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 ?



Determine $\alpha \in (0,1)$ such that $J_{\alpha}(\phi):=\int \psi/\phi^{\alpha}$ exists?
$\newcommand\al\alpha\newcommand\R{\mathbb R}$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. $\quad\Box$


