I'm trying to calculate the average of one half of the normal distribution curve.
$$ \lim_{a \to -\infty} \int_a^\mu \frac{1}{\sigma\sqrt{2\pi}(\mu-a)} \exp \left(-\left(\frac{z-\mu}{\sqrt{2}\sigma}\right)^2\right)\,dz $$
Using this as a reference, I realised that the integral is undefined in a domain that $\to \pm\infty$. For another quick reference I also checked on Wolfram. Now this makes sense to me, the average of 'infinity' numbers should be undefined, however I was wondering if there is a mathematical proof or research that I can use to backup this argument.
$$I=\int \frac{1}{\sigma\sqrt{2\pi}(\mu-a)} \exp \left(-\left(\frac{z-\mu}{\sqrt{2}\sigma}\right)^2\right)\,dz=\frac{\text{erf}\left(\frac{z-\mu }{\sqrt{2} \sigma }\right)}{2 (\mu -a)}$$ $$J=\int_a^\mu \frac{1}{\sigma\sqrt{2\pi}(\mu-a)} \exp \left(-\left(\frac{z-\mu}{\sqrt{2}\sigma}\right)^2\right)\,dz=\frac{\text{erf}\left(\frac{a-\mu }{\sqrt{2} \sigma }\right)}{2 (a-\mu )}$$ Let $$\frac{a-\mu }{\sqrt{2} \sigma}=x\implies J=\frac{\text{erf}(x)}{2 \sqrt{2} \sigma x}$$ When $x\to \pm\infty$, $\text{erf}(x)\to \pm 1$ and then $J\sim \pm\frac{1}{2 \sqrt{2} \sigma x}\to 0^\pm$