What is the actual form of Noether current in field theory?

1. Eq. (5) is (up to factors of the infinitesimal parameter $\varepsilon$) the standard expression for the full Noether current. Here:

   • $\delta x^{\mu}$ is the so-called horizontal component of the infinitesimal variation;
   • $\delta \phi -\frac{\partial \phi}{\partial x^{\mu}} \delta x^{\mu} $ is the so-called vertical component of the infinitesimal variation;
   • $F^{\mu}$ is an improvement term in case of quasisymmetry.

2. The main point is that Schweber (7), Peskin & Schroeder (6) are only considering situations with purely vertical transformations, i.e. situations where $\delta x^{\mu}=0$.

3. Let us mention that the last term in eq. (4) gets cancelled by the Jacobian contributions from the integration measure. Hence it is not present in eq. (5).

4. Finally, it seems relevant to mention that OP's boundary condition (2) is often not fulfilled in important applications, such as the canonical stress-energy-momentum (SEM) tensor, which is the Noether current for spacetime translations. See e.g. this Phys.SE post. Therefore the boundary condition (2) should be relaxed appropriately. Similarly, the improvement term $F^{\mu}$ is not some arbitrary field that vanishes on the boundary, as OP claims (v3) under eq. (5). Instead the improvement term $F^{\mu}$ is dictated by the quasisymmetry, which fixes $F^{\mu}$ up to a divergence-free term.

The issue is that there are two ways to write an infinitesimal field transformation. As a simple example, let's consider a triplet of fields $\phi_i$ which transform as a vector in space, and suppose we're dealing with a rotational symmetry. We can write this symmetry in two ways:

• Your method: the rotation changes the spatial coordinates (your $\delta x^\mu$) and changes the value of the field by rotation (your $\delta \phi^i$).
• The more common method: the rotation only changes the value of the field while holding spatial coordinates constant, i.e. $\delta x^\mu = 0$.

While it looks like your method is more general, the second method works equally well, as any shift in the coordinates by a small $\delta x^\mu$ is equivalent to a shift in the field value by $\partial_\mu \phi^i \delta x^\mu$.

Setting $\delta x^\mu = 0$ in Peskin and Schroeder's answer gives yours, so they agree with you, except that their $\delta \phi$ will be more complicated. The Schweber book is a little more basic and probably dropped the total derivative just to simplify things.