Expected value of a ratio of squared normal and linear combination of squared normal Given two positive constants $c_1,c_2$ and two independent standard normal random variables $a,b$, how to calculate the following expected value
$$
\mathbb{E}\left[\frac{a^2}{c_1a^2+c_2b^2}\right]
$$
Thank you.
 A: $\newcommand\E{\mathscr E}$The expectation in question is
$I/c_1$, where
$$I:=E\frac{a^2}{a^2+cb^2},\quad c:=c_2/c_1>0.$$
In turn, using polar coordinates, we get
$$I=\frac1{2\pi}\int_0^{2\pi}\frac{\cos^2 t\, dt}{\cos^2 t+c\sin^2 t},\quad c:=c_2/c_1>0.$$
Further, writing
$$I=\frac4{2\pi}J$$
for
$$J:=\int_0^{\pi/2}\frac{\cos^2 t\, dt}{\cos^2 t+c\sin^2 t}$$
and using the standard substitution $t=\arctan u$, so that $\cos^2 t=1/(1+u^2)$ and $\sin^2 t=u^2/(1+u^2)$, we have
$$J=\int_0^\infty\frac{du}{(1+cu^2)(1+u^2)}.$$
Using partial fraction decomposition to compute $J$ and collecting the pieces, we finally get that the expectation in question is
$$\E(c_1,c_2):=\frac1{c_1+\sqrt{c_1c_2}}.$$

"Sanity" checks: (i) $\E(tc_1,tc_2)=\E(c_1,c_2)/t$ for all real $t>0$; (ii) $\E(1,0)=1$; (iii) $\E(1,1)=1/2$; (iv) $\E(c_1,c_2)\to\infty$ as $c_1\downarrow0$.
A: In view of the identity
$$\frac{c_1a^2}{c_1a^2+c_2b^2}=1-\frac{c_2b^2}{c_1a^2+c_2b^2}$$
and the exchangeability of $a$ and $b$,
we may assume that $c_2\le c_1$.
The expectation in question is
$I/c_1$, where
$$I:=E\frac{a^2}{a^2+cb^2},\quad c:=\frac{c_2}{c_1}\in(0,1].$$
Using the identity $1/A=\int_0^\infty du\, e^{-Au}$ for real $A>0$ and the Fubini--Tonelli theorem, we have
$$I=\int_0^\infty du\, Ea^2 e^{-(a^2+cb^2)u}
=\int_0^\infty du\, Ea^2 e^{-u a^2}\,
Ee^{-cu b^2}.$$
Next, for real $u>0$,
$$Ee^{-u b^2}=(1+2u)^{-1/2}$$
and hence
$$Ea^2 e^{-u a^2}=-((1+2u)^{-1/2})'=(1+2u)^{-3/2}.$$
So,
$$I=\int_0^\infty du\, (1+2u)^{-3/2} (1+2cu)^{-1/2}.$$
Using now the substitution $u=(x^2-1)/2$, we get
$$I=\frac1{\sqrt c}\int_1^\infty \frac{dx}{x^2\,\sqrt{x^2+r^2}},$$
where $r:=\sqrt{\dfrac{1-c}c}$. So, $I$ is easily found if $c=1$. If $c\in(0,1)$, use the standard substitution $x=r\tan t$ to find $I$. Collecting all the pieces, we get that the expectation in question is
$$\frac1{c_1+\sqrt{c_1c_2}}.$$

Similarly and more generally, the calculation of
$$E\frac{Z_1^2}{c_1^2Z_1^2+\cdots+c_k^2Z_k^2}$$
for any natural $k\ge2$ reduces to the calculation of
$$I_k:=\int_1^\infty \frac{dx}{x^2\,\sqrt{x^2+R_2}\cdots\sqrt{x^2+R_k}}$$
for real $R_2,\dots,R_k>-1$.
For $k=3$, the integral $I_k$ can be expressed in terms of elliptic functions, as seen from the following image of a Mathematica notebook:

