with [itex]A^{\mu}[/itex] being the vector potential of the muon, as obtained above, and [itex]j_{\mu}^{fi}[/itex] being the 4-current of the electron, given by

Note that the [itex]j^{\mu}_{(2)}[/itex] which appears in their expression is not the Fourier Transformed version. While the final expression might just seem to turn out right, what I do not understand is how they could write the solution to [itex]A^{\mu}[/itex] quite so simply in position space.

EDIT: Perhaps you are trying to point out that their solution is correct. I got that part, but only after going to Fourier space and taking an inverse Fourier transform, in which the delta function essentially collapsed the integral to that 1/q^2 term.

which gives a + sign, as correctly stated in my post.

The problem was that if you take the fourier transform of [itex]\Box A^\mu(x) = j^\mu(x)[/itex] you get

[tex]A^\mu(q) = -\frac{1}{q^2}j^\mu(q)[/tex]

where the argument q denotes the Fourier Transformed version. In the book, it appears that 4.15 has been written in position space directly, which was confusing initially, until I realized that the delta function term plays a role and one gets a similar expression in position space as well.

Yes, you have (a-b) in the argument of the exponential, and (a+b) in the coefficient of the exponential term. You get that if you use the standard expression for the 4-current, and a asymptotic plane wave for the initial and final states of the (spinless) electron and muon.

Just use the definition of the d'Alembertian: [itex]\Box = \partial_\mu\partial^\mu = \frac{\partial^2}{\partial t^2}-\nabla^2[/itex].

As for the second part, it is not justified directly, but rather, one computes the Fourier transform of the 4-current, multiplies it by -1/q^2, takes the inverse Fourier transform, and ends up getting the expression mentioned in the book. The conclusion is that the final expression given in the book (solution of the differential equation) is correct.

Another way of looking at it (and thanks for making me think) is that the right hand side of the differential equation contains an exponential, and the left hand side is a linear differential operator acting on an undetermined function. So we can expand the solution as a series in the exponential function on the right hand side (the usual way a particular integral of an ODE is found..except that here we are dealing with uncoupled partial derivatives, so the method goes through). Comparing coefficients and rearranging gives the same result.

Finally, the property that the position space solution ends up 'looking' like the Fourier domain solution holds due to the presence of a single complex exponential on the RHS. If the 4-current were more complicated, life wouldn't be quite so simple (e.g. the general retarded potential solution for the 4-potential, which is up to quadrature).