1/r-1}\,dt,\quad 1/r-1<0. $$ Now $e^{-it}$ has a bounded primitive and $t^{1/r-1}$ is decreasing and converges to $0$. By Dirichlet's criterion (which amounts to integration by parts) the integral converges.

The argument generalizes to the case in which $x^r$ is replaced by a monic polynomial of degree $n\ge2$: $p(x)=x^n+\dots$. Let $z$ be such that $p$ is increasing and convex on $[z,\infty)$. The change of variables $t=p(z)$ gives $$ \int_z^\infty e^{-ip(x)}\,dx=\int_{p(z)}^\infty e^{-it}\,t^{1/r-1}\,\frac{dt}{p'(p^{-1}(t))},\quad 1/r-1<0. $$ It is easy to check that $1/p'(p^{-1}(t))$ is decreasing and converges to $0$, so that the integral is convergent.

Julián Aguirre
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