We know that the square of a Rayleigh random variable has exponential distribution, i.e.,
Let the random variable $X$ have Rayleigh distribution with PDF $$f_X(x)=\frac{2x}{\alpha}e^{-x^2/{\alpha}}.$$
Then the random variable $Y=X^2$ has the PDF given by $$f_Y(y)=\frac{1}{\alpha}e^{-y/{\alpha}}.$$
For an exponentially distributed r.v. $Y$ with mean $\mathbb{E}[Y]=1$
$$\mathbb{E}[Y^{\delta}]=\Gamma[1+\delta].$$
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Now, if the random variable $X$ has Rician distribution (unit power in direct and scattered paths), whose PDF is given by
$$f_X(x)=\frac{2x}{\alpha}\text{exp}\left(\frac{-(x^2+v^2)}{\alpha}\right)I_0\left(\frac{2xv}{\alpha}\right)$$ with $\frac{v^2}{\alpha}=1$ and $I_0(z)$ is the modified Bessel function of the first kind with order zero.
what is the PDF of $Y=X^2$?
And what is $\mathbb{E}[Y^{\delta}]$ when $\delta<1.$
Note: when $v^2=0$, $X$ has Rayleigh distribution.