Mean and variance of a non-standard pdf I have tried to compute the variance and the  mean for $\mu=0.5$ of the following PDF using Wolfram cloud but I failed 
$$ F(z,\mu,\sigma)=\frac{2 (z-\sigma )^2 \exp \left(-\frac{(z-\sigma )^2
   \sqrt{\left(1+0.25 \mu ^2\right) 2 \pi } \text{erf}\left(\frac{(z-\sigma )^2
   \sqrt{\left(1+0.25 \mu ^2\right) 2 \pi }}{1+0.25 \mu ^2}\right)}{1+0.25 \mu
   ^2}\right)}{\pi ^2 \sqrt{\left(1+0.25 \mu ^2\right) 2 \pi }} $$
Note: $\mu \in (0,1)$, $z , \sigma$ are reals and the integrand of that PDF is over $\sigma $
I have doubts this is a valid PDF formulation, However I have confirmed many times that is correct , the message I have got from Mathematica is " The integrand has evaluated to non numerical value for a sampling points in the region of boundaries region $(-\infty,0)$. Maybe the integrand does not converge under the conditions which I have assumed.
 A: Simplifications:


*

*replace 
$$\frac{(z-\sigma )^2
   \sqrt{\left(1+0.25 \mu ^2\right) 2 \pi }}{1+0.25 \mu ^2}$$
by
$$\frac{(z-\sigma )^2
   \sqrt{2 \pi }}{\sqrt{1+0.25 \mu ^2}}$$

*replace$$z-\sigma$$by $y$

*replace $\text{Erf}(x)$ with
$$\text{Erf}(x) = 2\Phi(x\sqrt{2}) - 1$$
This leads to consider instead
$$\frac{2 y^2}{\pi^2\sqrt{2\pi}} \exp\left\{-\sqrt{2\pi}y^2 \text{erf}\left(y^2
   \sqrt{2 \pi }\right)\right\}=\frac{2 y^2}{\pi^2\sqrt{2\pi}} \exp\left\{-\sqrt{2\pi}y^2 \left[2\Phi(2y^2\sqrt{\pi}) - 1\right]\right\}$$
Operate the change of variable $w=y^2\in(0,\infty)$, leading to
$$\frac{w^{1/2}}{\pi^2\sqrt{2\pi}} \exp\left\{-\sqrt{2\pi}w \left[2\Phi(2w\sqrt{\pi}) - 1\right]\right\}$$
which integrates to something of the order $10^{-2}$ over $(0,\infty)$ and is thus improperly normalised.
A: Not sure if this should be a comment or an answer but because this question was simultaneously posted on multiple forums, I placed an answer at Mathematica StackExchange.
In short the pdf is approximately
$$1.0105750026505362 \times \frac{\sqrt[4]{2 \pi }}{\sqrt[4]{1+ \mu^2/4}} \frac{(z-\sigma )^2 \sqrt{2 \pi }}{\sqrt{1+ \mu^2/4}}
\exp\left(-\frac{(z-\sigma )^2 \sqrt{2 \pi }}{\sqrt{1+ \mu^2/4}}
\text{erf}\left(\frac{(z-\sigma )^2 \sqrt{2 \pi }}{\sqrt{1+ \mu^2/4}}\right) \right)$$
The mean is $\sigma$ and because of the symmetry of the pdf all odd central moments are zero.  The variance only depends on $\mu$:
$$0.282617 \sqrt{\mu ^2+4}$$
(Not that it's wrong but I have no idea why the parameters are labeled $\sigma$ and $\mu$, respectively, rather than the other way around.)
