Gradient wrt a parameter of a parametrized integral

Question

I am unable to understand how the operator is taken out of the integral. Is this some standard practice?

$$ \begin{aligned} &=\int_{-\infty}^{\infty} \nabla_{\theta} p(y ; \theta) d y \\ &=\nabla_{\theta} \int_{-\infty}^{\infty} p(y ; \theta) d y \\ &=0 \end{aligned} $$

Answer 1

The operator is the gradient with respect to $\theta$, while the integral is with respect to $y$, so this is valid as long as the two variables are independent. For more information see the Leibniz integral rule.