Determine $\alpha \in (0,1)$ such that $J_{\alpha}(\phi):=\int \psi/\phi^{\alpha}$ exists? Fix $\alpha \in (0,1)$ and  $\psi\in C^{\infty}_{c}(\mathbb{R}\to \mathbb{R})$. For a smooth function $\phi\geq 0$ define the integral $$J_{\alpha}(\phi):=\int \frac{\psi}{\phi^{\alpha}}.$$
If $|\phi^{\prime}|$ is bounded below away from zero, then
$J_{\alpha}$ exists for all $\alpha<1$. Indeed, one can simply integrate by parts to get
$$J_{\alpha}=-\frac{1}{1-\alpha}\int \phi^{1-\alpha}\left(\frac{\psi}{\phi^{\prime}}\right)^{\prime}.$$
My question is about the case where $\phi^{\prime}$ changes sign finitely many times in the support of $\psi$. Using a smooth partition of unity, one may assume $\phi^{\prime}(x)=0$ at exactly one point $x_{0}$ in the support of $\psi$.
For simplicity, let us further assume that $|\phi^{\prime\prime}|$ is bounded below away from zero near $x_{0}$.
Obviously, we necessarily have $\alpha<1/2$. Just take $\phi$ to be a quadratic function.
I figured this out when $\phi^{\prime\prime}>C>0$: Write
$$\phi(x)=\phi(x_{0})+(x-x_{0})^{2}\int_{0}^{1}(1-t) \phi^{\prime\prime}(x_{0}+t(x-x_{0}))dt\geq \frac{C}{2}(x-x_{0})^{2}$$
In this case we have
$$|J_{\alpha}(\phi)|\leq \frac{2}{C}\int \frac{|\psi|}{|x-x_{0}|^{2\alpha}}$$
which exists for all $\alpha<1/2$.
What is left is the case $\phi^{\prime\prime}<c<0$.
Summary :
Fix $\alpha \in (0,1/2)$ and  $\psi\in C^{\infty}_{c}(\mathbb{R}\to \mathbb{R})$. Let $\phi\geq 0$ be a smooth function with a unique stationary point $x_{0}$ in the support of $\psi$. Assume that $\phi^{\prime\prime}<c<0$.
Show that
$J_{\alpha}(\phi)=\int \frac{\psi}{\phi^{\alpha}}$ is finite ?
 A: $\newcommand\al\alpha\newcommand\R{\mathbb R}$Let us prove the following generalization of your desired statement:

Let $\psi\colon\R\to\R$ be a continuous function with compact support $S$. Let $\phi\colon\R\to\R$ be a nonnegative twice differentiable function with the following property: for any $x_0\in S$ such that $\phi'(x_0)=0$ we have $\phi''(x_0)<0$. Then $J_\al(\phi):=\int\frac{\psi}{\phi^\al}$ is finite for any real $\al>0$.

The proof is based on a comment by Christian Remling. Indeed, take any $x_0\in S$ such that $\phi(x_0)=0$. Since $\phi$ is nonnegative, it follows that $\phi'(x_0)=0$ and hence $\phi''(x_0)<0$, which implies that $\phi<0$ in some punctured neighborhood of $x_0$. This contradicts the condition that $\phi$ is nonnegative.
So, there is no $x_0\in S$ such that $\phi(x_0)=0$. So, $\phi>0$ on $S$. Since $\phi$ is continuous and $S$ is compact, we have $\phi\ge b$ on $S$ for some real $b>0$. Thus, for any real $\al>0$
$$|J_\al(\phi)|\le\int_S\frac{|\psi|}{b^\al}<\infty,$$
since $\psi$ is continuous and $S$ is compact.
We conclude that $J_\al(\phi)$ is finite for any real $\al>0$, as claimed. $\quad\Box$
