Some delta distribution physicist calculus. Assume there is given
$$ \int_{\mathbb{R}^3} \sum_i f(\mathbf{x}) \delta^{(3)}(\mathbf{x}-\mathbf{a}_i) \ d^3x $$
with $f$ vanishing at infinity and $\delta^{(3)}=\delta_x\delta_y\delta_z$ and I want to approximate the integrand by "expanding" the delta distribution in the following way
$$ \sum_i \delta(\mathbf{x}-\mathbf{a}_i) = \sum_i \sum_{n=0}^{\infty} \frac{(-\mathbf{a}_i \cdot \nabla)^n}{n!} \delta(\mathbf{x})$$
then I can change the order of integration and sum and I get
$$\sum_i \sum_{n=0}^{\infty} \int_{\mathbb{R}^3} \left[ \frac{(-\mathbf{a}_i \cdot \nabla)^n}{n!} \delta(\mathbf{x})\right] \ f(\mathbf{x}) \ d^3x$$
I now want to integrate by parts and my question is how exactly this is done. Is it formally correct or at least legitimate to proceed
$$(-\mathbf{a}\cdot\nabla)^n \delta(\mathbf{x}) = (-\mathbf{a}\cdot\nabla)(-\mathbf{a}\cdot\nabla)...(-\mathbf{a}\cdot\nabla\delta(\mathbf{x}))$$
With other words, can I simply denote
$$\sum_i \sum_{n=0}^{\infty} \int_{\mathbb{R}^3} \underbrace{\left[ \frac{(-\mathbf{a}_i \cdot \nabla)^n}{n!} \delta(\mathbf{x})\right]}_{g'(x)} \ \underbrace{f(\mathbf{x})}_{h(x)} \ d^3x$$
and follow the rule $\int g'(x)\cdot h(x) \ dx = g(x) \cdot h(x) \big|_{-\infty}^{+\infty} - \int g(x) \cdot h'(x) \ dx$?
What will the result be and at which point do I use the delta distribution's definition
$$\int\delta(\mathbf{x})f(\mathbf{x}) \ d^3x=f(\mathbf{0})$$
and its property
$$\int \left[ \nabla \delta(\mathbf{x}) \right] \ f(\mathbf{x}) \ d^3x = -(\nabla f)(\mathbf{0}) ~?$$
Instead of expanding the $\delta $-function you can also rewrite the integral as \begin{eqnarray*} &&\int_{\mathbb{R}^{3}}d\mathbf{x}\sum_{i}f(\mathbf{x})\delta ^{3}(\mathbf{% x-a}_{i})=\sum_{i}\int_{\mathbb{R}^{3}}d\mathbf{x}f(\mathbf{x+a}_{i})\delta ^{3}(\mathbf{x}) \\ &=&\sum_{i}\int_{\mathbb{R}^{3}}d\mathbf{x}f(\mathbf{x+a}_{i})\delta ^{3}(% \mathbf{x})=\sum_{n=0}^{\infty }\int_{\mathbb{R}^{3}}d\mathbf{x}\frac{1}{n!}% \{(-\mathbf{a}_{i}\cdot \partial _{\mathbf{x}})^{n}f\}(\mathbf{x})\delta ^{3}(\mathbf{x}) \\ &=&\sum_{n=0}^{\infty }\frac{1}{n!}\{(-\mathbf{a}_{i}\cdot \partial _{% \mathbf{x}})^{n}f\}(\mathbf{0}) \end{eqnarray*} This presupposes infinite differentiability of the test function $f$ and convergence of the series but the result is the same. Actually derivatives of $\delta $-distributions can be defined in this way.