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In the paper "Scattering into the Fifth Dimension of $\mathcal{N}=4$ Super Yang-Mills", the authors give the following result for an integral:

$$\begin{align} I^{(1)}(x_{13}^2,x_{24}^2,m) =& \left( x_{13}^2 +m_{13}^2 \right) \left( x_{24}^2 + m_{24}^2 \right)\cr & \times\int d^4 x_5 \frac{1}{(x_{15}^2+m^2) (x_{25}^2+m^2) (x_{35}^2+m^2) (x_{45}^2+m^2)} \notag \\ =& 2 \ln \left( \frac{m^2}{x_{13}^2} \right) \ln \left( \frac{m^2}{x_{24}^2} \right) - \pi^2 + \mathcal{O}(m^2) \tag{1} \end{align}$$

where $x_{ij} := x_i - x_j$, $m_{ij} := m_i - m_j$. I would like to take the limit $m \to 0$ of this integral and compare it to a computation done with dimensional regularization. The following relation between cutoff and dim reg is given in the paper "Cutoff Regularization Method in Gauge Theories" :

$$\ln \Lambda^2 = \frac{1}{\epsilon} - \gamma + \ln(4\pi^2 \mu^2) + 1 \tag{2}$$

Can I use this relation for translating the "mass cutoff" of eq.(1)?

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A few things:

$\bullet$ Your definition of the $I^{(1)}$ integral, and then the expression involving $m$ that follows, is missing information. The paper you quote says that the latter expression holds only in the case where all the masses are the same ($m_i = m$), and also only in the limit where $m$ is small compared to $\hat{x}_{13}^2$ and $\hat{x}^2_{24}$

$\bullet$ As I think you are aware, it is in this sense that the latter expression already has the limit $m \to 0$ taken. That is what the $\mathcal{O}(m^2)$ means.

$\bullet$ In general, I don't think there is a consistent mapping between different regularization schemes: ie. Although it is often true that a $1/\epsilon$ in dim reg corresponds to a $\log(\Lambda)$ in a cutoff scheme, I think this is only true for simple loop integrals and I don't think you can always trust this.

$\bullet$ In the second paper, your relation between cutoff $\Lambda$ and dim-reg $\epsilon$ seems to be for an ultraviolet cutoff $\Lambda$ (which is supposed to be large compared to the scales in your integral). The mass $m$ in your loop integral parametrizes an infrared divergence (in that $m$ is small compared to the other scales in your integral). This further undermines using your last expression. I would definitely be wary of using it.

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  • $\begingroup$ Mm okay, yeah that makes sense. I will try to think of another way to proceed. Thanks for the clarification! $\endgroup$
    – Pxx
    Nov 2, 2019 at 0:31

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