derivative of integral: clarify step in paper I`m reading a paper and it came in it the following: 
$$J(t) := \int_{R^N} \vert x \vert^2  \vert u (t,x)\vert^2 dx $$
$$ J'(t)=4 Re \int x. \nabla u \overline u dx$$
and $$u$$ satisfy the Cauchy equation : 
$$ iu_t + \Delta  u + \vert x\vert^b \vert u \vert^{p-2} u =0  , t\geq0, x \in R^N $$
$$ u(0,x)=u_\circ $$
Can anyone please explain to me how to get the derivative this way? 
 A: Hint: Let us just consider $i\partial_t u+\Delta u = 0$, then observe
\begin{align}
J'(t) =&\ \frac{d}{dt}\int |x|^2|u(t, x)|^2\ dx\\
=&\ \int |x|^2 \left\{\partial_t u(t, x)\overline{u(t, x)}+u(t, x)\overline{\partial_t u(t, x)}\right\}\ dx\\
=&\ \int |x|^2 \left\{i\Delta u(t, x) \overline{u(t, x)}+u(t, x)\overline{i\Delta u(t, x)}\right\}\ dx\\
=&\ i\int |x|^2 \left\{\Delta u(t, x) \overline{u(t, x)}-u(t, x)\overline{\Delta u(t, x)}\right\}\ dx.
\end{align}
Next, note that
\begin{align}
\int |x|^2\overline{u(t, x)} \Delta u(t,x )\ dx =&\ -\int \nabla(|x|^2\overline{u(t, x)})\cdot \nabla u(t, x)\ dx\\
=&\ -\int \{2x\overline{u(t, x)}+|x|^2\overline{\nabla u(t, x)}\}\cdot \nabla u(t, x)\ dx \\
=&\ -2 \int [x\cdot \nabla u(t, x)]\overline{u(t, x)}\ dx -\int |x|^2|\nabla u(t, x)|^2\ dx
\end{align}
and
\begin{align}
\int |x|^2u(t, x)\overline{\Delta u(t, x)}\ dx =  -2 \int [x\cdot \overline{\nabla u(t, x)}]u(t, x)\ dx -\int |x|^2|\nabla u(t, x)|^2\ dx
\end{align}
then it follows
\begin{align}
J'(t) =&\  -2i\int [x\cdot\nabla u(t, x)]\overline{u(t, x)}-[x\cdot\overline{\nabla u(t, x)}] u(t, x)\ dx \\
=&\ 4\operatorname{Im}\left(\int[x\cdot \nabla u(t, x)] \overline{u(t, x)}\ dx\right).
\end{align}
Note: The paper made a typo. It should take the imaginary part instead of the real part. 
