= & \Re S+O(\frac{1}{16\pi^{2}}).\end{aligned}$$ It is important to note that in the heterotic case the classical 4 dimensional string coupling is defined as $$\frac{1}{g_{string}^{2}}=\Re S=e^{-2\phi}{\cal V}$$ where ${\cal V}$ is the volume of the internal space in string units and $e^{\phi}$ is the 10 dimensional string coupling. Then we get from $$\begin{aligned}
\frac{1}{g_{a}^{2}(\Phi,\bar{\Phi};\mu^{2})} & = & k_{a}\Re S+\frac{b_{a}}{16\pi^{2}}(\ln\frac{\Lambda^{2}}{\mu^{2}}-\ln\Re S)\nonumber \\
 &  & +\frac{1}{16\pi^{2}}[\Re f_{a}^{(1)}(M)+c_{a}\hat{K}(M,\bar{M})-\sum_{r}2T_{a}(r)\ln\det Z_{(r)}^{(0)}(M,\bar{M})].\label{eq:g1loopHet}\end{aligned}$$ This is to be compared with the string 1 loop calculation [@Kaplunovsky:1995jw] [^3] where $$\frac{1}{g_{string}^{2}(\Phi,\bar{\Phi})}=\Re S+\frac{\Delta^{univ}(M,\bar{M})}{16\pi^{2}}\label{eq:gstring}$$ Now in the Heterotic string $M_{string}^{2}=M_{P}^{2}/\Re S$ up to $S,M$ independent constants. So it appears that the $S$ dependence of the two formulae agrees. It should be stressed here that although in [@Kaplunovsky:1995jw] the cut-off $\Lambda$ is identified with $M_{P}$ this is merely a matter of convenience, and as the authors observed all that is required is that in the field theory expression the cutoff $\Lambda$ should be chosen in an $S,M$ independent manner.

As is well known the natural superspace variable that corresponds to the axio-dilaton of string theory is the linear multiplet (since the four dimensional axion is originally a two-form field). Suppose for the moment we ignore the (non-universal) second line as well as the last 