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In dimensional regularization, we replace a momentum integral $I= \int d^nk f(|k|)$ with the family of regularized integrals $$\mu^{n-d}\int d^dk f(|k|) = \mu^{\epsilon}\Omega_d \int p^{d-1} f(p)dp.\tag{1}$$ Here $\Omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the volume of the $d$-sphere.

There's a related regularization where instead we replace $I$ with $$\int d^nk \big(\frac{\mu^2}{|k|^2}\big)^{\epsilon/2} f(|k|) = \mu^{\epsilon}\Omega_n \int p^{n-\epsilon-1} f(p)dp.\tag{2}$$ These two expressions differ only in the angular contribution, which is momentum independent and regular in $\epsilon = n-d$. If the momentum dependence of $I$ is also regular in $\epsilon$ (and it really should be), this alters neither the $\epsilon$ pole structure nor the momentum dependence of the final result. Instead, it shifts the constant-in-$\epsilon$ terms by quantities that don't depend on the momenta.

For example, in the integral $I=(4\pi)^2\int \frac{d^4k}{(2\pi)^4} \frac{1}{(k^2+\Delta)^2}$, one finds that the "Macaroni & Pie" term $-\gamma + \log(4\pi)$ drops out, leaving behind $\frac{2}{\epsilon} + \log(\frac{\mu^2}{\Delta})$. (Proof: Instead of cancelling the $\Gamma(d/2)$'s and using Stirling's Formula, apply Euler's Reflection formula $\Gamma(d/2)\Gamma(1-d/2) = \pi \csc(\pi d/2)$. The Laurent expansion of $\csc(z)$ is $\frac{1}{z} + \frac{1}{6}z + ...$.)

My question: Are these two regularizations really equivalent?

It looks to me like they are, and I've spot checked in QED at 1-loop.

In dimensional regularization, we replace a momentum integral $I= \int d^nk f(|k|)$ with the family of regularized integrals $$\mu^{n-d}\int d^dk f(|k|) = \mu^{\epsilon}\Omega_d \int p^{d-1} f(p)dp.\tag{1}$$ Here $\Omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the volume of the $d$-sphere.

There's a related regularization where instead we replace $I$ with $$\int d^nk \big(\frac{\mu^2}{|k|^2}\big)^{\epsilon/2} f(|k|) = \mu^{\epsilon}\Omega_n \int p^{n-\epsilon-1} f(p)dp.\tag{2}$$ These two expressions differ only in the angular contribution, which is momentum independent and regular in $\epsilon = n-d$. It seems to me that multiplying all diagrams with the same number of loops by the same regular-in-$\epsilon$ factor should not affect the final result.

For example, in the integral $I=(4\pi)^2\int \frac{d^4k}{(2\pi)^4} \frac{1}{(k^2+\Delta)^2}$, one finds that the "Macaroni & Pie" term $-\gamma + \log(4\pi)$ drops out, leaving behind $\frac{2}{\epsilon} + \log(\frac{\mu^2}{\Delta})$. (Proof: Instead of cancelling the $\Gamma(d/2)$'s and using Stirling's Formula, apply Euler's Reflection formula $\Gamma(d/2)\Gamma(1-d/2) = \pi \csc(\pi d/2)$. The Laurent expansion of $\csc(z)$ is $\frac{1}{z} + \frac{1}{6}z + ...$.)

My question: Are these two regularizations really equivalent?

It looks to me like they are, and I've spot checked in QED at 1-loop. But I do not have much confidence in this claim. Can someone point to a specific computation where this regularization fails to give the same answer as dimensional regularization?

To complete the specification, let's say that traces of Dirac matrix products have their usual 4-dimensional form and that $g_{\mu\nu}g^{\mu\nu} = n$.

In dimensional regularization, we replace a momentum integral $I= \int d^nk f(|k|)$ with the family of regularized integrals $$\mu^{n-d}\int d^dk f(|k|) = \mu^{2\epsilon}\Omega_d \int p^{d-1} f(p)dp.\tag{1}$$ Here $\Omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the volume of the $d$-sphere.

There's a related regularization where instead we replace $I$ with $$\int d^nk \big(\frac{\mu^2}{|k|^2}\big)^{\epsilon/2} f(|k|) = \mu^{\epsilon}\Omega_n \int p^{n-\epsilon-1} f(p)dp.\tag{2}$$ These two expressions differ only in the angular contribution, which is momentum independent and regular in $\epsilon = n-d$. If the momentum dependence of $I$ is also regular in $\epsilon$ (and it really should be), this alters neither the $\epsilon$ pole structure nor the momentum dependence of the final result. Instead, it shifts the constant-in-$\epsilon$ terms by quantities that don't depend on the momenta.

For example, in the integral $I=(4\pi)^2\int \frac{d^4k}{(2\pi)^4} \frac{1}{(k^2+\Delta)^2}$, one finds that the "Macaroni & Pie" term $-\gamma + \log(4\pi)$ drops out, leaving behind $\frac{2}{\epsilon} + \log(\frac{\mu^2}{\Delta})$. (Proof: Instead of cancelling the $\Gamma(d/2)$'s and using Stirling's Formula, apply Euler's Reflection formula $\Gamma(d/2)\Gamma(2-d/2) = \pi \csc(\pi d/2)$. The Laurent expansion of $\csc(z)$ is $\frac{1}{z} + \frac{1}{6}z + ...$.)

My question: Are these two regularizations really equivalent?

It looks to me like they are, and I've spot checked in QED at 1-loop.

In dimensional regularization, we replace a momentum integral $I= \int d^nk f(|k|)$ with the family of regularized integrals $$\mu^{n-d}\int d^dk f(|k|) = \mu^{\epsilon}\Omega_d \int p^{d-1} f(p)dp.\tag{1}$$ Here $\Omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the volume of the $d$-sphere.

There's a related regularization where instead we replace $I$ with $$\int d^nk \big(\frac{\mu^2}{|k|^2}\big)^{\epsilon/2} f(|k|) = \mu^{\epsilon}\Omega_n \int p^{n-\epsilon-1} f(p)dp.\tag{2}$$ These two expressions differ only in the angular contribution, which is momentum independent and regular in $\epsilon = n-d$. If the momentum dependence of $I$ is also regular in $\epsilon$ (and it really should be), this alters neither the $\epsilon$ pole structure nor the momentum dependence of the final result. Instead, it shifts the constant-in-$\epsilon$ terms by quantities that don't depend on the momenta.

For example, in the integral $I=(4\pi)^2\int \frac{d^4k}{(2\pi)^4} \frac{1}{(k^2+\Delta)^2}$, one finds that the "Macaroni & Pie" term $-\gamma + \log(4\pi)$ drops out, leaving behind $\frac{2}{\epsilon} + \log(\frac{\mu^2}{\Delta})$. (Proof: Instead of cancelling the $\Gamma(d/2)$'s and using Stirling's Formula, apply Euler's Reflection formula $\Gamma(d/2)\Gamma(1-d/2) = \pi \csc(\pi d/2)$. The Laurent expansion of $\csc(z)$ is $\frac{1}{z} + \frac{1}{6}z + ...$.)

My question: Are these two regularizations really equivalent?

It looks to me like they are, and I've spot checked in QED at 1-loop.

In dimensional regularization, we replace a momentum integral $I= \int d^nk f(|k|)$ with the family of regularized integrals $\mu^{n-d}\int d^dk f(|k|) = \mu^{2\epsilon}\Omega_d \int p^{d-1} f(p)dp.$ Here $\Omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the volume of the $d$-sphere.

There's a related regularization where instead we replace $I$ with $\int d^nk \big(\frac{\mu^2}{|k|^2}\big)^{\epsilon/2} f(|k|) = \mu^{\epsilon}\Omega_n \int p^{n-\epsilon-1} f(p)dp.$ These two expressions differ only in the angular contribution, which is momentum independent and regular in $\epsilon = n-d$. If the momentum dependence of $I$ is also regular in $\epsilon$ (and it really should be), this alters neither the $\epsilon$ pole structure nor the momentum dependence of the final result. Instead, it shifts the constant-in-$\epsilon$ terms by quantities that don't depend on the momenta.

For example, in the integral $I=(4\pi)^2\int \frac{d^4k}{(2\pi)^4} \frac{1}{(k^2+\Delta)^2}$, one finds that the "Macaroni & Pie" term $-\gamma + \log(4\pi)$ drops out, leaving behind $\frac{2}{\epsilon} + \log(\frac{\mu^2}{\Delta})$. (Proof: Instead of cancelling the $\Gamma(d/2)$'s and using Stirling's Formula, apply Euler's Reflection formula $\Gamma(d/2)\Gamma(2-d/2) = \pi \csc(\pi d/2)$. The Laurent expansion of $\csc(z)$ is $\frac{1}{z} + \frac{1}{6}z + ...$.)

My question: Are these two regularizations really equivalent?

It looks to me like they are, and I've spot checked in QED at 1-loop.

In dimensional regularization, we replace a momentum integral $I= \int d^nk f(|k|)$ with the family of regularized integrals $$\mu^{n-d}\int d^dk f(|k|) = \mu^{2\epsilon}\Omega_d \int p^{d-1} f(p)dp.\tag{1}$$ Here $\Omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the volume of the $d$-sphere.

There's a related regularization where instead we replace $I$ with $$\int d^nk \big(\frac{\mu^2}{|k|^2}\big)^{\epsilon/2} f(|k|) = \mu^{\epsilon}\Omega_n \int p^{n-\epsilon-1} f(p)dp.\tag{2}$$ These two expressions differ only in the angular contribution, which is momentum independent and regular in $\epsilon = n-d$. If the momentum dependence of $I$ is also regular in $\epsilon$ (and it really should be), this alters neither the $\epsilon$ pole structure nor the momentum dependence of the final result. Instead, it shifts the constant-in-$\epsilon$ terms by quantities that don't depend on the momenta.

For example, in the integral $I=(4\pi)^2\int \frac{d^4k}{(2\pi)^4} \frac{1}{(k^2+\Delta)^2}$, one finds that the "Macaroni & Pie" term $-\gamma + \log(4\pi)$ drops out, leaving behind $\frac{2}{\epsilon} + \log(\frac{\mu^2}{\Delta})$. (Proof: Instead of cancelling the $\Gamma(d/2)$'s and using Stirling's Formula, apply Euler's Reflection formula $\Gamma(d/2)\Gamma(2-d/2) = \pi \csc(\pi d/2)$. The Laurent expansion of $\csc(z)$ is $\frac{1}{z} + \frac{1}{6}z + ...$.)

My question: Are these two regularizations really equivalent?

It looks to me like they are, and I've spot checked in QED at 1-loop.