(1) What would $\int_{-\infty }^{\infty} \frac{\sqrt{\left | x \right |} * sign(x)}{\sqrt{2\pi}\sigma}e^{-0.5*\left ( \frac{x-\mu}{\sigma} \right )^{2}}dx$ evaluate to?
This is expectation of $\sqrt{\left | x \right |} * sign(x)$ for a normal distribution with a non-zero mean
I found formulas for $E[x^p]$ and $E[|x|^p]$, but couldn't find with $sign(x)$. Tried to derive it by recursion, but wasn't able to solve it.
(2) Similarly, what would $\int_{-\infty }^{\infty} \frac{log(|x|)}{\sqrt{2\pi}\sigma}e^{-0.5*\left ( \frac{x-\mu}{\sigma} \right )^{2}}dx$ evaluate to? If this is hard, if someone can point me to the solution when $\mu = 0$, that would be greatly appreciated too.
Any pointers, references, sources or approximations would be greatly appreciated (not sure why someone downgraded this post!).
Thanks
* EDIT *
Further to @MichaelChirico's request, consider the first problem
$I = \int_{-\infty }^{\infty} \frac{\sqrt{\left | x \right |} * sign(x)}{\sqrt{2\pi}\sigma}e^{-0.5*\left ( \frac{x-\mu}{\sigma} \right )^{2}}dx$
can be written as
$I = \frac{1}{\sqrt{2\pi}\sigma} (J - K)$
where
$J = \int_{0}^{\infty} \sqrt{x} e^{-0.5*(\frac{x-\mu}{\sigma})^2} dx $
and
$K = \int_{0}^{\infty} \sqrt{x} e^{-0.5*(\frac{x+\mu}{\sigma})^2} dx $
Solving $J$ can give the solution of $K$ by replacing $\mu$ by $-\mu$.
Define:
$J_n = \int_{0}^{\infty} x^{\frac{n}{2}} e^{-0.5*(\frac{x-\mu}{\sigma})^2} dx $
Then integrating by parts,
$J_n = \int_{0}^{\infty} x^{\frac{n+2}{2}} . \frac{2}{n+2} e^{-0.5*(\frac{x-\mu}{\sigma})^2} * 0.5*2*\frac{(x-\mu)}{\sigma^2} dx $
$J_n = \frac{2}{(n+2)\sigma^2} \int_{0}^{\infty} x^{\frac{n+2}{2}} e^{-0.5*(\frac{x-\mu}{\sigma})^2} (x-\mu) dx $
$J_n = \frac{2}{(n+2)\sigma^2} (J_{n+4} - \mu * J_{n+2})$
I don't know how to proceed from here. Typically in a recursion, I would get $J_n$ as a function of only one $J_{n+\alpha}$ and then I can work backwards. I don't know how to work with recursions where $J_n$ is a function of two $J_{n+\alpha_1}$ and $J_{n+\alpha_2}$ variables.