as modification of the coupling $G(q^2)$. However, we will see that there still remains the crossing problem because $q^2$ can carry either sign.

The graviton propagator at lowest order is given by $$iD_F^{\alpha\beta\gamma\delta} = \frac {i{\cal
P}^{\alpha\beta\gamma\delta}}{q^2+i\epsilon}\,,$$ where $${\cal P}^{\alpha\beta\gamma\delta} =
\frac12\left[\eta^{\alpha\gamma}\eta^{\beta\delta} +
\eta^{\beta\gamma}\eta^{\alpha\delta}
-\eta^{\alpha\beta}\eta^{\gamma\delta}\right]\,.$$ The inclusion of the vacuum polarization diagram modifies the propagator $${i{\cal
P}^{\alpha\beta\lambda\xi}\over
q^2}i\Pi_{\lambda\xi\mu\nu}(q){i{\cal P}^{\mu\nu\gamma\delta}\over
q^2}~.$$

The vacuum polarization diagram is divergent and the required counter terms have the form obtained by ‘t Hooft and Veltman [@'tHooft:1974bx] $$\begin{aligned}
\Delta {\cal L}=\frac{\sqrt{g}}{16\pi^2\epsilon}\left[\frac{1}{120}R^2+\frac{7}{20}R_{\alpha\beta}R^{\alpha\beta} \right]\end{aligned}$$ with $\epsilon=(d-4)/2$. Because such terms appear in the most general effective Lagrangian, one can absorb these divergences into the renormalized values of their coefficients. Expressing $R^2$ and $R_{\alpha\beta}R^{\alpha\beta}$ in terms of the Fourier-space momenta, we find upon symmetrizing the indices $$\begin{aligned}
\sqrt{g}R^2=h^{\mu\nu}\left[ q^4\eta_{\alpha\beta}\eta_{\mu\nu}-q^2\left(\eta_{\mu\nu} q_\alpha q_\beta+ \eta_{\alpha\beta} q_\mu q_\nu \right) +q_{\alpha}q_{\beta}q_{\mu}q_{\nu} \right]h^{\alpha\beta}\,,\end{aligned}$$ and $$\begin{aligned}
\nonumber
\sqrt{g}R_{\alpha\beta}R^{\alpha\beta}&=&\frac{1}{4}h^{\mu\nu}\left[q^4\eta_{\alpha\beta}\eta_{\mu\nu}-\frac{q^2}{2}\left(q_{\alpha}q_\mu\eta_{\nu\beta}+q_{\alpha}q_{\nu}\eta_{\mu\beta}+q_{\beta}q_{\mu}\eta_{\nu\alpha}+q_{\beta}q_{\nu}\eta_{\mu\alpha}\right) \right.\\
&&\left.\quad\quad\quad -q^2(\eta_{\alpha\beta}q_\mu q_\nu+\eta_{\mu\nu}q_\alpha q_\beta)+q^4I_{\alpha\beta,\mu\nu}+2q_\alpha q_\beta q_\mu q_\nu\right]h^{\alpha\beta}\,,\end{aligned}$$ where $I_{\alpha\beta,\mu\nu}=(\eta_{\alpha\mu}\eta_{\beta\nu}+\eta_{\alpha\nu}\eta_{\beta\mu})/2$. In addition, one can use the presence of the $1/\epsilon$ terms to read out the dependence on $\ln q^2$ $$\begin{aligned}
\nonumber
\Pi_{\alpha\beta,\mu\nu}(q)&=&-\frac{2G}{\pi}\ln\left(-\frac{q^2}{\mu_1^2} \right) \left[\frac{q^4}{60}\eta_{\alpha\beta}\eta_{\mu\nu}-\frac{q^2}{60}\left(\eta_{\mu\nu} q_\alpha q_\beta+ \eta_{\alpha\beta} q_\mu q_\nu \right) +\frac{1}{60}q_{\alpha}q_{\beta}q_{\mu}q_{\nu} \right]\\
\nonumber
&&-\frac{2G}{\pi}\ln\left(-\frac{q^2}{\mu_2^2}\right)\left[\frac{7}{40}q^4\eta_{\alpha\beta}\eta_{\mu\nu}-\frac{7}{80}q^2\left(q_{\alpha}q_\mu\eta_{\nu\beta}+q_{\alpha}q_{\nu}\eta_{\mu\beta}+q_{\beta}q_{\mu}\eta_{\nu\alpha}+q_{\beta}q_{\nu}\eta_{\mu\alpha}\right) \right.\\
&&\left.\quad\quad\quad\quad\quad\quad\quad\quad -\frac{7}{40}q^2(\eta_{\alpha\beta}q_\mu q_\nu+\eta_{\mu\nu}q_\alpha q_\beta)+\frac{7}{40}q^4I_{\alpha\beta,\mu\nu}+\frac{7}{20}q_\alpha q_\beta q_\mu q_\nu\right]\,,\end{aligned}$$ where we have assigned $\ln(-q^2/\mu_1^2)$ and $\ln(-q^2/\mu_2^2)$ 