# Why is coulomb integral is called local while the corresponding exchange integral is called non-local

In the book "A chemist's guide to DFT" page 27-29 they said that coulomb integral is local interaction term while exchange integral is non-local. I can visualize, to some extent, that repulsion between two orbital/wave function at two different position (say x1 and x2) is called coulomb integral and that is why it's local. But I am having problem with the exchange term. It is also local, I feel, just the difference is two orbital can interchange their position between them. Can anyone please shed light on these concepts?

The definitions of local and non-local operators are given at p. 12 of the book. Basically, an operator $\hat{A}$ is called local if the value of $\hat{A} f(x)$ at some point $x=x'$ depends only on the value of $f(x)$ at the very same point $x=x'$ and not on the values of $f(x)$ at any other points. Otherwise an operator is called non-local.
For instance, any $j$-th Coulomb operator in Hartree-Fock theory is local since the value of $\hat{J}_{j} \psi_{i}(q)$ at $q=q'$ depends only on the value of $\psi_{i}(q)$ at $q=q'$, while any $j$-th exchange operator is non-local since the value of $\hat{K}_{j} \psi_{i}(q)$ at $q=q'$ depends on the values of $ψ_{i}(q)$ at all points $q$.