I have this formula to calculate the period of a motion in the phase space (plan, in this case) along a phase curve. \begin{equation} T(E)=\int_{x_1}^{x_2}\frac{dx}{\sqrt{2(E-U(x))}} \end{equation} where $E$ is the total energy of the system, $U(x)$ is the potential energy and $x_1$ and $x_2$ are the points of my phase curve that intercept the $x$-axis of the phase plan (points of inversion). Now, I'm asked to calculate the period of the small oscillations in a neighborhood of the point $x_0$, which is the point of potential energy minimum. So, basically, I have to find: \begin{equation} T_0=\lim_{E\to E_0}{T(E)} \end{equation} where $E_0=U(x_0)$.
My attempt: I have used Taylor to evaluate U(x) in a neighborhood of $x_0$. So, considering that E tends to $U(x_0)$ I got: \begin{equation} T(E)=\int_{x_1}^{x_2}\frac{dx}{\sqrt{2(U(x_0)-U(x_0)-U'(x_0)(x-x_0)-\frac{U''(x_0)(x-x_0)^2}{2} }} \end{equation} and, since $U'(x_0)=0$ (because $x_0$ is a minimum for $U(x)$) \begin{equation} T(E)=\int_{x_{1}}^{x_{2}}\frac{dx}{\sqrt{-U''(x_{0})(x-x_{0})^2} } \end{equation} So at this point I'm puzzled because $U''(x_{0})$ is positive for sure, and that leaves me with a negative number under a square root. I'm sure that somewhere i have applied some theorem where I shouldn't have. I feel like it's a mathematical issue.
My book (Arnold) says that the solution is $2\pi$/$\sqrt{U''(x_{0})}$ but I got totally stuck here. Can you help me? Where's my mistake?
P.S. Sorry for my lack in formalism, like omitting little-o notation, etc. but I think you got my point anyways :)