perform an endpoint subtraction, $$\begin{aligned}
\Delta F_i & = & \frac{\alpha_s}{\pi}~\Gamma(1+\epsilon)
\frac{\Gamma(1-\epsilon)}{\Gamma(1-2\epsilon)}
\left(\frac{4\pi\mu_0^2}{m_t^2}\right)^{2\epsilon}  \left[ \Delta
F_{i,1} + \Delta F_{i,2} \right] \, , \nonumber \\
\Delta F_{i,1} & = & \int_0^1 \frac{d^6 x}{s} \left\{ \frac{H_i(\vec
x)}{N^3(\vec x)} (1+\epsilon L) - \left. \frac{H_i(\vec x)}{N^3(\vec x)}
\right|_{s=0} (1+\epsilon L_0) \right\} \, , \nonumber \\
\Delta F_{i,2} & = & -\frac{1}{\epsilon} \int_0^1 d^5 x \left.
\frac{H_i(\vec x)}{N^3(\vec x)} \right|_{s=0} \left[1+\epsilon L_1 +
\epsilon^2 \left( \frac{L_1^2}{2} + 3\zeta_2 \right) \right] \nonumber \\
\mbox{with} \qquad L & = & \log\frac{x(1-x)}{s} - 2 \log N(\vec x) \, ,
\nonumber \\
L_0 & = & \log\frac{x(1-x)}{s} - 2 \log N(\vec x)|_{s=0} \, , \nonumber \\
L_1 & = & \log [x(1-x)] - 2 \log N(\vec x)|_{s=0} \, ,\end{aligned}$$ where in the second term $\Delta F_{i,2}$ the integration over $s$ has been performed analytically and the integration measure is given by $d^5
x = dx\,dy\,dz\,dr\,dt$. It should be noted that in the terms $L,
L_0, L_1$ the logarithms of the denominator $N$ need to be linear in $N$ to be consistent with the analytical continuation along the proper Riemann sheet. We have checked numerically that the first (subtracted) part $\Delta F_{i,1}$ is finite for each order in the dimensional 