\nonumber
&&\left.\quad\quad\quad\quad\quad-\frac{t(t^2+2s^2+2u^2)}{8}I_3(t)-\frac{u(u^2+2t^2+2s^2)}{8}I_3(u)+\frac{(163u^2+163t^2+43tu)}{960}I_2(s)  \right.\\
\label{AAAAScattering}
&&\left.\quad\quad\quad\quad\quad +\frac{(163u^2+163s^2+43us)}{960}I_2(t)+\frac{(163s^2+163t^2+43ts)}{960}I_2(u) \right]\,,\end{aligned}$$ where the $I_4(s,t)$, $I_3(s)$, and $I_2(s)$ are respectively the scalar box, triangle and bubble integrals: $$\begin{aligned}
\nonumber
I_4(s,t)&=&\frac{1}{st}\left\{\frac{2}{\epsilon^2}\left[(-s)^{-\epsilon}+(-t)^{-\epsilon} \right]-\ln^2\left(\frac{-s}{-t}\right)-\pi^2  \right\}\\
\nonumber
&&=\frac{1}{st}\left\{\frac{4}{\epsilon^2}-\frac{2\ln(-s)+2\ln(-t)}{\epsilon}+2\ln(-s)\ln(-t)+\mbox{finite} \right\}\,,\\
\nonumber
I_3(s)&=&\frac{1}{\epsilon^2}\left(-s\right)^{-1-\epsilon}=-\frac{1}{s}\left(\frac{1}{\epsilon^2}-\frac{\ln(-s)}{\epsilon}+\frac{\ln^2(-s)}{2} \right)\,,\\
I_2(s)&=&\frac{1}{\epsilon(1-2\epsilon)}\left(-s\right)^{-\epsilon}=\left(\frac{1}{\epsilon}-\ln(-s)+\mbox{finite}\right)\,.\end{aligned}$$

We follow Ref. [@Dunbar:1995ed] in removing the infrared divergences by use of $$\label{removing IR}
{\cal M}_{IR} =\frac{\kappa^2}{2(4\pi)^2}\frac{\left((-s)^{1-\epsilon}+(-t)^{1-\epsilon}+(-u)^{1-\epsilon}\right)}{\epsilon^2}{\cal M}_{tree}\,,$$ where the residual hard part is defined via $${\cal M}_h = {\cal M}_{1-loop} - {\cal M}_{IR}\,.$$ With the renormalization of the higher order operator, the one loop hard amplitude is $$\begin{aligned}
\nonumber
{\cal M}_{h}&=&i\frac{\kappa^4}{\left(4\pi\right)^2}\left\{\frac{(s^4+t^4)}{8st}\ln(-s)\ln(-t)+\frac{(s^4+u^4)}{8su}\ln(-s)\ln(-u)+\frac{(u^4+t^4)}{8tu}\ln(-t)\ln(-u) \right.\\
\nonumber
&&\left.\quad\quad\quad\quad+\frac{(s^2+2t^2+2u^2)}{16}\ln^2(-s) +\frac{(t^2+2s^2+2u^2)}{16}\ln^2(-t)+\frac{(u^2+2t^2+2s^2)}{16}\ln^2(-u)  \right.\\
\nonumber
&&\left.\quad\quad\quad\quad+\frac{1}{16}\left(\frac{st}{u}+\frac{tu}{s}+\frac{us}{t}  \right)\left(s\ln^2(-s)+t\ln^2(-t)+u\ln^2(-u)  \right)\right.\\
\nonumber
&&\left.\quad\quad\quad\quad+\left[ -\frac{(163u^2+163t^2+43tu)}{960}\ln\left(\frac{-s}{\mu^2}\right)-\frac{(163u^2+163s^2+43us)}{960}\ln\left(\frac{-t}{\mu^2}\right)\right.\right.\\
\label{AAAAhardScattering}
&&\left.\left.\quad\quad\quad\quad-\frac{(163s^2+163t^2+43ts)}{960}\ln\left(\frac{-u}{\mu^2}\right)+d_1^{ren}(\mu)(s^2+t^2+u^2) \right]\right\}\,,\end{aligned}$$ where $\mu$ is an infrared scale. In this result, we have grouped the single logs with the higher order operator, because those logs are the ones that pick up the scale dependence when you shift the scale associated with the higher order operator $ d_1^{ren}(\mu)$.

We again evaluate the matrix element at the central kinematic point $s=2E^2,~t=u=-E^2$. The result is $${\cal M}_{total}={\cal M}_{tree}+{\cal M}_h=  i\frac{9\kappa^2 E^2}{8}\left[1-\frac{\kappa^2E^2}{360\left(4\pi\right)^2} \left(609\ln\frac{E^2}{\mu^2}+\left(340\pi^2+\left(123-340\ln 2\right)\ln2  \right)  \right)   \right]\,.$$ If we were to use this to identify a running coupling the result would be $$G(E) = G\left[1-\frac{\kappa^2E^2}{360\left(4\pi\right)^2} \left(609\ln\frac{E^2}{\mu^2}+\left(340\pi^2+\left(123-340\ln 2\right)\ln2  \right)  \right)   \right]\,.
\label{running2}$$ The single log term 