Expectation of a chi-squared distribution I am to find out the value of this expectation : $$E \bigg(\frac{U^p}{U+V} \bigg),$$ where U $\sim$ $\chi^2_1$ and V $\sim$ $\chi^2_n$. U and V are independent.
Can anyone give me any hints about how to start this problem ?
 A: The PDFs are
$$f_U(u) = C(1)u^{-1/2} e^{-u/2}$$
and
$$f_V(v) = C(n)v^{n/2-1}e^{-v/2}$$
where
$$C(k) = \frac{1}{2^{k/2}\Gamma(\frac{k}{2})}$$
are the normalizing constants.  Use polar-like coordinates $u=(r\cos(\theta))^2$ and $v=(r\sin(\theta))^2$ to evaluate the expectation, after first computing
$$\eqalign{du\wedge dv &= (2 r \cos(\theta)^2 dr - 2r^2 \sin(\theta)\cos(\theta)d\theta)\wedge (2 r \sin(\theta)^2 dr + 2 r^2 \sin(\theta)\cos(\theta)d\theta) \\
&= 4r^3\sin(\theta)\cos(\theta) dr\wedge d\theta}$$
and
$$u+v = r^2(\cos(\theta)^2 + \sin(\theta)^2) = r^2,$$
so that (provided $n+2p \gt 1$) it splits into a Beta integral involving $\theta$ and a Gamma integral involving $r^2$ and a great deal of cancellation occurs:
$$\eqalign{\mathbb{E}\left(\frac{U^p}{U+V}\right) &= C(1)C(n)\int_0^\infty\int_0^\infty \frac{u^p}{u+v} u^{-1/2} v^{n/2-1} e^{-(u+v)/2}\, du\, dv,\\
&= 4C(1)C(n)\color{blue}{\int_0^{\pi/2}\sin(\theta)^{n-1}\cos(\theta)^{2p} d\theta}\color{red}{ \int_0^\infty r^{2p+n-2} e^{-r^2/2} dr} \\
&= 2^2 \frac{1}{2^{1/2}\Gamma(1/2)} \frac{1}{2^{n/2}\Gamma(n/2)} \color{blue}{\frac{\Gamma(n/2)\Gamma(p+1/2)}{2\Gamma(p+n/2-1/2)}}\; \color{red}{2^{p+n/2-3/2} \Gamma(p+n/2-1/2)} \\
&= \frac{2^p \Gamma(p+1/2)}{\sqrt{\pi}(n+2p-1)}.
}$$
Otherwise, if $n + 2p \le 1$, the integral diverges as $r\to 0$.
