# Momentum replacement in the axial anomaly calculation in dimensional regularisation (‘t Hooft prescription)

I have been studying the axial anomaly and everywhere I see the calculation of the triangle loop using dimensional regularisation (see for example pages 661-664 of section 19.2 of Peskin). In the ‘t Hooft prescription for the $$\gamma^5$$ they divide the Lorentz space into the usual 4 dimensional one and the rest of dimensions (inside the integration), so the loop momentum can be written as (eq. 19.53) $$$$l=l_\parallel+l_\perp$$$$

“ Where the first term has nonzero components in dimensions 0,1, 2, 3 and the second term has nonzero components in the other d—4 ($$-2\epsilon$$) dimensions.”

Then, we arrive to this integral

$$$$\int \frac{d^dl}{(2\pi)^d}\frac{l_\perp^2}{(l^2-\Delta)^3}$$$$

and what I do not understand is the following replacement for it (eq 19.57) $$$$l_\perp^2\to\frac{d-4}{d}l^2$$$$ “under the symmetrical integration”.

I do understand when we do similar things for an even integral for which any odd term is zero so

$$$$l_\mu l_\nu\to\frac{1}{d}l^2g_{\mu\nu}$$$$

But in the case with the $$l_\perp$$ I do not see a proper derivation, could someone help me with that? Thanks in advance.

• Just a side note, “the symmetrical integration” sort of sleight of hands only works when the Feynman integral is either convergent or logarithmically divergent. – MadMax Aug 5 '19 at 14:19
• As manifested in OP's example, whenever the pseudo scalar $\gamma_5$ is in the picture, dimensional regularization feels as awkward as Uncle Joe talking about internet. – MadMax Aug 5 '19 at 14:28

As the integral has rotational invariance in $$d$$ dimension, each $$\ell$$-component should yield the same value. There are $$d-4$$ non-zero components in $$\ell_\perp$$ and $$d$$ non-zero components in $$\ell$$, we should thus have $$$$\frac{1}{d-4}\int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}_\perp^2}{(\ell^2-\Delta)^3} = \frac{1}{d}\int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}^2}{(\ell^2-\Delta)^3}$$$$ Therefore $$$$\mathcal{I} = \frac{d-4}{d} \int \frac{d^4 \ell}{(2\pi)^4} \frac{{\ell}^2}{(\ell^2-\Delta)^3}$$$$