Let $\Omega$ be a region $\subset \subset \mathbb{R}^{n}$, and suppose that both $u_{\infty}$ and $\varphi$ $\in C_{0}^{\infty}(\Omega)$. Then, I want to perform integration by parts on the following:
$\displaystyle \int_{\Omega}\sum_{j=1}^{n} \frac{\partial u_{\infty}}{\partial x_{j}} \frac{\partial \varphi}{\partial x_{j}}$
Now, if I follow all the steps of integration by parts as taught to me back in the day by my Calc II professor, I would let $\displaystyle u = \frac{\partial u_{\infty}}{\partial x_{j}}$ and $\displaystyle dv = \frac{\partial \varphi }{\partial x_{j}}$. Then, I would calculate $\displaystyle du = \frac{\partial u_{\infty}^{2}}{\partial x_{j}^{2}}$ and $v = \varphi$.
Plugging all this into the integration by parts formula, I would obtain
$\displaystyle uv - \int_{\Omega} v\, du = \frac{\partial u_{\infty}}{\partial x_{j}}\varphi - \int_{\Omega}\sum_{j=1}^{n} \frac{\partial u_{\infty}^{2}}{\partial x_{j}^{2}}\varphi$.
This is part of a problem that we did in my analysis class. In class, however, the $\displaystyle \frac{\partial u_{\infty}}{\partial x_{j}}\varphi$ part was missing from our solution. In other words, we got that $\displaystyle \int_{\Omega}\sum_{j=1}^{n} \frac{\partial u_{\infty}}{\partial x_{j}} \frac{\partial \varphi}{\partial x_{j}} = - \int_{\Omega}\sum_{j=1}^{n} \frac{\partial u_{\infty}^{2}}{\partial x_{j}^{2}}\varphi$ only.
For the purposes of the problem we were working on, it makes sense to get this part only; however, my question is what happened to the $\displaystyle \frac{\partial u_{\infty}}{\partial x_{j}}\varphi$ part? Is it somehow $=0$?
I am extremely confused by this, and would very much value someone else's insight! Thanks in advance! :)