Let $0<a<d/2$, let $B$ be the unit ball in $\mathbb{R}^{d}$ centered at the origin, and let $f:B \to [0,\infty[$ be a a smooth function such that
(1) $f(x)\geq f(0).$
(2) $\nabla f(x)\neq 0,\quad \forall x\neq 0.$
(3) The Hessian matrix $D^2 f(0)$ is positive definite.
I am trying to prove $f^{-a}$ is integrable on $B$. I am obviously interested in the case $f(0)=0$.
Since $0$ is a minimum of $f$ we necessarily have $\nabla f(0)=0$. So there exists $\xi \in B$ such that $$f(x)=f(0)+x^T D^2 f(0) x+ \sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi),\quad x \in B.$$
Since $D^2f(0)$ is positive definite, there exists $C>0$ such that $x^T D^2f(0)x\geq C|x|^2$. And, for $x \in B^{\prime}$, a small enough ball centered at $0$, we have $$|\sum_{|\alpha|=3}\frac{x^{\alpha}}{\alpha !} D^{\alpha}f(\xi)|<\frac{1}{2} C|x|^2. \qquad (1)$$ So, for $x \in B^{\prime}$, we have $$f(x)\geq \frac{1}{2} C|x|^2.$$
So we have the integral
$$\int_{B^{\prime}}f^{-a} \lesssim \int_{B^{\prime}}|x|^{-2 a}dx.$$ We are done with this part by the assumption $2a<d$.
What about the easier integral $$\int_{B\setminus B^{\prime}}f^{-a}$$
Edit: I think the answer to my question is simple. This is also inspired by the helpful comments of @Fedor Petrov and @Bazin
Since $D^{2}f(0)$ is positive definite then $f$ is strictly convex in some neighborhood of the origin, say $B(\delta_{0})\subseteq B^{\prime}$. We claim that if $|x|>\delta_{0}$ then $f(x)>\frac{1}{4} C\delta_{0}^2$. (Recall that we have shown $f(y)\geq \frac{1}{2}C|y|^2$ for all $y\in B^{\prime}$). Otherwise $f(x)\leq f(y)$ for some $y \in B(\delta_{0})$. This can not occur since $f$ has no critical points outside $B(\delta_{0})$, by the condition $(2)$.
