Let $A(t)$ be uniformly positive definite for any $t$, and $x(t)$ is a vector, and $x(0)$ is a finite.

since $\int_0^t x(\tau)\dot{x}(\tau)\,\mathrm d\tau = \frac{1}{2}[x(t)^2-x(0)^2]$ is finite, I wonder whether $\int_0^t x(\tau) A(\tau) \dot{x}(\tau)\,\mathrm d\tau > -\infty$ ?



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