in the center of the box at $x=0$, and at the edge at $x=\pm\pi$, where $\psi_{eq0}'=0$.

In this Section we study an elementary current corrugation, $\delta J_{eq}= A K^2 \cos(Kx+\alpha)$, generated by a “zonal” magnetic field [@Diamond2005] of given shape: $\delta
\psi_{eq}= A \cos(Kx+\alpha)$. Here $A$ is the (normalized) amplitude of the “zonal” field, $K$ its wavelength, and $\alpha$ its phase with respect to the resonant surface. In order to investigate its effect on the central resonance, we reduce Eq.\[7\] to a form that is appropriate to slab geometry: $$\label{8} \delta \Delta'=P\int_{-\infty}^{\infty} dx \left(
\frac{\delta
J_{eq}'}{\psi_{eq0}'}\frac{\widetilde{\psi}_0}{\widetilde{\psi}_{r_s}}\right).$$ Note that $x$ is normalized to a typical length scale and so is $\delta \Delta'$. Furthermore, slab geometry requires symmetric boundaries far from the reconnecting surface, so that the lower extreme in Eq.\[7\], $r=0$, is replaced by $x=-\infty$ in Eq.\[8\].

For the equilibrium and the corrugation studied here, it is straightforward to obtain: $$\label{8a} \delta \Delta'= AK^3\left\{\cos(\alpha)
\int^{\pi/2}_{-\pi/2}dx\left[\frac{\sin(Kx)}{\sin(x)}\frac{\widetilde{\psi}_0}{\widetilde{\psi}_{r_s}}\right]+\sin(\alpha)
P
\int^{\pi/2}_{-\pi/2}dx\left[\frac{\cos(Kx)}{\sin(x)}\frac{\widetilde{\psi}_0}{\widetilde{\psi}_{r_s}}\right]\right\}.$$ As a consequence of the parity of the integrand, the second term on the right hand side of the previous equation is equals to zero. At the same time, the first term on the right hand side does not contain any singularity at the 