The tree level 4-fermion amplitude you get when you collapse the propagator to a point is a $J_\mu^+ J^{\mu ~-}$ contact term, the 4-Fermi interaction of the 30s.

With the advent of functional integrals to the SM, the relevant part of the SM action contributing to this charged current amp is 
$$
{\cal L}_{eff}= m^2_W W_{\mu}^+ W^{\mu ~-}+ \frac{g}{\sqrt 2}( W_{\mu}^+ J^{\mu -} +W^{\mu ~-} J_\mu^+) +O(p^2/M_W^2).
$$
Complete the complex square,
$$
{\cal L}_{eff} = m^2_W \left (W_{\mu}^+  +\frac{g}{\sqrt {2} m_W^2} J^{\mu~+}\right) \left(W_{\mu}^-  +\frac{g}{\sqrt {2} m_W^2}J^{\mu~-} \right ) -\frac{g^2}{2m_W^2} J_{\mu}^{+} J^{\mu~-} .
$$
Now observe the first term represents a shift in the definition of the *W*s;
integrated over spacetime and stuck in the exponent of the functional integral, it amounts to two Gaussians, when you resolve it into the original "new", shifted  *W<sub>1</sub>, W<sub>2</sub>*; functional integration of these Gaussians w.r.t. the shifted *W*s leaves no trace of the *W*s in this low energy part of the path integral. They have been "integrated out", as per your question. 

The sole usable residue of their presence, is the "constant" (as far as *W* degrees of freedom are concerned) second term, the current-current interaction, $-\frac{2G_F}{\sqrt 2}  J_{\mu}^{+} J^{\mu~-} $, where one defined 
$G_F\sqrt 2 \equiv g^2/4m_w^2$. Note you would get the very same answer from the merely algebraic equations of motion of ${\cal L}_{eff}$, namely $W_\mu^{\pm}=-g J_\mu^{\pm}/\sqrt{2} m_w^2$; using these to eliminate the *W*s would result in the same current-current residual interaction.

(By the way, in 1933, this was essentially the first application of QFT: its linchpin  feature of creation and annihilation of fermion species.) 

A very analogous procedure, naturally, occurs for the neutral current amplitudes involving *Z* exchange.