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[itex]\int f(x) \delta(x-a) dx=f(a) [/itex]

This is fair enough. But in ym notes there is a step that goes like the following:

[itex]\mathbf{\nabla} \wedge \mathbf{B}(\mathbf{r})=-\frac{\mu_0}{4 \pi} \int_V dV' \nabla^2(\frac{1}{|\mathbf{r}-\mathbf{r'}|}) \mathbf{J}(\mathbf{r'}) = \mu_0 \mathbf{J}(\mathbf{r})[/itex]

where we have used that [itex]\nabla^2(\frac{1}{|\mathbf{r}-\mathbf{r'}|}) =-4 \pi \delta(\mathbf{r}-\mathbf{r'})[/itex]

clearly the minus signs and the [itex]4 \pi[/itex]'s cancel so it's now just

[itex]\mathbf{\nabla} \wedge \mathbf{B}(\mathbf{r})=\mu_0 \int_V dV' \mathbf{J}(\mathbf{r'}) \delta(\mathbf{r}-\mathbf{r'})[/itex]

i don't see how that goes from there to give [itex]\mu_0 \mathbf{J}(\mathbf{r})[/itex] as we are integrating with respect to [itex]V'[/itex] are we not? and so i would assume that the delta function would need to be of the form [itex]\delta(\mathbf{r'}-\mathbf{r})[/itex] in order to give the desired answer.

the onnly explanation i can come up with is that [itex]\delta(\mathbf{r'}-\mathbf{r})=\delta(\mathbf{r}-\mathbf{r'})[/itex] since the delta function is symmetric about the spike. however the spike would be at different positions in these two cases. i'm kind of lost-any advice?