equivalent and the advantage of using one or the other only depends on which quantity one is interested in. Since the expansions are likely to be asymptotic some resummation scheme, such as Padè discussed in Section \[SecPad\], are needed. Finally in Section \[SecNum\] we present a new integration procedure and compare the analytical results with those obtained from a direct numerical solution of the $\infty$-replica symmetry breaking equations.

Expansion of the free energy functional {#SecExp}
=======================================

The Parisi’s free energy $f$ for a the SK model in an external field $h$ at temperature $T$ is [@P80]:

$$-f = \frac{\beta}{4}\,\Bigl( 
                      1 - 2\,q(1) + \int_0^1dx\, q^2(x)
                         \Bigr) 
+ \int_{-\infty}^{+\infty} \frac{d y}{\sqrt{2 \pi q(0)}}
\exp\left(-\frac{(y-h)^2}{2\,q(0)}\right)\phi(0,y)
\label{eqfree}$$

where $\phi(0,y)$ is the solution evaluated at $x=0$ of the the Parisi’s equation $$\dot\phi(x,y)=-\frac{\dot{q}(x)}{2}\,
    \Bigl[
               \phi''(x,y)+\beta\,x\,\phi'(x,y)^2 
        \Bigr]
\label{eqPhi}$$ with the boundary condition $$\phi(1,y)= \beta^{-1}\log\left(2\cosh \beta y\right)
\label{Phi1}$$ where we have used the standard notation and denote derivatives with respect to $x$ by a dot and derivatives with respect to $y$ by a prime. The order parameter $q(x)$ at temperature $T$ is obtained by the stationarity condition of (\[eqfree\]) with respect to variations of $q(x)$, while the value of (\[eqfree\]) at the stationarity point gives the free 