Expectation of rational formula I have two independent normal random variables $x$ & $y$ that are zero mean and unit variance. $a$ & $b$ are positive.
I need to find the mean of $$z=\frac{ax^2y^2}{1 + bx^2}.$$
Any help please?!
 A: I symbolically evaluated the double integral for the desired expectation in MAPLE as follows (note that I executed expand just so that I could get the image to display better).
expand(int(int(a*x^2*y^2/(1+b*x^2)*1/(2*Pi)*exp(-1/2*x^2)*exp(-1/2*y^2),x=-infinity..infinity),y=-infinity..infinity));

Here is the result:

I checked this result against stochastic simulation for several values of b, and it matches, so I have confidence it is correct.
Here are the numerical results for a = 1 and a few values of b:
b = 0.1 --> 0.79214855
b = 1   --> 0.34432046
b = 5   --> 0.11888698

Edit: I am bumping this because even if this was a homework problem, by now the OP has probably graduated or flunked out.
A: This has an elementary solution.  It employs a technique often found to be useful when integrating exponentials: a fraction can be expressed in terms of an integral of an exponential function.

Because $X$ and $Y$ are independent, the expectation splits into the product of $\mathbb{E}(Y^2)=1$ and 
$$\mathbb{E}\left(\frac{aX^2}{1+bX^2}\right) = \frac{a}{b}\left(1 - \frac{1}{b}\mathbb{E}\left(\frac{1}{1/b + X^2}\right)\right).\tag{1}$$
Write $1/b = 2s$ (so that $s=1/(2b)$ is positive, too) and note that for any $x$,
$$\frac{1}{2s + x^2} = \int_0^\infty \exp(-(2s + x^2)t)\mathrm{d}t.$$
Apply Fubini's theorem to compute the expectation before performing this integral:
$$\eqalign{
\mathbb{E}\left(\frac{1}{1/b+X^2}\right) &= \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}\exp(-x^2/2)\int_0^\infty \exp(-(2s + x^2)t)\mathrm{d}t \mathrm{d}x \\
&= \frac{1}{\sqrt{2\pi}}\int_0^\infty\int_{-\infty}^\infty \exp\left(-\frac{1}{2}\left(x^2+2t(2s + x^2)\right)\right) \mathrm{d}x\mathrm{d}t \\
&= \int_0^\infty\frac{e^{-2 s t}}{\sqrt{1+2t}} \left[\frac{\sqrt{1+2t}}{\sqrt{2\pi}}\int_{-\infty}^\infty \exp\left(-\frac{1}{2}\left((1+2t)x^2)\right)\right) \mathrm{d}x\right]\mathrm{d}t \\
&=\int_0^\infty\frac{e^{-2 s t}}{\sqrt{1+2t}} \mathrm{d}t.\tag{2}
}$$
The last equality follows by observing that the integral in brackets is the total probability of a Normal variable (with mean $0$ and variance $1/(1+2t)$), which is just $1$.
A substitution with $1+2t$ as the variable easily expresses this in terms of a $\chi^2(1)$ tail probability (computed as an incomplete Gamma function).  Alternatively--to bring the ideas full circle and get back to Normal distribution probabilities--let's apply an aggressive substitution to clear the denominator in the integrand: letting $x^2 =2s(1+2t)$, deduce $dt = x dx/(2s)$, whence (writing $\Phi$ for the standard normal CDF)
$$\int_0^\infty\frac{e^{-2 s t}}{\sqrt{1+2t}} \mathrm{d}t = \frac{e^{s}\sqrt{2\pi}}{\sqrt{2s}}\frac{1}{\sqrt{2\pi}}\int_\sqrt{2s}^\infty\exp(-x^2/2)\mathrm{d}x = \frac{e^{s}\sqrt{2\pi}}{\sqrt{2s}}\left(1 - \Phi(\sqrt{2s})\right).$$
Plugging this into $(2)$ and then into $(1)$ and re-expressing $s$ as $1/(2b)$ yields
$$\mathbb{E}\left(\frac{aX^2Y^2}{1+bX^2}\right) = \frac{a}{b}\left(1 - e^{1/(2b)}\sqrt{\frac{2\pi}{b}}\left(1 - \Phi\left(\frac{1}{\sqrt{b}}\right)\right)\right).$$
A: No real answer, but it makes things more simple:
$$\mathbb{E}Z=\mathbb{E}\frac{aX^{2}Y^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}Y^{2}\mathbb{E}\frac{bX^{2}}{1+bX^{2}}=\frac{a}{b}\mathbb{E}\left[1-\frac{1}{1+bX^{2}}\right]=\frac{a}{b}-\frac{a}{b}\mathbb{E}\frac{1}{1+bX^{2}}$$
So actually to be found is $$\mathbb{E}\frac{1}{1+bX^{2}}$$
where $X$ has standard normal distribution.
