How to solve this kind of integrals? Derivation of probability current operator

Question

I am actually trying to derive expression for probability current operator using continuity equation $\nabla \cdot J = -\partial_t (c_s^+c_s)=-i/\hbar [H_0,c_s^+ c_s]$ where $$H_0=\frac{\hbar^2}{2m}\int dr (\nabla_r^+ c_r)(\nabla_r c_r) $$ After using commutation relation $[\nabla_r c_r,c_s^+ ]=\nabla_r[ c_r,c_s^+ ]=\nabla_r \delta(s-r)$, I arrive at expression:

$$ \nabla\cdot J = \frac{-i\hbar}{2m}\bigg(-\int dr c_s^+(\nabla_r \delta(s-r)) (\nabla_rc_r) + \int dr (\nabla_r c_r^+)(\nabla_r \delta(s-r)) c_s \bigg) $$ I need help in solving this integral.

I know the correct answer is $$ \nabla\cdot J = \frac{-i\hbar}{2m}\bigg(c_r^+(\nabla_r^2c_r)-(\nabla_r^2c_r^+) c_r\bigg) $$ but I am confused how $\nabla_r\delta(s-r)$ simplify to solve integration

Answer 1

By using distributional derivatives (as suggested by @Jakob in comments), I was able to solve above integrals. According to Wikipedia, we have identity:

$$ \int [\nabla_x \delta(x)] \phi(x) dx = -\int \delta(x) [\nabla_x \phi (x)]dx $$

So, I get: $$ \nabla\cdot J = \frac{-i\hbar}{2m}\bigg(+\int dr \delta(s-r) \nabla_r (c_s^+ (\nabla_rc_r)) - \int dr \delta(s-r) \nabla_r ( (\nabla_r c_r^+) c_s )\bigg) \\ \nabla\cdot J = \frac{-i\hbar}{2m}\bigg(+\int dr \delta(s-r) c_s^+ (\nabla_r^2c_r) - \int dr \delta(s-r) (\nabla_r^2 c_r^+) c_s \bigg) \\ \nabla\cdot J = \frac{-i\hbar}{2m}\bigg( c_r^+ (\nabla_r^2c_r) - (\nabla_r^2 c_r^+) c_r \bigg) $$