The correct answer here is no.
Suppose we let "x" go to infinity,
then x-Mu2 for any fixed Mu would go to infinity.
So if x with a -infinity would go to zero--
because it is a constant--
Therefore, we know that in the limit of x goes to infinity--
this expression must be zero.
However, in this graph--it stays at alpha, and doesn't go to zero--
So, therefore, there can't be a valid Mu at sigma square.
If a deep improbability, you know that the area in the Gaussian
has to integrate into one,
and this area diverges, it is actually infinite in size,
so it's not even a valid execution.