= Solution
Differentiate the <Green-function representation>, valid away from the charged sheet:
$$
E_y=\frac1{2\pi\epsilon_0}\int_0^L\frac{y\,ds}{(x-s)^2+y^2}
=\frac{\arctan(x/y)-\arctan((x-L)/y)}{2\pi\epsilon_0}.
$$
For $0<x<L$, the two <inverse tangent> limits at $y\downarrow0$ are $\pi/2$ and $-\pi/2$; approaching from below reverses both signs. Thus
$$
\boxed{E_y(x,0^+)=\frac1{2\epsilon_0},\quad E_y(x,0^-)=-\frac1{2\epsilon_0},\quad |\Delta E_y|=\frac1{\epsilon_0}.}
$$
This agrees with the normal <electric field> jump obtained by applying <Gauss's law> to a thin pillbox enclosing unit <surface charge density>.
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