# Which order do you take derivative of delta function?

1. Jul 21, 2012

### geoduck

If you have I=∫∫dxdy [∇xy δ(x-y)] f(x)g(y)

where ∇x is the derivative with respect to x (and similarly for y), then doesn't it matter which order you take the derivatives? For example:

I=∫∫dxdy f(x) ∇x [∇y δ(x-y)] g(y)
=∫dx f(x) ∇x[-g'(x)]=∫dx f(x) [-g''(x)]

whereas if you take the other order:

I=∫∫dxdy ∇y [∇x δ(x-y)] f(x)g(y)
=∫∫dxdy g(y)∇y [∇x δ(x-y)] f(x)
=∫dx g(x) [-f''(x)]

2. Jul 21, 2012

### Hurkyl

Staff Emeritus
I don't follow your calculation. Assuming notation isn't misleading, and writing things mathematics-style rather than physics-style, the calculation would go

\begin{align*} \int_{\mathbb{R}} \int_{\mathbb{R}} (\nabla_x \nabla_y \delta(x-y)) f(x) g(y) \, dx \, dy &= - \int_{\mathbb{R}} \int_{\mathbb{R}} (\nabla_y \delta(x-y)) \nabla_x(f(x) g(y)) \, dx \, dy \\& = - \int_{\mathbb{R}} \int_{\mathbb{R}} (\nabla_y \delta(x-y)) \nabla_x(f(x) g(y)) \, dy \, dx \\& = \int_{\mathbb{R}} \int_{\mathbb{R}} \delta(x-y) \nabla_y \nabla_x(f(x) g(y)) \, dy \, dx \\& = \cdots \end{align*}

Of course, I think the end result is the same -- just continue using integration by parts to switch between the result you get from this and the two results you got.

EDIT: I see how your calculation works now: if the inner integral is in terms of y, you are factoring the $\nabla_x$ outside of the inner integral.