= Solution
The <gamma reflection formula>, $\Gamma(a)\Gamma(1-a)=\pi/\sin(\pi a)$, at $a=1/3$ gives $\Gamma(1/3)\Gamma(2/3)=2\pi/\sqrt3$. Consequently, with $\omega=e^{2\pi i/3}$ and $C=\Gamma(1/3)/3$,
$$
\boxed{\frac{2\pi i e^{i\pi/3}}{3\Gamma(2/3)}=C(\omega-1).}
$$
Multiplication by $[1+\operatorname{erf}(\mu/\sqrt2)]/2$ 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.
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