
The correct answer here is no.

Suppose we let "x" go to infinity,

then xMu2 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 graphit 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.