= Local subtraction with a sharp momentum cutoff
{title2=$I_R(M;\mu)=\log(\mu^2/M^2)/(16\pi^2)$}
The following scalar toy isolates one step in <renormalization>: removing a local divergence while retaining a finite dependence on a physical <mass>. For $M,\mu>0$ and a sharp <cutoff regularization> in <Euclidean space> with momentum bound $\Lambda$, define the dimensionless <Feynman integral>
$$
I_\Lambda(M)=\int_{|k_E|<\Lambda}\frac{d^4k_E}{(2\pi)^4}\frac1{(k_E^2+M^2)^2}.
$$
This is a zero-external-<momentum> <scalar bubble integral> stripped of coupling and combinatorial factors. Using the three-sphere area $2\pi^2$ and $t=k_E^2$ reduces the <integral> to
$$
I_\Lambda(M)=\frac1{16\pi^2}\int_0^{\Lambda^2}\frac{t\,dt}{(t+M^2)^2}
=\frac1{16\pi^2}\left[\log\left(1+\frac{\Lambda^2}{M^2}\right)-\frac{\Lambda^2}{\Lambda^2+M^2}\right].
$$
The <antiderivative> follows by writing $t=(t+M^2)-M^2$. At large $\Lambda$ the divergent part is $\log\Lambda^2/(16\pi^2)$, with the logarithm's dimensionful argument understood only within the displayed ratios. It has no external-<momentum> dependence. In a four-<scalar>-field vertex this structure can therefore be removed by a <local operator> proportional to $\phi^4$, rather than a momentum-dependent nonlocal interaction.
Choose the fixed reference <mass> $\mu$ and subtract $I_\Lambda(\mu)$. If the loop contribution to a chosen four-point vertex coefficient is $c\lambda^2 I_\Lambda(M)$, choose the additive <counterterm> in that same coefficient as $\delta\lambda_\Lambda=-c\lambda^2I_\Lambda(\mu)$, where $c$ includes its stated diagram factors. This defines a finite subtraction prescription:
$$
I_R(M;\mu)=\lim_{\Lambda\to\infty}\left[I_\Lambda(M)-I_\Lambda(\mu)\right]
=\frac1{16\pi^2}\log\frac{\mu^2}{M^2}.
$$
The <renormalization condition> for this toy loop contribution is $I_R(\mu;\mu)=0$. The reference <mass> stays fixed when $M$ is varied; finite dependence on $M$ survives. A different finite subtraction changes the <renormalization scheme>, and the defined <renormalized coupling> must change consistently.
\Image[/cutoff-subtraction-local-counterterm.png]
{title=A local <counterterm> cancels a logarithmic <cutoff regularization> divergence}
{description=Original scalar toy with $M=2\mu$: the two separately growing terms leave $16\pi^2I_R=-\log4$. The subtraction is constant in external momentum; it does not remove all mass dependence.}
For $M=2\mu$, the limiting result is $16\pi^2I_R=-\log4\approx-1.38629436$. At $\Lambda=10\mu$, the exact regulated difference is about $-1.32846343$; at $\Lambda=100\mu$, it is about $-1.38569459$. Expanding the exact expression gives the leading residual $2(M^2-\mu^2)/\Lambda^2$ in $16\pi^2[I_\Lambda(M)-I_\Lambda(\mu)]$. This explicitly separates finite-cutoff error from the chosen finite <renormalization condition>.
A sharp <cutoff regularization> is convenient for this scalar demonstration, but shifts of a divergent <loop momentum> can change its integration domain. A <gauge theory> calculation cannot assume that it preserves the <Ward identities>. The toy contains no interacting <gauge bosons>, <ghost loops> or overlapping <ultraviolet subdivergences>, so its successful subtraction does not prove <Non-Abelian gauge theory> <renormalizability>.
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