 Solution

The gamma reflection formula, $\Gamma (a)\Gamma (1-a)=\pi /\sin (\pi a)$, at $a=1/3$ gives $\Gamma (1/3)\Gamma (2/3)=2\pi /\sqrt{3}$. Consequently, with $\omega =e^{2\pi i/3}$ and $C=\Gamma (1/3)/3$,Multiplication by $[1+\mathrm{erf}(\mu /\sqrt{2})]/2$ therefore changes the algebraic coefficient by exactly the required amount, from $C$ to $C\omega$. This confirms both the magnitude and phase of the smoothed jump, rather than only its error-function shape.