Proof plan. The integral $I(x) = \int_{\mathbb{R}} |\xi|^\beta e^{i(x\xi + \xi^3)}\, d\mathcal{L}^1(\xi)$ is an oscillatory integral with phase $\varphi_x(\xi) = x\xi + \xi^3$. The strategy is to split into low and high frequencies using a smooth cutoff, then analyse the high-frequency part in two spatial regimes: when $x \ge -1$ (no stationary points in the support) and when $x < -1$ (stationary points present). The first regime uses integration by parts (non-stationary phase), while the second uses the Van der Corput lemma.

Step 1: Frequency decomposition. Let $\psi \in C_c^\infty(\mathbb{R})$ with $\psi(\xi) = 1$ on $[-1, 1]$ and $\psi(\xi) = 0$ outside $[-2, 2]$. Write
\begin{align*} I(x) = \underbrace{\int_{\mathbb{R}} |\xi|^\beta e^{i(x\xi + \xi^3)} \psi(\xi)\, d\mathcal{L}^1(\xi)}_{=:\, I_{\mathrm{low}}(x)} + \underbrace{\int_{\mathbb{R}} |\xi|^\beta e^{i(x\xi + \xi^3)} (1 - \psi(\xi))\, d\mathcal{L}^1(\xi)}_{=:\, I_{\mathrm{high}}(x)}. \end{align*}
Since $\psi$ has compact support and $\beta \ge 0$, the integrand of $I_{\mathrm{low}}$ is absolutely integrable, giving $|I_{\mathrm{low}}(x)| \le C$ uniformly in $x$.

Step 2: High frequencies, Case 1 ($x \ge -1$). The phase derivatives are $\varphi_x'(\xi) = x + 3\xi^2$ and $\varphi_x''(\xi) = 6\xi$. On the support of $1 - \psi$, we have $|\xi| \ge 1$, so $3\xi^2 \ge 3$. Since $x \ge -1$:
\begin{align*} |\varphi_x'(\xi)| = |x + 3\xi^2| \ge 3 - 1 = 2. \end{align*}
There are no stationary points, and $|\varphi_x'(\xi)| \gtrsim |x| + \xi^2$. Integrating by parts once (as in the non-stationary phase lemma):
\begin{align*} I_{\mathrm{high}}(x) = i \int_{|\xi| \ge 1} \partial_\xi\left(\frac{|\xi|^\beta(1 - \psi(\xi))}{x + 3\xi^2}\right) e^{i\varphi_x(\xi)}\, d\mathcal{L}^1(\xi). \end{align*}
The derivative acting on the amplitude produces terms bounded by $O(|\xi|^{\beta - 3})$. Since $\beta \le \frac{1}{2}$, the integrand decays as $|\xi|^{-5/2}$, making it absolutely integrable. Hence $|I_{\mathrm{high}}(x)| \le C$ uniformly for $x \ge -1$.

Step 3: High frequencies, Case 2 ($x < -1$). Here $\varphi_x$ has stationary points at $\xi = \pm\xi_0$, where $\xi_0 = \sqrt{|x|/3} \ge 1/\sqrt{3}$. We isolate these using a secondary cutoff $\rho \in C_c^\infty(\mathbb{R})$ with $\rho(s) = 1$ for $|s| \le \frac{1}{2}$ and $\rho(s) = 0$ for $|s| \ge \frac{3}{4}$. Define
\begin{align*} \rho_{\mathrm{near}}(\xi) := \rho\!\left(\frac{3\xi^2 - |x|}{|x|}\right), \qquad \rho_{\mathrm{away}}(\xi) := 1 - \rho_{\mathrm{near}}(\xi), \end{align*}
and split $I_{\mathrm{high}} = I_{\mathrm{near}} + I_{\mathrm{away}}$.

Away from stationary points. On the support of $\rho_{\mathrm{away}}$, $|3\xi^2 - |x|| \gtrsim |x|$, which implies $|\varphi_x'(\xi)| = |x + 3\xi^2| \gtrsim |x| + \xi^2$. As in Case 1, integration by parts gives $|I_{\mathrm{away}}(x)| \le C$.

Near the stationary points. On the support of $\rho_{\mathrm{near}}$, we have $|3\xi^2 - |x|| \le \frac{3}{4}|x|$, so $\frac{1}{4}|x| \le 3\xi^2 \le \frac{7}{4}|x|$. In particular $|\xi| \sim \sqrt{|x|}$, and the second derivative satisfies
\begin{align*} |\varphi_x''(\xi)| = 6|\xi| \sim \sqrt{|x|}. \end{align*}
The amplitude $|\xi|^\beta(1 - \psi(\xi))\rho_{\mathrm{near}}(\xi)$ is a $C^1$ function supported where $|\xi| \sim \sqrt{|x|}$. Applying the Van der Corput lemma with $k = 2$:
\begin{align*} |I_{\mathrm{near}}(x)| \lesssim \frac{\sup_{\mathrm{supp}(\rho_{\mathrm{near}})} |\xi|^\beta}{\sqrt{\inf_{\mathrm{supp}(\rho_{\mathrm{near}})} |\varphi_x''(\xi)|}} \lesssim \frac{|x|^{\beta/2}}{|x|^{1/4}} = |x|^{\beta/2 - 1/4}. \end{align*}
This remains bounded as $x \to -\infty$ precisely when $\beta/2 - 1/4 \le 0$, i.e. $\beta \le \frac{1}{2}$.

Step 4: Conclusion. Combining all contributions: $|I(x)| \le |I_{\mathrm{low}}(x)| + |I_{\mathrm{away}}(x)| + |I_{\mathrm{near}}(x)| \le C$ uniformly in $x$, for all $\beta \in [0, \frac{1}{2}]$.