# Dirac delta and derivative inside an integral

In development of a calculus (in GR but it doesn't matter here) I have seen one dubious substitution: in an integral of the form:

$$\int dx ~\delta(x-x_0)~\partial_x F(x)$$

The author substitutes $\partial_x$ for $-\partial_{x_0}$ and I honestly don't see where this minus sign is coming from. The integral only takes account when $x - x_0 = 0$ so $dx - dx_0 = 0$ implying $dx = dx_0$. Finally giving something of the form $\frac{d}{dx}=\frac{d}{dx_0}$ and the same for partial derivatives.

• Most weird things that happen with derivatives and delta functions inside integrals can be demonstrated using integration by parts. This is one of those things.
– wsc
Commented Feb 22, 2012 at 16:04

To expand on the comment by wsc, we could write $$\int\,dx\,\delta(x-x_0) \partial_x F(x) = -\int dx \partial_x \delta(x-x_0) F(x)\,,$$ and then pick our favorite representation of the $\delta$ function. Let's say $$\lim_{\epsilon \searrow 0} \frac{1}{\sqrt{2\pi\epsilon}}\,e^{-(x-x_0)^2/2\epsilon} = \delta(x - x_0)\,,$$ and then write $$\int dx \delta(x-x_0) \partial_x F(x) = -\lim_{\epsilon \searrow 0} \int dx \frac{1}{\sqrt{2\pi\epsilon}}\,\partial_x e^{-(x-x_0)^2/2\epsilon} F(x)\,.$$ Since we see that $$\partial_x e^{-(x-x_0)^2/2\epsilon} = -\partial_{x_0} e^{-(x-x_0)^2/2\epsilon}\,,$$ we can write $$\int dx \delta(x-x_0) \partial_x F(x) = \lim_{\epsilon \searrow 0} \int dx \frac{1}{\sqrt{2\pi\epsilon}}\,\partial_{x_0} e^{-(x-x_0)^2/2\epsilon} F(x)\,,$$ and conclude $$\int dx \delta(x-x_0) \partial_x F(x) = \partial_{x_0} \lim_{\epsilon \searrow 0} \int dx \frac{1}{\sqrt{2\pi\epsilon}} e^{-(x-x_0)^2/2\epsilon} F(x) = \partial_{x_0} \int dx \delta(x-x_0) F(x) = F^\prime(x_0)\,.$$