$\int t^2 p(x,t) dt$ , where $p(x,t)$ is the joint density function $t^2 > 0$ $\forall t$ and $p(x,t) > 0$ everywhere, thus the answer must greater than $0$. However, using the integration by parts, I cancel out everything. Can someone tell me what's wrong with this integration by parts?
$$\int t^2 p(x,t) dt = t^2\cdot p(x) - 2\cdot \int tp(x) dt = t^2\cdot p(x) - t^2\cdot p(x) = 0$$