Show that $det(\frac{\partial X_i}{\partial \alpha_j}) \ge 0$ where $X(\alpha,t)$ is the flow of the vector field $F(X,t)$

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Let $F(x,t)$ be a smooth vector field in $R^N$ , $X(\alpha,t)$ be the flow of $$\frac{dX}{dt}=F(X,t)$$ with $X(\alpha,0)=\alpha$ . Assume also that $X(\alpha,t)$ can be defined for all $\alpha \in R^N$ and $t \in R$ .
Let $J(\alpha,t)=det(\frac{\partial X_i}{\partial \alpha_j})$, can we show that $$\int_{X(\Omega,t)}f(x,t)\,dx =\int_{\Omega}f(X(\alpha,t),t)J(\alpha,t)\,d\alpha$$
Where $f$ is a smooth function , $\Omega$ is a bounded domain with smooth boundary , $X(\Omega,t)$ consist of all point of the form $X(\alpha,t)$ with $x \in \Omega$ .

Note that $J(\alpha,0)=1$ , so $J \ge 0$ in a neighborhood of $(\Omega,0)$ . If $J\ge 0$ for all $(\alpha,t)$ , then we have $$\int_{X(\Omega,t)}f(x,t)\,dx =\int_{\Omega}f(X(\alpha,t),t)|J(\alpha,t)|\,d\alpha=\int_{\Omega}f(X(\alpha,t),t)J(\alpha,t)\,d\alpha$$ When $t$ is large enough , I want to show $J(\alpha,t)$ still $\ge0$ . I tried to construct some examples in one dimensional , when $X(\alpha,t)$ is of the form $f(t)\alpha$ with $f(0)=1$, then $F(X,t)=\frac{Xf'(t)}{f(t)}$ . So $f(t)\gt 0$ , $J\gt 0$ .