$\displaystyle T\overline{T}^{\frac{1}{3}}_{0}|_{\text{flat}}$
$\displaystyle=\frac{1}{2^{\frac{5}{3}}}\bigg{[}\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\bigg{]}^{\frac{1}{3}}\,,$ (20) $\displaystyle
T\overline{T}^{\frac{1}{3}}_{1}|_{\text{flat}}$
$\displaystyle=\frac{-\dot{\varphi}^{i}\dot{\varphi}^{i}+\varphi^{\prime
i}\varphi^{\prime i}}{9\cdot
2^{\frac{1}{3}}\bigg{[}\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\bigg{]}^{\frac{1}{3}}}\,,$ $\displaystyle
T\overline{T}^{\frac{1}{3}}_{2}|_{\text{flat}}$
$\displaystyle=-\frac{\left(\dot{\varphi}^{i}\dot{\varphi}^{i}\right)^{2}+(\varphi^{\prime
i}\varphi^{\prime i})^{2}+12(\dot{\varphi}^{i}\varphi^{\prime
i})^{2}-14\dot{\varphi}^{i}\dot{\varphi}^{i}\varphi^{\prime j}\varphi^{\prime
j}}{216\bigg{[}\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime j}\right)\bigg{]}}\,.$

For this deformation, it does not seem possible to find an all-orders closed-
form solution to the flow equation, but the perturbative
$T\overline{T}^{\frac{1}{3}}$-deformed action to
$\mathcal{O}\left(\lambda^{3}\right)$ is

$\displaystyle S^{(\lambda)}$ $\displaystyle=\int
dt\,d\theta\,\bigg{(}\frac{1}{2}\left(-\dot{\varphi}^{i}\dot{\varphi}^{i}+\varphi^{\prime
i}\varphi^{\prime i}\right)$ (21)
$\displaystyle+\frac{\lambda}{2^{\frac{5}{3}}}\bigg{[}\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\bigg{]}^{\frac{1}{3}}$ $\displaystyle+\frac{\lambda^{2}}{9\cdot
2^{\frac{4}{3}}}\frac{-\dot{\varphi}^{i}\dot{\varphi}^{i}+\varphi^{\prime
i}\varphi^{\prime i}}{\bigg{[}\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\bigg{]}^{\frac{1}{3}}}$
$\displaystyle-\frac{\lambda^{3}}{648}\frac{\left(\dot{\varphi}^{i}\dot{\varphi}^{i}\right)^{2}+(\varphi^{\prime
i}\varphi^{\prime i})^{2}+12(\dot{\varphi}^{i}\varphi^{\prime
i})^{2}-14\dot{\varphi}^{i}\dot{\varphi}^{i}\varphi^{\prime j}\varphi^{\prime
j}}{\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime j}\right)}+\cdots\bigg{)}\,.$

_$f(T^{\alpha}{}_{\alpha},T^{\alpha\beta}T_{\alpha\beta})$ Perturbative Flow
for Multiple Bosons_

To conclude this appendix, we note that one can also study the perturbative
solution to the flow driven by the
$f(T^{\alpha}{}_{\alpha},T^{\alpha\beta}T_{\alpha\beta})=f(z,x)$ operator of
equation (1) for an _arbitrary_ function $f$. We again take the initial
condition for the flow to be the action (6) for $N$ free massless bosons. The
first few terms in the perturbative expansion are

$\displaystyle
f(T\alpha{}_{\alpha},T^{\alpha\beta}T_{\alpha\beta})_{0}|_{\text{flat}}$
$\displaystyle=f\left(x\right)\,,\quad
f(T^{\alpha}{}_{\alpha},T^{\alpha\beta}T_{\alpha\beta})_{1}|_{\text{flat}}=4xy\left(\frac{\partial
f(x)}{\partial x}\right)^{2}\,,$ (22)

where

$\displaystyle x$
$\displaystyle=\frac{1}{2}\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{i}-\varphi^{\prime
i}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime
j}\right)\left(\dot{\varphi}^{j}+\varphi^{\prime j}\right)$ (23)
$\displaystyle=\frac{1}{2}\left(2\left(\partial_{\mu}\varphi^{i}\partial^{\nu}\varphi^{i}\right)\left(\partial_{\nu}\varphi^{j}\partial^{\mu}\varphi^{j}\right)-\left(\partial_{\mu}\varphi^{i}\partial^{\mu}\varphi^{i}\right)^{2}\right)\,,$
$\displaystyle y$
$\displaystyle=-\dot{\varphi}^{i}\dot{\varphi}^{i}+\varphi^{\prime
i}\varphi^{\prime i}=\partial_{\mu}\varphi^{i}\partial^{\mu}\varphi^{i}\,.$

The perturbative action at $\mathcal{O}(g^{2})$ is

$\displaystyle S^{(g)}$ $\displaystyle=\int
dtd\theta\bigg{[}\frac{y}{2}+gf\left(x\right)+2g^{2}xy\left(\frac{\partial
f(x)}{\partial x}\right)^{2}+\cdots\bigg{]}\,.$ (24)

Furthermore, to summarize in the table below, one can check equation (24)
recovers the correct coefficients at $\mathcal{O}(g^{2})$ for the perturbative
actions describing $N$ free massless bosons considered in this appendix.

In principle, one could also study the perturbative quantization of these more
general $f(T^{\alpha}{}_{\alpha},T^{\alpha\beta}T_{\alpha\beta})$-deformed
scalar models. For instance, one could use the background field expansion and
determine their Feynman rules as done in section 5 for the Modified Scalar
theory, or study canonical quantization following section 4.

##  Appendix 11.B Details of Feynman Diagram Calculations

In this appendix, we collect the technical details of certain evaluations of
Feynman diagrams which occur in the analysis of section 5.

####  1 One-loop, $2$-vertex calculation

Let us first focus on the divergence structure of the diagram
$\mathcal{D}_{2}$ of equation (260), which we repeat here for convenience:

$\displaystyle\mathcal{D}_{2}\;=\;\raisebox{-0.5pt}{\includegraphics[width=164.77771pt]{diagrams/divergent_contribution.pdf}}\;.$
(25)

As we mentioned around equation (263), the value of this diagram can be
expressed in terms of the simpler quantity

$\displaystyle{\mathcal{I}}^{\mu\nu\rho\tau}_{2}=\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\frac{(\ell+q)^{\nu}\ell^{\mu}(\ell+q)^{\tau}\ell^{\rho}}{\ell^{2}\left(\ell+q\right)^{2}}\,.$
(26)

All of the dependence on loop momenta is encoded within
${\mathcal{I}}^{\mu\nu\rho\tau}_{2}$, which we will also write as
${\mathcal{I}}_{2}$ with indices suppressed for simplicity. From the value of
$\mathcal{I}_{2}$, the original diagram $\mathcal{D}_{2}$ is recovered from
the expression (262), which only involves additional dependence on the
classical background via the tensor $P_{\mu\nu}^{ij}$ and an additional
integration over the momentum $q$. Therefore, in order to study the
divergences arising from the loop, it suffices to perform dimensional
regularization of the quantity $\mathcal{I}_{2}$.

Expanding out the products and introducing a Feynman parameter $x$ in order to
resolve the denominator, we find

$\displaystyle\mathcal{I}_{2}$
$\displaystyle=\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\frac{\ell^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}+\ell^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}}{\ell^{2}\left(\ell+q\right)^{2}}$
$\displaystyle=\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\int_{0}^{1}d^{d}x\frac{\ell^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}+\ell^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}}{\left[\ell^{2}\left(1-x\right)+x\left(\ell+q\right)^{2}\right]^{2}}$
$\displaystyle=\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\int_{0}^{1}d^{d}x\frac{\ell^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}+\ell^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}}{\left[\ell^{2}+x\left(2\ell_{\mu}q^{\mu}+q^{2}\right)\right]^{2}}$
$\displaystyle=\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\int_{0}^{1}d^{d}x\frac{\ell^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}+\ell^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}}{\left[\left(\ell^{\mu}+xq^{\mu}\right)^{2}+x\left(1-x\right)q^{2}\right]^{2}}\,.$
(27)

In the final step, we have completed the square in the denominator by adding
and subtracting $q^{2}x^{2}$. We now shift the integration variable from
$\ell^{\mu}$ to

$\displaystyle\ell^{\prime\mu}=\ell^{\mu}-xq^{\mu}\,,$ (28)

which causes the denominator to become even in $\ell^{\prime}$, and thus terms
in the numerator which are odd in $\ell^{\prime\mu}$ will vanish by symmetry.
We immediately drop the primes on $\ell^{\prime\mu}$ and write the surviving
terms as

$\displaystyle{\mathcal{I}}_{2}$
$\displaystyle=\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\int_{0}^{1}d^{d}x\Bigg{[}\frac{\ell^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}}{\left(\ell^{2}+q^{2}x\left(1-x\right)\right)^{2}}+\frac{x^{2}\ell^{\nu}q^{\mu}\ell^{\tau}q^{\rho}}{\left(\ell^{2}+q^{2}x\left(1-x\right)\right)^{2}}$
$\displaystyle+\frac{\left(x^{2}-2x+1\right)q^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}}{\left(\ell^{2}+q^{2}x\left(1-x\right)\right)^{2}}$
$\displaystyle+\frac{\left(x^{2}-x\right)\left(q^{\nu}q^{\mu}\ell^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}\ell^{\tau}q^{\rho}+\ell^{\nu}\ell^{\mu}q^{\tau}q^{\rho}+\ell^{\nu}q^{\mu}q^{\tau}\ell^{\rho}\right)}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}$
$\displaystyle+\frac{\left(x^{4}-2x^{3}+x^{2}\right)q^{\nu}q^{\mu}q^{\tau}q^{\rho}}{\left(\ell^{2}+q^{2}x\left(1-x\right)\right)^{2}}\Bigg{]}\,$
$\displaystyle=\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\int_{0}^{1}d^{d}x\Bigg{[}\frac{\ell^{\nu}\ell^{\mu}\ell^{\tau}\ell^{\rho}}{\left(\ell^{2}+q^{2}x\left(1-x\right)\right)^{2}}+\frac{x^{2}\ell^{\nu}q^{\mu}\ell^{\tau}q^{\rho}}{\left(\ell^{2}+q^{2}x\left(1-x\right)\right)^{2}}$
$\displaystyle+\frac{\left(1-x\right)^{2}q^{\nu}\ell^{\mu}q^{\tau}\ell^{\rho}}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}$
$\displaystyle+\frac{x\left(1-x\right)\left(q^{\nu}q^{\mu}\ell^{\tau}\ell^{\rho}+q^{\nu}\ell^{\mu}\ell^{\tau}q^{\rho}+\ell^{\nu}\ell^{\mu}q^{\tau}q^{\rho}+\ell^{\nu}q^{\mu}q^{\tau}\ell^{\rho}\right)}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}$
$\displaystyle+\frac{\left[x^{2}\left(1-x\right)^{2}\right]q^{\nu}q^{\mu}q^{\tau}q^{\rho}}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}\Bigg{]}\,.$
(29)

By a symmetry argument similar to the one discussed around equations (256) and
(258), within the integral we can replace products of loop momenta with
symmetrized combinations of metric tensors:

$\displaystyle\ell^{\mu}\ell^{\nu}$
$\displaystyle\to\frac{1}{d}\ell^{2}g^{\mu\nu}\,,$
$\displaystyle\ell^{\mu}\ell^{\nu}\ell^{\rho}\ell^{\tau}$
$\displaystyle\to\frac{1}{d\left(d+2\right)}\ell^{4}\left(g^{\mu\nu}g^{\rho\tau}+g^{\mu\rho}g^{\nu\tau}+g^{\mu\tau}g^{\nu\rho}\right)\,.$
(30)

Applying the replacements (1), the integral ${\mathcal{I}}_{2}$ becomes

$\displaystyle{\mathcal{I}}_{2}$
$\displaystyle=\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\int_{0}^{1}d^{d}x\Bigg{(}\frac{\ell^{4}}{d\left(d+2\right)}\frac{g^{\mu\nu}g^{\rho\tau}+g^{\mu\tau}g^{\nu\rho}+g^{\mu\rho}g^{\nu\tau}}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}$
$\displaystyle+\frac{\ell^{2}}{d}\frac{x^{2}g^{\nu\tau}q^{\mu}q^{\rho}}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}$
$\displaystyle+\frac{\ell^{2}}{d}\frac{\left(1-x\right)^{2}g^{\mu\rho}q^{\nu}q^{\tau}}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}$
$\displaystyle+\frac{\ell^{2}}{d}\frac{x\left(1-x\right)\left(q^{\nu}q^{\mu}g^{\tau\rho}+g^{\mu\tau}q^{\nu}q^{\rho}+g^{\nu\mu}q^{\tau}q^{\rho}+g^{\nu\rho}q^{\mu}q^{\tau}\right)}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}$
$\displaystyle\qquad+\frac{x^{2}\left(1-x\right)^{2}q^{\nu}q^{\mu}q^{\tau}q^{\rho}}{\left[\ell^{2}+q^{2}x\left(1-x\right)\right]^{2}}\Bigg{)}\,.$
(31)

It will be convenient to make use of the standard result

$\displaystyle\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\frac{\ell^{2\beta}}{\left(\ell^{2}-\Delta^{2}\right)^{\alpha}}=i\left(-1\right)^{\alpha+\beta}\frac{\Gamma\left(\beta+\frac{d}{2}\right)\Gamma\left(\alpha-\beta-\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(\alpha\right)\Gamma\left(\frac{d}{2}\right)}\Delta^{2\left(\frac{d}{2}-\alpha+\beta\right)}\,,$
(32)

which can be found, for instance, in equation (A.4) in [375]. Using (32) with

$\displaystyle\Delta^{2}=-q^{2}x\left(1-x\right)$ (33)

in equation (1), we find

$\displaystyle{\mathcal{I}}_{2}$
$\displaystyle=\frac{i}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(2\right)\Gamma\left(\frac{d}{2}\right)}\int_{0}^{1}d^{d}x\Bigg{(}\frac{\Delta^{d}\Gamma\left(2+\frac{d}{2}\right)\Gamma\left(-\frac{d}{2}\right)}{d\left(d+2\right)}\left(g^{\mu\nu}g^{\rho\tau}+g^{\mu\rho}g^{\nu\tau}+g^{\mu\tau}g^{\nu\rho}\right)$
$\displaystyle+\frac{\Delta^{d-2}\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)}{d}x\left(1-x\right)\left(q^{\nu}q^{\mu}g^{\tau\rho}+g^{\mu\tau}q^{\nu}q^{\rho}+g^{\nu\mu}q^{\tau}q^{\rho}+g^{\nu\rho}q^{\mu}q^{\tau}\right)$
$\displaystyle+\frac{\Delta^{d-2}\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)}{d}x^{2}g^{\nu\tau}q^{\mu}q^{\rho}$
$\displaystyle+\frac{\Delta^{d-2}\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)}{d}\left(1-x\right)^{2}g^{\mu\rho}q^{\nu}q^{\tau}$
$\displaystyle+\Delta^{d-4}\Gamma\left(\frac{d}{2}\right)\Gamma\left(2-\frac{d}{2}\right)\left[x^{2}\left(1-x\right)^{2}\right]q^{\nu}q^{\mu}q^{\tau}q^{\rho}\Bigg{)}\,.$
(34)

Using gamma function identities and some algebra, one can simplify this to

$\displaystyle{\mathcal{I}}_{2}$
$\displaystyle=\frac{i\Gamma\left(-\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}}\int_{0}^{1}d^{d}x\Bigg{(}\frac{\Delta^{d}}{4}\left(g^{\mu\nu}g^{\rho\tau}+g^{\mu\rho}g^{\nu\tau}+g^{\mu\tau}g^{\nu\rho}\right)-\frac{d\Delta^{d-2}}{4}x^{2}g^{\nu\tau}q^{\mu}q^{\rho}$
$\displaystyle-\frac{d\Delta^{d-2}}{4}\left(1-x\right)^{2}g^{\mu\rho}q^{\nu}q^{\tau}$
$\displaystyle-\frac{d\Delta^{d-2}}{4}x\left(1-x\right)\left(q^{\nu}q^{\mu}g^{\tau\rho}+g^{\mu\tau}q^{\nu}q^{\rho}+g^{\nu\mu}q^{\tau}q^{\rho}+g^{\nu\rho}q^{\mu}q^{\tau}\right)$
$\displaystyle+\frac{d\left(d-2\right)\Delta^{d-4}}{4}x^{2}\left(1-x\right)^{2}q^{\nu}q^{\mu}q^{\tau}q^{\rho}\Bigg{)}\,.$
(35)

After substituting in for $\Delta^{2}$ using the definition (33), we can now
evaluate the resulting integrals using the formula

$\displaystyle\int_{0}^{1}d^{d}xx^{\alpha-1}\left(1-x\right)^{\beta-1}$
$\displaystyle=\frac{\Gamma\left(\alpha\right)\Gamma\left(\beta\right)}{\Gamma\left(\alpha+\beta\right)}=\mathrm{B}(\alpha,\beta)\,,$
(36)

which we recognize as the definition of the beta function
$\mathrm{B}(\alpha,\beta)$. By doing this, we find

$\displaystyle{\mathcal{I}}_{2}=\frac{i\Gamma\left(-\frac{d}{2}\right)}{4\left(4\pi\right)^{\frac{d}{2}}}\int_{0}^{1}d^{d}x\Bigg{(}$
$\displaystyle
d\left(d-2\right)q^{d-4}\left[-x\left(1-x\right)\right]^{\frac{d}{2}}q^{\nu}q^{\mu}q^{\tau}q^{\rho}-dq^{d-2}\left(-x\right)^{\frac{d}{2}-1}\left(1-x\right)^{\frac{d}{2}+1}g^{\mu\rho}q^{\nu}q^{\tau}$
$\displaystyle+q^{d}\left[-x\left(1-x\right)\right]^{\frac{d}{2}}\left(g^{\mu\nu}g^{\rho\tau}+g^{\mu\rho}g^{\nu\tau}+g^{\mu\tau}g^{\nu\rho}\right)$
$\displaystyle-
dq^{d-2}\left[\left(-x\right)^{\frac{d}{2}+1}\left(1-x\right)^{\frac{d}{2}-1}\right]g^{\nu\tau}q^{\mu}q^{\rho}$
$\displaystyle+dq^{d-2}\left[-x\left(1-x\right)\right]^{\frac{d}{2}}\left(q^{\nu}q^{\mu}g^{\tau\rho}+g^{\mu\tau}q^{\nu}q^{\rho}+g^{\nu\mu}q^{\tau}q^{\rho}+g^{\nu\rho}q^{\mu}q^{\tau}\right)\Bigg{)}$
$\displaystyle=\frac{i\left(-1\right)^{\frac{d}{2}}\Gamma\left(-\frac{d}{2}\right)}{4\left(4\pi\right)^{\frac{d}{2}}}\Bigg{(}q^{d}\frac{\Gamma\left(\frac{d}{2}+1\right)^{2}}{\Gamma\left(d+2\right)}\left(g^{\mu\nu}g^{\rho\tau}+g^{\mu\rho}g^{\nu\tau}+g^{\mu\tau}g^{\nu\rho}\right)$
$\displaystyle+dq^{d-2}\frac{\Gamma\left(\frac{d}{2}\right)\Gamma\left(\frac{d}{2}+2\right)}{\Gamma\left(d+2\right)}\left[g^{\mu\rho}q^{\nu}q^{\tau}+g^{\nu\tau}q^{\mu}q^{\rho}\right]+d\left(d-2\right)q^{d-4}\frac{\Gamma\left(\frac{d}{2}+1\right)^{2}}{\Gamma\left(d+2\right)}q^{\nu}q^{\mu}q^{\tau}q^{\rho}$
$\displaystyle+dq^{d-2}\frac{\Gamma\left(\frac{d}{2}+1\right)^{2}}{\Gamma\left(d+2\right)}\left(q^{\nu}q^{\mu}g^{\tau\rho}+g^{\mu\tau}q^{\nu}q^{\rho}+g^{\nu\mu}q^{\tau}q^{\rho}+g^{\nu\rho}q^{\mu}q^{\tau}\right)\Bigg{)}\,.$
(37)

Note that each term in (1) scales as $q^{d}$, as expected. Factoring out the
gamma functions, we have found

$\displaystyle{\mathcal{I}}_{2}$
$\displaystyle=\frac{i\left(-1\right)^{\frac{d}{2}}\Gamma\left(-\frac{d}{2}\right)}{4\left(4\pi\right)^{\frac{d}{2}}}\frac{\Gamma\left(\frac{d}{2}+1\right)^{2}}{\Gamma\left(d+2\right)}\Bigg{[}q^{d}\left(g^{\mu\nu}g^{\rho\tau}+g^{\mu\rho}g^{\nu\tau}+g^{\mu\tau}g^{\nu\rho}\right)$
$\displaystyle+d\left(d-2\right)q^{d-4}q^{\nu}q^{\mu}q^{\tau}q^{\rho}+dq^{d-2}\left(q^{\nu}q^{\mu}g^{\tau\rho}+g^{\mu\tau}q^{\nu}q^{\rho}+g^{\nu\mu}q^{\tau}q^{\rho}+g^{\nu\rho}q^{\mu}q^{\tau}\right)$
$\displaystyle+\left(d+2\right)q^{d-2}\left(g^{\mu\rho}q^{\nu}q^{\tau}+g^{\nu\tau}q^{\mu}q^{\rho}\right)\Bigg{]}\,.$
(38)

Finally, to perform dimensional regularization, we set the spacetime dimension
to $d=2+2\epsilon$ and take $\epsilon\to 0$ using the limiting behavior

$\displaystyle\Gamma\left(-1-\epsilon\right)=\frac{1}{\epsilon}-\gamma+1+\mathcal{O}\left(\epsilon\right)$
(39)

for the gamma functions. Keeping only divergent terms, we arrive at the final
expression

$\displaystyle{\mathcal{I}}_{2}$
$\displaystyle=\left(\frac{1}{\epsilon}\right)\frac{-i}{24\left(4\pi\right)}\bigg{[}q^{2}\left(g^{\mu\nu}g^{\rho\tau}+g^{\mu\rho}g^{\nu\tau}+g^{\mu\tau}g^{\nu\rho}\right)$
$\displaystyle+2\left(q^{\nu}q^{\mu}g^{\tau\rho}+g^{\mu\tau}q^{\nu}q^{\rho}+g^{\nu\mu}q^{\tau}q^{\rho}+g^{\nu\rho}q^{\mu}q^{\tau}\right)$
$\displaystyle+4\left(g^{\mu\rho}q^{\nu}q^{\tau}+g^{\nu\tau}q^{\mu}q^{\rho}\right)\bigg{]}\,.$
(40)

This completes the evaluation of the divergent contribution from
${\mathcal{I}}_{2}$, which justifies the result (2) which was quoted in the
body of the paper.

####  2 One-loop, $n$-vertex calculation

The $n$-vertex diagram $\mathcal{D}_{n}$ can be computed via a generalization
of the method used in appendix 1. We again write $\ell$ for the loop momentum
and we label the external momenta as $q_{i}$, for $i=0,\ldots,n-1$, with
momentum conservation implying that

$\displaystyle q_{n-1}=-\sum_{i=0}^{n-2}q_{i}\,.$ (41)

As we did with $\mathcal{D}_{2}$ in equation (2), let us strip off various
factors of $P_{\mu\nu}{}^{ij}$ to write

$\displaystyle\mathcal{D}_{n}$
$\displaystyle=\left(-2\tanh(\gamma)\right)^{N}$
$\displaystyle\cdot\int\left(\frac{d^{d}q_{0}}{(2\pi)^{d}}P_{(\mu_{1}\mu_{2})}^{i_{1}i_{2}}(q_{0})\right)\left(\frac{d^{d}q_{0}}{(2\pi)^{d}}P_{(\mu_{3}\mu_{4})}^{i_{2}i_{3}}(q_{1})\right)\cdots\left(\frac{d^{d}q_{n-2}}{(2\pi)^{d}}P_{(\mu_{2n-3}\mu_{2n-2})}^{i_{n-1}i_{n}}(q_{n-2})\right)$
$\displaystyle\cdot
P_{(\mu_{2n-1}\mu_{2n})}^{i_{n}i_{1}}(q_{n-1})\left({\mathcal{I}}_{n}\right)^{\mu_{1}\mu_{2}\ldots\mu_{2n-1}\mu_{2n}}\,,$
(42)

where ${\mathcal{I}}_{n}$ is the simpler integral

$\displaystyle\left({\mathcal{I}}_{n}\right)^{\mu_{1}\mu_{2}\ldots\mu_{2n-1}\mu_{2n}}$
$\displaystyle=\int\frac{d^{d}\ell}{(2\pi)^{d}}\,\left(\prod_{i=0}^{n-1}\left(\ell+\sum_{j=0}^{i}q_{j}\right)^{-2}\right)\left(\prod_{k=1}^{n}\left(\ell+\sum_{j=0}^{k-1}q_{j}\right)^{\mu_{2k-1}}\left(\ell+\sum_{j=0}^{k-1}q_{j}\right)^{\mu_{2k}}\right)\,.$
(43)

We will further break up $\mathcal{I}_{n}$ into pieces and evaluate each piece
in turn. Let us write the integrand of (2) as a product

$\displaystyle\left({\mathcal{I}}_{n}\right)^{\mu_{1}\mu_{2}\ldots\mu_{2n-1}\mu_{2n}}=\int\frac{d^{d}\ell}{(2\pi)^{d}}\,\mathcal{P}_{n}\mathcal{V}^{\mu_{1}\mu_{2}\ldots\mu_{2n-1}\mu_{2n}}\,,$
(44)

where the symbol

$\displaystyle\mathcal{P}_{n}$
$\displaystyle=\prod_{i=0}^{n-1}\left(\ell+\sum_{j=0}^{i}q_{j}\right)^{-2}$
(45)

refers to the collection of all factors in $\mathcal{I}_{n}$ which come from
propagators, and the symbol

$\displaystyle\mathcal{V}^{\mu_{1}\mu_{2}\ldots\mu_{2n-1}\mu_{2n}}$
$\displaystyle=\prod_{k=1}^{n}\left(\ell+\sum_{j=0}^{k-1}q_{j}\right)^{\mu_{2k-1}}\left(\ell+\sum_{j=0}^{k-1}q_{j}\right)^{\mu_{2k}}\,,$
(46)

which we will sometimes abbreviate as $\mathcal{V}$, refers to the pieces
coming from vertex factors.

In order to highlight the divergence structure of the diagram
$\mathcal{D}_{n}$, we will focus on performing the loop momentum integral of
various terms appearing in the product $\mathcal{P}\mathcal{V}$ of equation
(44), and neglect the additional structure arising from the contraction with
the various tensors $P_{\mu\nu}^{ij}$ to obtain $\mathcal{D}_{n}$ in (2).

Let us begin by simplifying the product $\mathcal{P}$ of $n$ propagator
factors. In general, we can write the product of $n$ propagators using a
Feynman parameterization:

$\displaystyle\prod_{i=0}^{n-1}A_{i}^{-1}$
$\displaystyle=\int_{0}^{1}\left(\prod_{i=0}^{n-1}d^{d}x_{i}\right)\delta\left(\sum_{i=0}^{n-1}x_{i}-1\right)\frac{\left(n-1\right)!}{\left[\sum_{i}x_{i}A_{i}\right]^{n}}\,.$
(47)

The product of propagators inside the loop can thus be expressed as

$\displaystyle\mathcal{P}=\int_{0}^{1}\left(\prod_{i=0}^{n-1}d^{d}x_{i}\right)\delta\left(\sum_{i=0}^{n-1}x_{i}-1\right)\left[\sum_{i=0}^{n-1}x_{i}\left(\ell+\sum_{j=0}^{i}q_{j}\right)^{2}\right]^{-n}\,.$
(48)

As $\sum_{i}x_{i}=1$, we can expand and reduce the square bracketed term to

$\displaystyle\left[\sum_{i}x_{i}\left(\ell+\sum_{j=0}^{i}q_{j}\right)^{2}\right]^{-n}$
$\displaystyle=\left[\ell^{2}+\sum_{i}x_{i}\left(2\ell_{\mu}\sum_{j=0}^{i}q_{j}^{\mu}+\left(\sum_{j=0}^{i}q_{j}\right)^{2}\right)\right]^{-n}$
$\displaystyle=\left[\left(\ell+\sum_{i}\sum_{j=0}^{i}x_{i}q_{j}\right)^{2}-\left(\sum_{i}\sum_{j=0}^{i}x_{i}q_{j}\right)^{2}+\sum_{i}x_{i}\left(\sum_{j=0}^{i}q_{j}\right)^{2}\right]^{-n}\,.$
(49)

We change variables in the loop momentum by shifting

$\displaystyle\ell^{\mu}\to\ell^{\mu}-\sum_{i=0}^{n-1}\sum_{j=0}^{i}x_{i}q^{\mu}_{j}\,,$
(50)

so that the bracketed expression becomes

$\displaystyle\left[\sum_{i}x_{i}\left(\ell+\sum_{j=0}^{i}q_{j}\right)^{2}\right]^{-n}$
$\displaystyle=\left[\ell^{2}-\left(\sum_{i}\sum_{j=0}^{i}x_{i}q_{j}\right)^{2}+\sum_{i}x_{i}\left(\sum_{j=0}^{i}q_{j}\right)^{2}\right]^{-n}$
$\displaystyle=\left[\ell^{2}-\Delta^{2}\right]^{-n}\,,$ (51)

where we have defined

$\displaystyle\Delta^{2}$
$\displaystyle=\left(\sum_{i}\sum_{j=0}^{i}x_{i}q_{j}\right)^{2}-\sum_{i}x_{i}\left(\sum_{j=0}^{i}q_{j}\right)^{2}\,.$
(52)

Overall this allows us to write the propagators as

$\displaystyle\mathcal{P}=(n-1)!\int_{0}^{1}\left(\prod_{i=0}^{n-1}d^{d}x_{i}\right)\delta\left(\sum_{i=0}^{n-1}x_{i}-1\right)\left(\ell^{2}-\Delta^{2}\right)^{-n}\,.$
(53)

Next let us turn to the contributions from the vertices in equation (46),
which yield factors of momenta in the numerator of the integrand. Under the
change of variables (50) which renders the denominator of $\mathcal{P}$
quadratic in $\ell$, the vertex factor contribution becomes

$\displaystyle\mathcal{V}^{\mu_{1}\mu_{2}\ldots\mu_{2n-1}\mu_{2n}}$
$\displaystyle=\prod_{k=1}^{n}\left(\ell-\sum_{i=0}^{n-1}\sum_{j=0}^{i}x_{i}q_{j}+\sum_{j=0}^{k-1}q_{j}\right)^{\mu_{2k-1}}\left(\ell-\sum_{i=0}^{n-1}\sum_{j=0}^{i}x_{i}q_{j}+\sum_{j=0}^{k-1}q_{j}\right)^{\mu_{2k}}\,.$
(54)

We expand this product in descending powers of $\ell$ as only powers
$\ell^{2n}$ and $\ell^{2n-2}$ will lead to divergent terms. We have that

$\displaystyle\mathcal{V}^{\mu_{1}\mu_{2}\ldots\mu_{2n-1}\mu_{2n}}$
$\displaystyle=\prod_{k=1}^{n}\ell^{\mu_{2k-1}}\ell^{\mu_{2k}}+\sum_{a=1}^{2n}\sum_{b>a}^{2n}\left(\prod_{c\neq
a,b}^{2n}\ell^{\mu_{c}}\right)f^{\mu_{a}}\left(x,q,a\right)f^{\mu_{b}}\left(x,q,b\right)+\mathcal{O}\left(\ell^{2n-4}\right)\,,$
(55)

where we have defined for brevity

$\displaystyle
f^{\mu}\left(x,q,a\right)=\sum_{i=0}^{n-1}\sum_{j=0}^{i}x_{i}q_{j}^{\mu}+\sum_{j=0}^{\lfloor\frac{a-1}{2}\rfloor}q_{j}^{\mu}\,.$
(56)

Next we will replace products of loop momenta using the generalized
symmetrization rule of equation (258). To ease notation, let us write

$\displaystyle g^{\mu_{1}\ldots\mu_{n}}=g^{(\mu_{1}\mu_{2}}\cdots
g^{\mu_{n-1}\mu_{n})}\,$ (57)

for the symmetrized combination of derivatives appearing in this expression.
When no confusion is possible, we will also write $g^{\\{\mu\\}}$ for (57),
where $\\{\mu\\}$ is understood to refer to a multi-index
$\\{\mu\\}=\mu_{1}\ldots\mu_{n}$. With this notation, the replacement rule
becomes

$\displaystyle\prod_{i=1}^{n}\ell^{\mu_{2i-1}}\ell^{\mu_{2i}}\to\frac{\ell^{2n}\left(d-2\right)!!}{\left(d-2+2n\right)!!}g^{\mu_{1}\ldots\mu_{2n}}\,.$
(58)

This transforms the vertex factor contribution to

$\displaystyle\mathcal{V}^{\mu_{1}\mu_{2}\ldots\mu_{2n-1}\mu_{2n}}$
$\displaystyle=\frac{\ell^{2n}\left(d-2\right)!!}{\left(d-2+2n\right)!!}g^{\mu_{1}\cdots\mu_{2n}}$
$\displaystyle\quad+\frac{\ell^{2n-2}\left(d-2\right)!!}{\left(d-4+2n\right)!!}\sum_{a=1}^{2n}\sum_{b>a}^{2n}g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}f^{\mu_{a}}\left(x,q,a\right)f^{\mu_{b}}\left(x,q,b\right)+\mathcal{O}\left(\ell^{2n-4}\right)\,.$
(59)

In equation (2), we have written $g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}$ to refer
to a product of the form (57) in which the multi-index $\\{\mu\\}$ runs over
all possible values except for the two indices $\mu_{a}$ and $\mu_{b}$, which
are excluded.

Let us now combine the pieces and identify the divergent terms in
$\mathcal{D}_{n}$. We can evaluate

$\displaystyle\left({\mathcal{I}}_{n}\right)^{\mu_{1}\mu_{2}\ldots\mu_{2n}}$
$\displaystyle\quad=\int\frac{d^{d}\ell}{(2\pi)^{d}}\mathcal{P}\mathcal{V}^{\mu_{1}\mu_{2}\ldots\mu_{2n}}\,$
$\displaystyle\quad=\int\frac{d^{d}\ell}{(2\pi)^{d}}\left(n-1\right)!\int_{0}^{1}\left(\prod_{i=0}^{n-1}d^{d}x_{i}\right)\delta\left(\sum_{i=0}^{n-1}x_{i}-1\right)\left(\ell^{2}-\Delta^{2}\right)^{-n}$
$\displaystyle\cdot\Bigg{(}\frac{\ell^{2n}\left(d-2\right)!!}{\left(d-2+2n\right)!!}g^{\mu_{1}\ldots\mu_{2n}}+\frac{\ell^{2n-2}\left(d-2\right)!!}{\left(d-4+2n\right)!!}\sum_{a=1}^{2n}\sum_{b>a}^{2n}g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}f^{\mu_{a}}\left(x,q,a\right)f^{\mu_{b}}\left(x,q,b\right)$
$\displaystyle+\mathcal{O}\left(\ell^{2n-4}\right)\Bigg{)}\,,$ (60)

in terms of the known integral

$\displaystyle\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\frac{\ell^{2\beta}}{\left(\ell^{2}-\Delta^{2}\right)^{\alpha}}$
$\displaystyle=i\left(-1\right)^{\alpha+\beta}\frac{\Gamma\left(\beta+\frac{d}{2}\right)\Gamma\left(\alpha-\beta-\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(\alpha\right)\Gamma\left(\frac{d}{2}\right)}\Delta^{2\left(\frac{d}{2}-\alpha+\beta\right)}\,.$
(61)

First let us justify why the terms of order $\ell^{2n-4}$ and lower in
equation (2) will not give divergent contributions. A term proportional to
$\ell^{2n-4}$ in the parentheses of (2), after multiplying the propagator
factor $(\ell^{2}-\Delta^{2})^{-n}$, gives a term in the integrand of the form
(61) with $\alpha=n$ and $\beta=n-2$. Such a term gives a contribution

$\displaystyle\int\frac{d^{d}\ell}{\left(2\pi\right)^{d}}\frac{\ell^{2(n-2)}}{\left(\ell^{2}-\Delta^{2}\right)^{n}}$
$\displaystyle=i\left(-1\right)^{n+(n-2)}\frac{\Gamma\left(n-2+\frac{d}{2}\right)\Gamma\left(2-\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(n\right)\Gamma\left(\frac{d}{2}\right)}\Delta^{2\left(\frac{d}{2}-n+(n-2)\right)}\,.$
(62)

In the limit as $d\to 2$, the two factors of gamma functions in the numerator
of (62) tend to $\Gamma(n-1)$ and $\Gamma(1)$, which are both finite since
$n>2$. Since we are only interested in computing the divergent contributions
arising from these diagrams, we ignore these terms. Similarly, any terms of
lower order in $\ell$ can be evaluated in the same way but with even smaller
values of $\beta$, which also lead to finite contributions from the gamma
functions.

Let us therefore focus on the divergent terms. The term in the integrand
proportional to $\ell^{2n}$ in equation (2) takes the form (61) with
$\alpha=\beta=n$. Similarly, the term in the integrand that scales as
$\ell^{2n-2}$ is of the form (61) with $\alpha=n$ and $\beta=n-1$. Evaluating
the loop momentum integrals then gives

$\displaystyle\left({\mathcal{I}}_{n}\right)^{\mu_{1}\mu_{2}\ldots\mu_{2n}}$
$\displaystyle=\left(n-1\right)!\int_{0}^{1}\left(\prod_{i=0}^{n-1}dx_{i}\right)\delta\left(\sum_{i=0}^{n-1}x_{i}-1\right)$
$\displaystyle\quad\cdot\Bigg{(}\frac{i\left(d-2\right)!!}{\left(d-2+2n\right)!!}g^{\mu_{1}\cdots\mu_{2n}}\frac{\Gamma\left(n+\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(n\right)\Gamma\left(\frac{d}{2}\right)}\Gamma\left(-\frac{d}{2}\right)\Delta^{d}$
$\displaystyle\quad+\frac{i\left(d-2\right)!!}{\left(d-4+2n\right)!!}\sum_{a=1}^{2n}\sum_{b>a}^{2n}g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}f^{\mu_{a}}\left(x,q,a\right)f^{\mu_{b}}\left(x,q,b\right)$
$\displaystyle\quad\cdot\frac{\Gamma\left(n-1+\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(n\right)\Gamma\left(\frac{d}{2}\right)}\Gamma\left(1-\frac{d}{2}\right)\Delta^{d-2}\Bigg{)}\,.$
(63)

To better analyze the divergence structure, it is useful to define shorthand
notation for the two terms appearing in (2), which we call
$\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}$ and
$\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}$:

$\displaystyle\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle\equiv\frac{i\left(d-2\right)!!}{\left(d-2+2n\right)!!}g^{\mu_{1}\cdots\mu_{2n}}\frac{\Gamma\left(n+\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(n\right)\Gamma\left(\frac{d}{2}\right)}\Gamma\left(-\frac{d}{2}\right)\Delta^{d}\,\,$
$\displaystyle\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle\equiv\frac{i\left(d-2\right)!!}{\left(d-4+2n\right)!!}\sum_{a=1}^{2n}\sum_{b>a}^{2n}g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}f^{\mu_{a}}\left(x,q,a\right)f^{\mu_{b}}\left(x,q,b\right)$
$\displaystyle\qquad\cdot\frac{\Gamma\left(n-1+\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(n\right)\Gamma\left(\frac{d}{2}\right)}\Gamma\left(1-\frac{d}{2}\right)\Delta^{d-2}\,.$
(64)

Both contributions $\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}$ and
$\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}$ scale as $q^{d}$ and contain a
polynomial in $x_{i}$ of degree $d$. In
$\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}$, this $q^{d}$ dependence is
contained within $\Delta^{d}$, and for
$\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}$, the power of $q^{d-2}$ from
$\Delta^{d-2}$ is compensated by two factors of $q$, one of which sits in each
function $f^{\mu}$.

Therefore, we conclude that in the limit $d\to 2$, all such $n$ vertex
diagrams have the same general structure as the $2$-vertex diagram which we
saw in (1). In particular, both $\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}$ and
$\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}$ generate divergences of the form
$\frac{1}{\epsilon}$ because they are proportional to
$\Gamma\left(-\frac{d}{2}\right)$ and $\Gamma\left(1-\frac{d}{2}\right)$,
respectively.

We conclude that

$\displaystyle\mathbf{C}_{2n}^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle\sim\frac{1}{\epsilon}q^{2}g^{\mu_{1}\cdots\mu_{2n}}\,,$
$\displaystyle\mathbf{D}_{2n}^{\mu_{1}\ldots\mu_{2n}}$
$\displaystyle\sim\frac{1}{\epsilon}\sum_{a=1}^{2n}\sum_{b>a}^{2n}q^{\mu_{a}}q^{\mu_{b}}g^{\\{\mu\neq\mu_{a},\mu_{b}\\}}\,.$
(65)

####  3 $m$-loop, two-vertex calculation

In this appendix, we will show how to evaluate the integral over one of the
$m$ loop momenta $\ell_{i}$ which appear in the expression for the two-vertex,
$m$-loop diagram of equation (268). It suffices to integrate over the final
momentum $\ell_{m}$, since the result may then be iterated to evaluate the
other $m-1$ integrals.

Therefore, let us focus on performing the integration over $\ell_{m}$ in the
quantity $\mathfrak{L}_{m,2}$ of equation (274). Specifically, we will compute
the quantity

$\displaystyle L_{m,2}$ $\displaystyle=\int
d^{d}\ell_{m}\frac{\ell_{m}^{\nu_{m+1}}\ell_{m}^{\tau_{m+1}}\left(\ell_{m}-\ell_{m-1}\right)^{\nu_{m}}\left(\ell_{m}-\ell_{m-1}\right)^{\tau_{m}}}{\ell_{m}^{2}\left(\ell_{m}-\ell_{m-1}\right)^{2}}\,.$
(66)

This object $L_{m,2}$ is proportional to the remaining integrand that one
finds by performing the integral over $\ell_{m}$ in the definition of
$\mathfrak{L}_{m,2}$. As we will see, after obtaining an expression for
$L_{m,2}$, this result can be used recursively to evaluate
$\mathfrak{L}_{m,2}$ itself.

We notice the integral in equation (66) is exactly of the form of the one
appearing in the $1$-loop, $2$-vertex diagram which we evaluated in appendix
1. Proceeding in the same way, we introduce a Feynman parameter $x$ to write

$\displaystyle L_{m,2}$ $\displaystyle=\int
d^{d}\ell_{m}dx\frac{\ell_{m}^{\nu_{m+1}}\ell_{m}^{\tau_{m+1}}\left(\ell_{m}-\ell_{m-1}\right)^{\nu_{m}}\left(\ell_{m}-\ell_{m-1}\right)^{\tau_{m}}}{\left[\left(1-x\right)\ell_{m}^{2}+x\left(\ell_{m}-\ell_{m-1}\right)^{2}\right]^{2}}$
$\displaystyle=\int
d^{d}\ell_{m}dx\frac{\ell_{m}^{\nu_{m+1}}\ell_{m}^{\tau_{m+1}}\left(\ell_{m}-\ell_{m-1}\right)^{\nu_{m}}\left(\ell_{m}-\ell_{m-1}\right)^{\tau_{m}}}{\left[\ell_{m}^{2}+x\left(2\ell_{m}\cdot\ell_{m-1}-\ell_{m-1}^{2}\right)\right]^{2}}$
$\displaystyle=\int
d^{d}\ell_{m}dx\frac{\ell_{m}^{\nu_{m+1}}\ell_{m}^{\tau_{m+1}}\left(\ell_{m}-\ell_{m-1}\right)^{\nu_{m}}\left(\ell_{m}-\ell_{m-1}\right)^{\tau_{m}}}{\left[\left(\ell_{m}+x\ell_{m-1}\right)^{2}-x^{2}\ell_{m-1}^{2}+x\ell_{m-1}^{2}\right]^{2}}\,,$
(67)

or after shifting the integration variable as
$\ell_{m}\to\ell_{m}-x\ell_{m-1}$,

$\displaystyle L_{m,2}=\int d^{d}\ell_{m}dx$
$\displaystyle\frac{\left(\ell_{m}-x\ell_{m-1}\right)^{\nu_{m+1}}\left(\ell_{m}-x\ell_{m-1}\right)^{\tau_{m+1}}\left(\ell_{m}-\left(1-x\right)\ell_{m-1}\right)^{\nu_{m}}\left(\ell_{m}-\left(1-x\right)\ell_{m-1}\right)^{\tau_{m}}}{\left[\ell_{m}^{2}+x\left(1-x\right)\ell_{m-1}^{2}\right]^{2}}\,.$
(68)

We may keep only even powers of $\ell_{m}$ in the integrand, as odd powers
vanish by symmetry:

$\displaystyle L_{m,2}$ $\displaystyle=\int
d^{d}\ell_{m}dx\Bigg{(}\frac{\ell_{m}^{\nu_{m+1}}\ell_{m}^{\tau_{m+1}}\ell_{m}^{\nu_{m}}\ell_{m}^{\tau_{m}}+x^{2}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\ell_{m}^{\nu_{m}}\ell_{m}^{\tau_{m}}+x\left(1-x\right)\ell_{m}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\ell_{m}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}}{\left[\ell_{m}^{2}+x\left(1-x\right)\ell_{m-1}^{2}\right]^{2}}$
(69)
$\displaystyle+\frac{x\left(1-x\right)\ell_{m-1}^{\nu_{m+1}}\ell_{m}^{\tau_{m+1}}\ell_{m-1}^{\nu_{m}}\ell_{m}^{\tau_{m}}+\left(1-x\right)^{2}\ell_{m}^{\nu_{m+1}}\ell_{m}^{\tau_{m+1}}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}}{\left[\ell^{2}_{m}+x\left(1-x\right)\ell_{m-1}^{2}\right]^{2}}$
$\displaystyle+\frac{x^{2}\left(1-x\right)^{2}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}}{\left[\ell^{2}_{m}+x\left(1-x\right)\ell_{m-1}^{2}\right]^{2}}\Bigg{)}\,.$

We now replace products of $\ell_{m}^{\mu}$ with powers of $\ell_{m}$ and
symmetrized metric factors, following the generalized symmetrization rule
(258), which yields

$\displaystyle L_{m,2}$ $\displaystyle=\int
d^{d}\ell_{m}dx\Bigg{(}\frac{\frac{\ell_{m}^{4}}{d\left(d+2\right)}g^{(\nu_{m+1}\tau_{m+1}}g^{\nu_{m}\tau_{m})}+x^{2}\frac{\ell_{m}^{2}}{d}g^{\mu_{m}\nu_{m}}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}}{\left[\ell_{m}^{2}+x\left(1-x\right)\ell_{m-1}^{2}\right]^{2}}$
$\displaystyle+\frac{x\left(1-x\right)\frac{\ell_{m}^{2}}{d}g^{\nu_{m+1}\nu_{m}}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\tau_{m}}}{\left[\ell_{m}^{2}+x\left(1-x\right)\ell_{m-1}^{2}\right]^{2}}$
$\displaystyle+\frac{x\left(1-x\right)\frac{\ell_{m}^{2}}{d}g^{\tau_{m+1}\nu_{m}}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m}}+\left(1-x\right)^{2}\frac{\ell_{m}^{2}}{d}g^{{\nu_{m+1}}{\tau_{m+1}}}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}}{\left[\ell^{2}_{m}+x\left(1-x\right)\ell_{m-1}^{2}\right]^{2}}$
$\displaystyle+\frac{x^{2}\left(1-x\right)^{2}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}}{\left[\ell^{2}_{m}+x\left(1-x\right)\ell_{m-1}^{2}\right]^{2}}\Bigg{)}\,.$
(70)

Splitting the numerator up, we can once again apply the standard formula (61)
with

$\displaystyle\Delta^{2}=-x\left(1-x\right)\ell_{m-1}^{2}\,,$ (71)

which gives

$\displaystyle L_{m,2}$
$\displaystyle=\frac{i}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(\frac{d}{2}\right)}\int_{0}^{1}dx\Bigg{(}\frac{1}{d\left(d+2\right)}g^{(\nu_{m+1}\tau_{m+1}}g^{\nu_{m}\tau_{m})}\Gamma\left(2+\frac{d}{2}\right)\Gamma\left(-\frac{d}{2}\right)\Delta^{d}$
$\displaystyle+\frac{x^{2}}{d}g^{\mu_{m}\nu_{m}}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)\Delta^{d-2}$
$\displaystyle+\frac{x\left(1-x\right)}{d}g^{\nu_{m+1}\nu_{m}}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\tau_{m}}\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)\Delta^{d-2}$
$\displaystyle+\frac{x\left(1-x\right)}{d}g^{\tau_{m+1}\nu_{m}}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m}}\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)\Delta^{d-2}$
$\displaystyle+\frac{\left(1-x\right)^{2}}{d}g^{\nu_{m+1}\tau_{m+1}}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)\Delta^{d-2}$
$\displaystyle+x^{2}\left(1-x\right)^{2}\ell_{n-2}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}\Gamma\left(\frac{d}{2}\right)\Gamma\left(2-\frac{d}{2}\right)\Delta^{d-4}\Bigg{)}\,.$
(72)

Replacing $\Delta$ using its definition in equation (71), we see that each
term contains $d$ overall factors of loop momenta:

$\displaystyle L_{m,2}$
$\displaystyle=\frac{i}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(\frac{d}{2}\right)}\int_{0}^{1}dx\Bigg{(}\frac{\Gamma\left(2+\frac{d}{2}\right)\Gamma\left(-\frac{d}{2}\right)}{d\left(d+2\right)}g^{(\nu_{m+1}\tau_{m+1}}g^{\nu_{m}\tau_{m})}\left(x\left(1-x\right)\right)^{\frac{d}{2}}\ell_{m-1}^{d}$
$\displaystyle+\frac{\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)}{d}g^{\mu_{m}\nu_{m}}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}x^{\frac{d}{2}+1}\left(1-x\right)^{\frac{d}{2}-1}\ell_{m-1}^{d-2}$
$\displaystyle+\frac{\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)}{d}g^{\nu_{m+1}\nu_{m}}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\tau_{m}}\left(x\left(1-x\right)\right)^{\frac{d}{2}}\ell_{m-1}^{d-2}$
$\displaystyle+\frac{\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)}{d}g^{\tau_{m+1}\nu_{m}}\left(x\left(1-x\right)\right)^{\frac{d}{2}}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m}}\ell_{m-1}^{d-2}$
$\displaystyle+\frac{\Gamma\left(1+\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right)}{d}g^{\nu_{m+1}\tau_{m+1}}x^{\frac{d}{2}-1}\left(1-x\right)^{\frac{d}{2}+1}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}\ell_{m-1}^{d-2}$
$\displaystyle+\Gamma\left(\frac{d}{2}\right)\Gamma\left(2-\frac{d}{2}\right)\left(x\left(1-x\right)\right)^{\frac{d}{2}}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}\ell_{m-1}^{d-4}\Bigg{)}\,.$
(73)

We now use the identity $\Gamma\left(1+x\right)=x\Gamma\left(x\right)$ to
factor and cancel the gamma functions, and finally evaluate the Feynman
integrals, giving the result

$\displaystyle L_{m,2}$
$\displaystyle=\frac{i\Gamma\left(-\frac{d}{2}\right)}{\left(4\pi\right)^{\frac{d}{2}}\Gamma\left(d+2\right)}\Bigg{(}\frac{\left(1+\frac{d}{2}\right)\frac{d}{2}}{d\left(d+2\right)}g^{(\nu_{m+1}\tau_{m+1}}g^{\nu_{m}\tau_{m})}\Gamma\left(\frac{d}{2}+1\right)^{2}\ell_{m-1}^{d}$
$\displaystyle\qquad\qquad\qquad+\frac{\left(\frac{d}{2}\right)\left(-\frac{d}{2}\right)}{d}g^{\mu_{m}\nu_{m}}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\Gamma\left(\frac{d}{2}+2\right)\Gamma\left(\frac{d}{2}\right)\ell_{m-1}^{d-2}$
$\displaystyle\qquad\qquad\qquad+\frac{\frac{d}{2}\left(-\frac{d}{2}\right)}{d}g^{\nu_{m+1}\nu_{m}}\Gamma\left(\frac{d}{2}+1\right)^{2}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\tau_{m}}\ell_{m-1}^{d-2}$
$\displaystyle\qquad\qquad\qquad+\frac{\frac{d}{2}\left(-\frac{d}{2}\right)}{d}g^{\tau_{m+1}\nu_{m}}\Gamma\left(\frac{d}{2}+1\right)^{2}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m}}\ell_{m-1}^{d-2}$
$\displaystyle\qquad\qquad\qquad+\frac{\frac{d}{2}\left(-\frac{d}{2}\right)}{d}g^{\nu_{m+1}\tau_{m+1}}\Gamma\left(\frac{d}{2}+2\right)\Gamma\left(\frac{d}{2}\right)\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}\ell_{m-1}^{d-2}$
$\displaystyle\qquad\qquad\qquad+\left(1-\frac{d}{2}\right)\left(-\frac{d}{2}\right)\Gamma\left(\frac{d}{2}+1\right)^{2}\ell_{m-1}^{\nu_{m+1}}\ell_{m-1}^{\tau_{m+1}}\ell_{m-1}^{\nu_{m}}\ell_{m-1}^{\tau_{m}}\ell_{m-1}^{d-4}\Bigg{)}\,.$
(74)

We see that all terms scale as $\ell_{m-1}^{d}$ and the only divergent gamma
function is $\Gamma\left(-\frac{d}{2}\right)$. This establishes the result
used in the body of section 5, in the text above equation (276), which can
then be applied iteratively to evaluate the remaining loop integrals.

## References

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