assume $u_0\gg 1$.

The integral over $u_\delta$ is performed analytically, and the large-$u_0$ limit may eventually be taken: \_0\^[u\_0]{}du\_(1+)=(1+u\_0)(1+u\_0)-u\_0u\_0 u\_0. The remainder reads \_1\^+=, \[eq:mu1+I0I1\] where \_0=\_[t\_0\^/t]{}\^1 d\^[-3/2]{}(1-)\^[-1/2]{},  \_1=\_[t\_0\^/t]{}\^1 d\^[-3/2]{}(1-)\^[-1/2]{} (\^[3/2]{}). \[eq:defI0\] ${\cal I}_0$ and ${\cal I}_1$ can be performed with the help of the change of variable $\lambda=\sin^2\theta$: \_0=2\_\^ =2 () =2, while for ${\cal I}_1$, a further integration by parts is needed to get rid of the log. We eventually arrive at the following exact expression for (\[eq:mu1+I0I1\]): \_1\^[+]{}=2CC\_1{ - }. \[eq:c1+\] The term $\mu_1^-$ is the same as the term $\mu_1^+$ except for the replacement $t\rightarrow t_0$.

Since we are neglecting terms of relative order $t_0/t$, $t_0^\prime/t_0$ and higher, we may expand the expressions for $\mu_1^+$ and $\mu_1^-$. The difference $\mu_1^+-\mu_1^-$ then reads \_1\^+-\_1\^-=4CC\_1(-). Equation (\[eq:c1\]) eventually leads to the following expression for $\mu_1$: \_1=(4C C\_1-C\_[|Y]{})(-). \[eq:mu1\]

We note a very important property of this result: It does not depend on $t_0^\prime$. If it did, then we would loose predictivity because $t_0^\prime$ is the arbitrary time after which we declare that the fluctuations are small enough for our calculation to apply. (This would not be true at the next order in $1/\sqrt{t}$, $1/\sqrt{t_0}$).

\[sec:det\]Deterministic calculations
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In 