 Solution

Split the Fourier modes into $|p|<\Lambda /b$ and the shell $\Lambda /b<|p|<\Lambda$. The source is uniform, so it couples only to the zero mode, which belongs to the retained part. The Gaussian field theory has no couplings between the shell and retained modes. Integrating out the shell therefore contributes only a source-independent determinant to the free energy, without changing the quadratic coefficients of the retained Hamiltonian.

Restore the cutoff by $x=b{x}^{\prime }$ and choose the field normalization ${\varphi }_{<}(b{x}^{\prime })=b^{-(D-2)/2}{\varphi }^{\prime }(x^{\prime })$. Counting the measure, derivative and field factors explicitly givesThus in the convention that preserves the kinetic normalization, the Gaussian momentum-shell scaling isThe field has engineering dimension $(D-2)/2$ and zero anomalous dimension. The measure Jacobian and shell determinant supply the additive constant described in Q2. A different choice of field rescaling can redistribute factors into $\alpha$, but must give the same physical transformed correlation length ${\xi }^{\prime }=\xi /b$.