I was stuck for this problem whole night and I tried numerical solution using MATLAB and the following result seems hold
for x follow normal N(0,1) and for any positive number (not integer only) r>1, we have
$$\int_{-\infty}^{\infty} \! \frac{|x|^r}{x} \, \mathrm{d}\Phi({x}) =0$$ where $\Phi({x})$ is the CDF for standardize normal.
I tried various r = 1.5, 2.5 numerically, and they are all equal to zero.
Of course when r is positive even integer, this result is trivial. But I am stuck with the general r>0 case, especially for r is NOT integer.
Any help would be extremely appreciated!
Let $I = \int_{-\infty}^{\infty} \frac{|x|^r}{x} \phi(x) dx; \quad r>0$. Since the integrand is an even function through out the domain, we can write it as, $$ \begin{eqnarray} I &=& 2\int_{0}^{\infty} \frac{x^r}{x} \phi(x) dx\\ &=& 2\int_{0}^{\infty} x^{r-1} \phi(x) dx\\ &=& \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} x^{r-1} e^{-\frac{1}{2}x^2} dx\\ \end{eqnarray} $$ let $u = x^2 / 2$, so that $du = xdx$. Now, $$ \begin{eqnarray} I &=& \sqrt{\frac{2}{\pi}} \int_0^{\infty} (2u)^{\frac{r-2}{2}} e^{-u} du\\ &=& \sqrt{\frac{2}{\pi}} 2^{\frac{r-2}{2}} \int_0^{\infty} u^{\frac{r}{2} - 1} e^{-u} du\\ &=& \sqrt{\frac{2}{\pi}} 2^{\frac{r-2}{2}} \Gamma\left(\frac{r}{2}\right) \end{eqnarray} $$