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Is there a way in which one can assign the meaning $$ - \frac{\delta'(|\mathbf{x} - \mathbf{y}|)}{ 4 \pi | \mathbf{x} - \mathbf{y}|} \ \to \ \delta^{(3)}(\mathbf{x} - \mathbf{y}) \ $$ for $\mathbf{x}, \mathbf{y} \in \mathbb{R}^3$? It seems to me that this should be true. Notice that for some test function $f(\mathbf{x})$ we have (using spherical coordinates) $$ - \int_{\mathbb{R}^3} d^3\mathbf{x} \; \frac{\delta'(|\mathbf{x}|)}{ 4 \pi | \mathbf{x}|} f(\mathbf{x}) = - \frac{1}{4\pi} \int_{\mathbb{R}_3} dX \; d\theta \; d\phi \; X^2 \sin\theta \; \frac{\delta'(X)}{X} f(X,\theta, \phi) $$ and using integration by parts on the $\delta'$ yields $$ \cdots = \frac{1}{4 \pi} \int_{\mathbb{R}_3} dX \; d\theta \; d\phi \; \sin\theta \; \delta(X) \left[ f(X,\theta, \phi) + X \; \partial_X f(X,\theta, \phi) \right] $$ which evaluates to $$ \cdots = \frac{1}{4 \pi} f(0,0,0) \int_{S^2} d\theta \; d\phi \; \sin\theta = f(\mathbf{0})\ . $$ This seems to have the same effect as using $\delta^{(3)}(\mathbf{x})$. Is

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  • $\begingroup$ can I confirm - $\delta^{(3)}$ means the 3D dirac delta, not the third derivative of $\delta$? $\endgroup$ – Calvin Khor Dec 9 '20 at 5:39
  • $\begingroup$ Yes that's right. Sorry - I should have been more explicit about that. $\endgroup$ – QuantumEyedea Dec 9 '20 at 5:41
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Answering my own question, but the identity holds true: I found the following formula in Kanwal's book $$ \delta(x)\delta(y)\delta(z) = \frac{\delta(r)}{4\pi r^2} $$ which is equation (3.1.30), where $r = \sqrt{x^2 + y^2 + z^2}$. Noting the identity $r \delta'(r) = - \delta(r)$ yields the result I quoted above.

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