Can someone explain to me why that is an $\frac{1}{2}$ multiplying the result of this integral (I think I can understand the rest. I did the calculations, but can't find this $1/2$):
$$ \int \frac{1}{\sqrt{2\pi\sigma}} \ \ e^{ikx \ – \ x^{2}/2\sigma} \ dx = -\frac{1}{2}\ \ i \ \ e^{-k^2\sigma/2} \operatorname{erfi} \bigg( \frac{k\sigma + ix}{\sqrt2\sqrt\sigma} \bigg) $$
where the $ \operatorname{erfi} $ is the imaginary error function defined by:
$$ \operatorname{erfi} (z) = -i \operatorname{erf} (iz)$$
You can go to wolfram alpha to see what I'm talking about (copy and paste the following):
integrate (1/((2*pi*o)^(1/2)))e^(ik*x – (x^2/(2*o))) dx
Note: I know that are other ways to obtain the Characteristic Function of the Gaussian Distribution, this is just a curiosity that emerged from doing $ \ g(k) = \langle e^{ikx} \rangle$.
Best Answer
There are two things going on. First, the $\exp(-x^2)$ versus the $\exp(-x^2/2)$ in the integrands of the physicists' $\operatorname{erf}$ function and the probabalists' $\Phi$ function induces some change-of-variables scale factor. Second, the fact that $\sup_{x,y}|\operatorname{erf}(x)-\operatorname{erf}(y)|=2$ but $\sup_{x,y}|\Phi(x)-\Phi(y)|=1$, which is to say, the $\operatorname{erf}$ function is not a cumulative distribution function. Keeping track of these differences is a matter of careful bookkeeping, the bottom line of which is your $1/2$.
In summary, the $1/2$ is an artifact of using the $\operatorname{erf}$ function, part of the "cost of doing business" that way. See the Wikipedia page on the normal distribution for further details. Probabalists work with the normal density $\varphi(x)=\exp(-x^2/2)/\sqrt{2\pi}$ and its indefinite integral $\Phi(x)=\int_{-\infty}^x\varphi(t)\,dt$. You can check that $\Phi(x)=(1+\operatorname{erf}(x/\sqrt 2))/2;$ that's the source of your $1/2$.