I am asking this question because I don't know the Leibniz Rule thoroughly enough to answer the question that I have. Can $\frac{d}{dy}\int^{\infty}_{-\infty}f(y,\theta) dy = \int^{\infty}_{-\infty}\frac{d}{dy}f(y,\theta)dy$, providing that $f$ is smooth? (note: the integral is with respect to $y$ rather than $\theta$, and we are differentiating with respect to $y$ as well.)
Thank you,
Certainly it is not true as stated. The integral $$\int_{-\infty}^\infty f(y,\theta)dy$$ is constant with respect to $y$ since you've integrated that variable out; $y$ is just a dummy name, you could change it to $z$ without changing the value of the integral. The integral expression depends only on $\theta$. So $$\frac{d}{dy}\int_{-\infty}^\infty f(y,\theta)dy = 0.$$ However, you should have $$\frac{d}{d\theta} \int^\infty_{-\infty} f(y,\theta) dy = \int^\infty_{-\infty} \frac{\partial f}{\partial \theta} (y,\theta) dy$$ under some fairly mild smoothness and integrability conditions.