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Let $\lambda\in \mathbb{R}$, $\phi\in C^\infty(\mathbb{R})$. We define the Stein-Wainger oscillatory integral by $$I=p.v.\int_\mathbb{R} e^{i\lambda\phi(t)}\frac{dt}{t}.$$

Stein-Wainger [1] showed that if $\phi(t)$ is a polynomial of degree $d$, then $|I|\le C_d$, where $C_d$ is independent of $\lambda$ and the coefficients of the polynomial.

My question 1 is: Do we always have $|I|\le C_\phi$, where $C_\phi$ is a constant independent of $\lambda$?

The answer is no. See the counterexample by @Mateusz Kwaśnicki in the comments below.

My question 2 is: If $$I=p.v.\int_{-1}^1 e^{i\lambda\phi(t)}\frac{dt}{t},$$ do we always have $|I|\le C_\phi$, where $C_\phi$ is a constant independent of $\lambda$?

I cannot find a theorem or a counterexample for it, though I have searched almost all the related papers online. I guess the answer is "No" and the counterexample may be some "flat" function. Any help would be appreciated.

Remark: In [2], page 248, Theorem 4.1, Nagel and Wainger constructed a "flat" function $\gamma(t)$ such that $$m(x,y)=p.v.\int_{-1}^1e^{i(xt+y\gamma(t))}\frac{dt}{t}$$ is unbounded on $\mathbb{R}^2$.

[1]E. Stein and S. Wainger, The estimation of an integral arising in multiplier transformations, Studia Math. 35 (1970), 101-104.

[2]A. Nagel and S. Wainger, Hilbert Transforms Associated With Plane Curves, Trans. AMS. Vol. 223 (Oct., 1976), pp. 235-252

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  • $\begingroup$ Isn't $\phi(t) = \operatorname{arccot}(t)$ a counter-example? I believe $I$ is infinite in this case, unless $\lambda$ is an even integer. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 27, 2018 at 21:46
  • $\begingroup$ @MateuszKwaśnicki At the beginning we assume $\phi$ is smooth. I think the integral has finite principal value with this assumption. $\endgroup$
    – orange
    Commented Apr 27, 2018 at 22:00
  • $\begingroup$ I meant the continuous arccot, with values in $(0,\pi)$. Singularity at $t=0$ is not an issue, integrability at $\pm\infty$ causes problems: $e^{i \lambda \phi(t)} = (\tfrac{t+i}{\sqrt{1+t^2}})^\lambda$ has limit $(\pm 1)^\lambda$ as $t \to \pm\infty$, and so the integral is not convergent, unless $\lambda$ is an even integer. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 27, 2018 at 22:23
  • $\begingroup$ @MateuszKwaśnicki Thanks! You are right. What if we only consider $\phi$ with compact support? Namely, $I=p.v.\int_{-1}^{1}e^{i\lambda\phi(t)}t^{-1}dt$. $\endgroup$
    – orange
    Commented Apr 27, 2018 at 22:39

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