As $sigma>0$ we can divide the expression by $sigma$ without changing its sign. Then, defining $x=(p-v_0)/sigma$, the expression becomes

```
Sign(Sqrt(2) + E^(x^2/2)*Sqrt(π)*x*Erfc(-x/Sqrt(2)))
```

This expression seems to be positive for any $xinmathbb{R}$, so I’d say that the answer to your question is that your `Sign(...)`

is always 1:

```
LogPlot(Sqrt(2) + E^(x^2/2)*Sqrt(π)*x*Erfc(-x/Sqrt(2)), {x, -1000, 10},
WorkingPrecision -> 100, PlotRange -> All)
```

The asymptotes of your expression are

- $sqrt{2}/x^2$ for $xto-infty$, which is positive,
- $2xsqrt{pi} e^{frac{x^2}{2}}$ for $xto+infty$, which is positive.