Is it true that $\nabla_x \int_0^\infty f(t,0) dt = 0 \implies \nabla_x f(t,0) = 0 \ \forall t>0$? [closed]

Question

Let $f:\mathbb R_+ \times \mathbb R^N \to \mathbb R$ and $$F(x) = \int_0^\infty f(t,x) dt.$$ If $\nabla_x F(0) = 0$ do we have that $\nabla_x f(t,0) = 0$ for all $t \in \mathbb R_+$? If not, which assumptions do we need to have to make it true?

The motivation comes from the fact that if $f$ solves the heat equation with initial data $f_0$ and boundary data $0$ in a bounded domain $\Omega$, then $\int_0^\infty f(t,x) dt$ solves the Laplace equation with right-hand side $f_0$ and $0$ boundary data.

So, if needed, you can think of the function $f$ of the statement asked above as the solution of the heat equation.

Answer 1

No, the implication in the title is not true, as a simple counterexample, take a factorized $f(t,x)=a(t)b(x)$, with $\int_0^\infty a(t)dt=0$ but $a(t)$ not identically equal to zero and $\nabla_x b(0)\neq 0$; then $F(x)=0$ identically but $\nabla_x f(t,0)=a(t)\nabla_x b(0)\neq 0$.