Hartree-Fock: Coulomb integral Today I was wondering how to better understand the Coulomb integral in the Hartree-Fock approximation. 
Extracted from: Szabo & Ostlund, Modern Quantum Chemistry, p. 112

The Coulomb term has a simple interpretation. In an exact
  theory, the Coulomb interaction is represented by the two-electron
  operator $r_{ij}^{-1}$. In the Hartree-Fock approximation, an electron in
  a state $\chi_{a}$ experiences a one-electron Coulomb potential:
$$
    v_{a}^\text{coul}(1) = \sum_{b \neq a} \int \mathrm{d} \vec{x}_{2} | \chi_{b}(2) |^{2} r_{12}^{-1} \tag{3.7}
$$

Now, the Coulomb integral (CI) for a system of two interacting electrons, using Slater determinants, is:
$$\hat{J}=\frac{1}{2} \left( \int \psi_1^*\left( 1 \right) \psi_2^*\left( 2 \right) \frac{1}{r_{12}} \psi_2\left( 1 \right)\psi_1\left( 2 \right) + \psi_2^*\left( 1 \right) \psi_1^*\left( 2 \right) \frac{1}{r_{12}} \psi_1\left( 1 \right)\psi_2\left( 2 \right) \mathrm{d}x \right)$$
Now, the help I'm looking for how this single Slater determinant definition comes up from the mathematical point of view.
This is how I understand the HF approximation: Let's start only with the Coulomb integral. This integral, as I understand it, should be understood as follows for the first electron:
\begin{align}\text{CI} & =\int \psi_1^*\left( 1 \right) \psi_2^*\left( 2 \right) \frac{1}{r_{12}} \psi_2\left( 1 \right)\psi_1\left( 2 \right) \mathrm{d}x \\
& =\int \psi_1^*\left( 1 \right) \big( \sum_{n=2}^{n=2} \psi_2^*\left( 2 \right) \frac{1}{r_{12}} \psi_1\left( 2 \right) \big) \psi_2\left( 1 \right) \mathrm{d}x\\
& =\int \psi_1^*\left( 1 \right) \big( \sum_{n=2}^{n=2} \hat{J}_n \big) \psi_2\left( 1 \right)\end{align}
And, more general, the Coulomb integral for the first electron in a system of $n$ electrons would be:
$$\text{CI}= \int \psi_1^*\left( 1 \right) \big( \sum_{n \neq 1}^{n} \hat{J}_n \big) \psi_2\left( 1 \right) \mathrm{d}x$$
Is this correct?
If this is not correct, would someone enlighten me? 
To ACuriousMind, Thank so much for editing this post.
Specific help I'm looking for:
I'll try to explain myself better through these questions: 
i.) This operator $\frac{1}{r_{ij}}$ involves the spatial coordinates of electrons $i$ and $j$. So, taken this CI:
$$J_{2}(1) \chi_{1}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \Big] \chi_{1}(1)$$ 
Mathematically speaking, the first step is to integrate $\int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2)$ (which only the spatial coordinates of electron 2 are taken in account) and then multiply the result to $\chi_1 \left( 1 \right)$ ?
ii.) This question is more curiosity. For the same electron (Let's say the first one), How would it be with a system of N-electrons?
 A: 
Now, the Coulomb integral for a system of two interacting electrons,
  until Slater determinants, is:

I don't understand what the heck do you mean by "until Slater determinants", but for a system of two electrons you have a system of two HF equations (3.9): each HF equation for each spin-orbital $\chi_{a}$,
$$
    [h(1) + J_{2}(1) - K_{2}(1) ] \, \chi_{1}(1) = \varepsilon_{1} \chi_{1}(1) \, , \\
    [h(1) + J_{1}(1) - K_{1}(1) ] \, \chi_{2}(1) = \varepsilon_{2} \chi_{2}(1) \, , \\
$$
where 
$$
    J_{2}(1) \chi_{1}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \Big] \chi_{1}(1) \, , \\
    K_{2}(1) \chi_{1}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{1}(2) \Big] \chi_{2}(1) \, , \\
    J_{1}(1) \chi_{2}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{1}^{*}(2) r_{12}^{-1} \chi_{1}(2) \Big] \chi_{2}(1) \, , \\
    K_{1}(1) \chi_{2}(1) := \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{1}^{*}(2) r_{12}^{-1} \chi_{2}(2) \Big] \chi_{1}(1) \, . \\
$$
So, you have two different Coulomb terms $J_{2}(1) \chi_{1}(1)$ and $J_{1}(1) \chi_{2}(1)$, as well as two different exchange terms $K_{2}(1) \chi_{1}(1)$ and $K_{1}(1) \chi_{2}(1)$, entering each and every HF equation.
Coulomb integrals are expression of the following form 
$$
    \int \mathrm{d} \vec{x}_{1} \chi_{1}^{*}(1) J_{2}(1) \chi_{1}(1) \, , \\
    \int \mathrm{d} \vec{x}_{1} \chi_{2}^{*}(1) J_{1}(1) \chi_{2}(1) \, ,
$$
which contribute to the HF electronic energy $\langle \Phi\,|\,H\,|\,\Phi\rangle$.
By the definition of Coulomb operator $J_{2}(1)$ above could be written as follows
$$
    \int \mathrm{d} \vec{x}_{1} \chi_{1}^{*}(1) \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \Big] \chi_{1}(1) \, , \\
    \int \mathrm{d} \vec{x}_{1} \chi_{2}^{*}(1) \Big[ \int \mathrm{d} \vec{x}_{2} \, \chi_{1}^{*}(2) r_{12}^{-1} \chi_{1}(2) \Big] \chi_{2}(1) \, ,
$$
or as follows
$$
    \iint \mathrm{d} \vec{x}_{1} \mathrm{d} \vec{x}_{2} \chi_{1}^{*}(1) \chi_{2}^{*}(2) r_{12}^{-1} \chi_{2}(2) \chi_{1}(1) \, , \\
    \iint \mathrm{d} \vec{x}_{1} \mathrm{d} \vec{x}_{2} \chi_{2}^{*}(1) \chi_{1}^{*}(2) r_{12}^{-1} \chi_{1}(2) \chi_{2}(1) \, .
$$
