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

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

• Hint: It has something to do with the notion of a distributional derivative, cf. e.g. here. Sep 5 '21 at 8:48
• Integration by parts Sep 5 '21 at 8:58
• @Jakob thank you very much. It worked when I used identity $\int (\nabla\delta)\phi dx=-\int \delta (\nabla\phi) dx$. I have put this as an answer to this question. At the same time, I have an other question. I have an other expression that has two derivatives of delta function $\int dr (\nabla_s c_s^+) (\nabla_r c_r)[\nabla_r(\nabla_s \delta(r-s))]$. I know its solution is $(\nabla_r^3 c_r^+) (\nabla_r c_r)$ but I am unable to apply the same distributive derivative theory on this integral. Could you please help me in solving this one. Sep 5 '21 at 11:24

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