# Delta distribution - integration by parts of its differentiation

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

$$\int \left[ \nabla \delta(\mathbf{x}) \right] \ f(\mathbf{x}) \ d^3x = -(\nabla f)(\mathbf{0}) ~?$$

## migrated from physics.stackexchange.comMay 15 '14 at 6:24

This question came from our site for active researchers, academics and students of physics.

• Even if it was legitimate at all to Taylor-expand Dirac Peaks (it isn't; but perhaps you could handwave some conditions as to when the result might nevertheless be meaningful), then this would still be no use as an approximation, because you definitely would need to carry out the series to infinity. Any finite subseries has nothing to do with the Delta function. – leftaroundabout May 14 '14 at 21:35
• @leftaroundabout truncating the taylor expansion for the delta function gives you the same answer for the integral as truncating the taylor expansion for $f$ around zero when trying to compute $f(a_i)$, so I would say taylor expanding the delta function is reasonable. On a side note, should this be migrated to math.se? – Brian Moths May 15 '14 at 1:54

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.