Exchange Integral and Derivative respect to a parameter of a Dirac delta-function

I'm trying to solve the 6.2 problem of Jackson's Classical Electrodynamics textbook. At some point, to get the desired solution, I have to exchange a derivative applied to a Dirac delta-function with the integral operator:

$$\int_{\mathbb{R^3}} \frac{\partial \delta(\mathbf{x}-\mathbf{x_0}(t))}{\partial t}f(\mathbf{x})\,d^3x=\frac{\partial}{\partial t}\int_{\mathbb{R^3}} \delta(\mathbf{x}-\mathbf{x_0}(t))f(\mathbf{x})\,d^3x=\frac{\partial}{\partial t}f(\mathbf{x_0}(t))$$

Under which hypothesis can I do something like that (i.e. exchange the order of differentiation and integration)? I expect that known theorems of real analysis do not apply in this case, since the $\delta$ is not even a proper function.

• Would Mathematics be a better home for this question? – Qmechanic Feb 19 '17 at 11:23
• @Qmechanic, I think it's a transversal question, since this result is needed to obtain the Feynman-Heaviside electromagnetical fields. – Alessandro Zunino Feb 19 '17 at 11:27

By definition, if $T$ is a distribution, then $$\langle \partial_x T, f \rangle := - \langle T, \partial_x f \rangle\tag{1}$$ for every test function $f=f(x)$ in $C_0^\infty(\mathbb R^n)$ (or also $C^\infty(\mathbb R^n)$ if $T$ has compact support as the delta function). Here the derivative is just a bit more complicated. However, since the chain rule for taking the derivative of distributions composed with smooth functions is valid also for distributions (it is a general theorem) we have $$\partial_t \delta(x-x_0(t)) = \frac{dx_0}{dt}|_{x_0}\cdot \nabla_{x_0} \delta(x-x_0(t)) = -\frac{dx_0}{dt}|_{x}\cdot \nabla_x \delta(x-x_0(t))\:. \tag{2}$$ As a consequence, for every function $f \in C^\infty(\mathbb R^n)$, applying (2), $$\int \partial_t \delta(x-x_0(t)) f(x) d^nx = -\int \frac{dx_0}{dt}|_{x_0}\cdot \nabla_x \delta(x-x_0(t)) f(x) d^nx$$ $$= -\frac{dx_0}{dt}|_{x_0}\cdot\int \nabla_x\delta(x-x_0(t)) f(x) d^nx \:.$$ Applying (1) $$\int \partial_t \delta(x-x_0(t)) f(x) d^nx = + \frac{dx_0}{dt}|_{x_0}\cdot \int \delta(x-x_0(t)) \nabla_x f(x) d^nx$$ $$= \frac{dx_0}{dt}|_{x_0}\cdot \nabla_x f(x)|_{x_0(t)} = \frac{d}{dt}f(x_0(t))\:.$$
• Yes, that is nothing but the standard chain rule which is assumed to be valid also for distributions if the internal function ($x_0$) is smooth. Regardind the second issue, I used the obvious fact that taking the derivative with respect to $x$ or to $-x_0$ is the same as they appear summed in the argument. This is again an application of chain rule. – Valter Moretti Feb 19 '17 at 11:51
• If the poles of your function are far from the singularity of $\delta$ you can simply ignore them... – Valter Moretti Feb 19 '17 at 16:24