Kaplunovsky Paper: Deriving eqn (5) from eqn (4) (One-loop gauge coupling correction)

Question

I am reading through a paper by Kaplunovsky, `One Loop Threshold Effects...', the link is http://inspirehep.net/record/253723?ln=en. But right at the beginning I can't reproduce going from eqn (4) to (5).
\begin{equation}\frac12 \text{Str} ( \log L) = \int_0^\infty \frac{dt}{2t} C_{\Lambda}(t) \text{Str}(e^{-tL}) \hspace{4cm} (4)\end{equation}
where $L$ is first quantized Lagrangian, so $L_\phi = M^2 - D^2$ for scalars, $L_{\psi} = M^2 - D^2 - \frac12 \gamma_\mu \gamma_\nu F^{\mu \nu}$ for spinors, etc, and supertrace is over all fields. $D_\mu = \partial_\mu - iQ^a A_\mu^a$ where $Q^a$ generate gauge group $G$.  $C_{\Lambda}(t)$ is just a UV regulator (for $t \rightarrow 0$). We work in a background gauge: $F_{\mu \nu} = \text{const}, A_\mu = -\frac12 F_{\mu \nu} x^\nu$. Equation 5 is
\begin{equation}\left(\substack{\text{coefficient of } \frac14 F_{\mu \nu}F^{\mu \nu}\\ \text{in the above}} \right) = \frac{1}{16 \pi^2} \int_0^\infty \frac{dt}{t} C_{\Lambda}(t) \cdot 2 \text{str} \left( Q_a^2 (\frac{1}{12}-\chi^2)\right)e^{-tM^2} \hspace{1cm} (5)\end{equation}
where the str is now over all physical states (eg no ghosts). I completely understand eqn (4). Kaplunovsky says to expand the exponential to second order in $F^2$ and compute the $x$ integral (implicit in str).
The result, eqn (5), is fairly standard I think, but have not seen this method before. Can someone provide a reference, or add in a few steps to make it clearer?

Comments

did you ask Vadim himself?