When taking the integral over a gaussian distribution, coincidentally the distribution describing a single particle at minimum uncertainty, you get an answer in terms of pi.
Importantly, the 'minimum uncertainty' is not just about the width of the wavefunction in position space---the uncertainty principle (roughly speaking) says that the variance in position space times the variance in momentum space has a lower bound. A gaussian saturates that bound.
I'm guessing that is what is meant by minimum uncertainty.
https://physicspages.com/pdf/Griffiths%20QM/Griffiths%20Prob...