Hint: you can integrate over x first of all

@Daren after the Higgs mechanism you are still left with a $U(1)$ gauge symmetry, subgroup of the original $SU(2)\times U(1)$, that is the one for the photon.

@Daren gauge freedom means that you have redundant degrees of freedom in your description, so before you can do any computation you have to explicitly choose a specific gauge where you don't have this freedom anymore. The unitary gauge is just a choice, but there are other possibilities (see the next chapter in P&S)

They will always be back-to-back, no matter what particle X is since you will be boosting to the $b \bar{b}$ rest frame.

Any source for saying that it is not observed?

I'm a bit ignorant on CFTs, but wouldn't spontaneous breaking require some (relevant?) deformations? If conformal symmetry is exact how can you generate the scale at which the breaking happens?

If you do perturbation theory on derivatives of delta function potentials you find divergences exactly like in QFT: look for the renormalization lecture notes by Luty.

I'm not sure I understand your question. Do you want the theory with the $\psi$ to reduce to $\mathcal{L}_{UV}$ when you integrate out $\psi$ or do you just want to find a renormalizable theory that reduces to E-H?

You should add for clarity that the Pauli principle only applies to fermions, not to all particles.

@LibertarianFeudalistBot I got it from the second to last line of the second to last equation in the question. OP was comparing that line with the last equation that he found on some lecture notes and wondering whether they are equal or not. He was not wondering whether the boundary term was zero.

@LibertarianFeudalistBot The Integral of the total derivative is not zero in general. What OP was wondering is whether the terms $\int \epsilon^{\mu\nu\rho} A_\rho \partial_\mu \partial_\nu \omega$ in the second to last line were zero or not. Are you saying that also those terms are not zero in general?

Biot-Savart is only valid in the static limit. If you have a time-dependent current you have to use the full Maxwell equations and you'll see that the propagation of the EM field travels at $c$.