Find $I=\int_{x\in R}\int_{y\in R}x\frac{(x^2+y^2)^{\frac{k}{2}-1}e^{-\frac{1}{2}(x^2+y^2)}}{\Gamma(\frac{k}{2})2^{\frac{k}{2}}}dy\Phi(\alpha x)dx$

Just a long comment. First of all, as I said, you can solve the middle round bracket via the standard integral I wrote you in the comments section.

This being said, notice that $$\Phi(ax) = \int_{-\infty}^{ax} \frac{1}{\sqrt{2\pi}} e^{-z^2/2}\ \text{d}z = \frac{1}{2}\left(1 + \text{erf}\left(\frac{ax}{\sqrt{2}}\right)\right)$$

This is well know as it is a representation of the Error Function.

This leads you to have, in the right round bracket term:

$$\frac{1}{2}\int_{\mathbb{R}} x^{2m+1}e^{-x^2/2}\text{d}x + \frac{1}{2}\int_{\mathbb{R}} x^{2m+1}e^{-x^2/2}\text{erf}\left(\frac{ax}{\sqrt{2}}\right)\text{d}x$$

The first term is again standard, it follows indeed:

$$\frac{1}{2}\int_{\mathbb{R}} x^{2m+1}e^{-x^2/2}\text{d}x = -2^m \left((-1)^{2 m}-1\right) \Gamma (m+1) ~~~~~~~~~~~~~~~~~ \Re(m) > -1$$

The last term is rather more complicated to obtain, but a good knowledge of special function will help. It can be expressed in terms of hypergeometric function (and Gamma too):

$$\frac{1}{2}\int_{\mathbb{R}} x^{2m+1}e^{-x^2/2}\text{erf}\left(\frac{ax}{\sqrt{2}}\right)\text{d}x = \frac{a 2^{m+1} \left((-1)^{2 m}+1\right) \Gamma \left(m+\frac{3}{2}\right) \, _2F_1\left(\frac{1}{2},m+\frac{3}{2};\frac{3}{2};-a^2\right)}{2\sqrt{\pi }}$$

What you have to do then is to put this all together, multiply by the result I wrote in the comment and obtain a Series result which won't be fun to evaluate (if possible).

At least you can manage it to get the first terms, explore the behaviour a bit. Have fun!