What is the expected value of $\frac{X}{X+Y}$? I am trying to find the expected value of 
$\displaystyle E\Bigg[\frac{X}{X+Y}\Bigg]$.
I started with writing 
$\displaystyle E\Bigg[\frac{X}{X+Y}\Bigg] = 
               E\Bigg[X\cdot\frac{1}{X+Y}\Bigg]
$.
I then noticed that 
$E[XY] = \text{cov}(X,Y) + E[X]E[Y]$
which follows from the definition of covariance. So, I have
$\displaystyle E\Bigg[X\cdot\frac{1}{X+Y}\Bigg] =
               \text{cov}\Big(X, \frac{1}{X+Y}\Big)+E[X]E\Bigg[\frac{1}{X+Y}\Bigg]$
but I don't know how to proceed from here.
The variables $X$ and $Y$ are both normally distributed and positively correlated.
 A: If $(X,Y)$ is binormal, then so is $(X,Z) = (X,X+Y)$.  The ratio $X/Z$ is the tangent of the slope of the line through the origin and the point $(Z,X)$.  When $X$ and $Z$ are uncorrelated with zero means, it is well known (and easy to compute) that $X/Z$ has a Cauchy distribution.  Cauchy distributions have no expectations.  This should lead us to suspect $X/Z$ might not have a mean, either.  Let's see whether it does nor not.
For any angle $0 \lt \theta \lt \pi/2$, consider the event
$$E_\theta = \{(Z,X)\,|\, X \ge Z\cot(\theta\}.$$
This is of interest because its probability is the chance that $X/Z$ exceeds $\cot(\theta)$: the survival function of $X/Z$.  It carries all the information of the distribution function of $X/Z$.
$E_\theta$ is a (closed) cone in the plane consisting of all points on all lines making an angle of $\theta$ or less to the right of the vertical ($X$) axis. Let's underestimate the probability of $E_\theta$.  To do so, we will work in polar coordinates.  Consider any possible radius $\rho$.  Among all points of this radius within the set $E_\theta$, the density $f$ of $(Z,X)$ will achieve a minimum value $f_\theta(\rho)$.  This minimum must be nonzero provided the density does not degenerate.  (More about this possibility later.)  Use this to bound the probability
$$\eqalign{
\Pr(E_\theta) &= \int_{\pi/2-\theta}^{\pi/2}\int_0^\infty f(\phi,\rho) \rho d\rho d\phi  \\
&\ge \int_{\pi/2-\theta}^{\pi/2}\int_0^\infty \rho f_\theta(\rho) d\rho d\phi \\
&=\theta \int_0^\infty \rho f_\theta(\rho) d\rho \\
&= C(\theta) \theta
}$$
where I have written $C(\theta)$ for the integral, which is some positive number depending on $\theta$.  Moreover, for $0\lt\theta\lt\pi/2$, $C(\theta)$ has a nonzero lower bound $C \gt 0$.
By definition, the expectation of $X/Z$ is the sum of two parts: one integral for the positive part when $X/Z \ge 0$ and another for the negative part when $X/Z \lt 0$.  Let's tackle the positive part.  For any positive random variable $W$ with distribution function $F$, integration by parts shows its expectation equals the integral of its survival function $1-F$, since
$$\mathbb{E}(W) = \int_0^\infty w dF(w) = (w(1-F(w))|_0^\infty + \int_0^\infty (1-F(w)) dw = \int_0^\infty (1-F(w)) dw.$$
Applying this to $W = X/Z$ and substituting $w=\cot(\phi)$ gives for the positive part of the integral
$$\eqalign{
\int_0^\infty (1 - F(w)) dw &= \int_0^{\pi/2} (1 - F(\cot(\phi))) \csc^2(\phi) d\phi \\
&= \int_0^{\pi/2} \Pr(E_\phi) \csc^2(\phi) d\phi \\
&\ge C \int_0^\theta \phi \csc^2(\phi) d\phi \\
&\gt C \int_0^\theta \frac{d\phi}{\phi}. 
}$$
(The final inequality is a simple consequence of the well-known inequalities $0 \lt \sin(\phi) \lt \phi$ for $0 \lt \phi \lt \pi$, which upon taking the $-2$ power gives $\csc^2(\phi) \gt 1/\phi^2$.)
For any $\theta \gt 0$, the last term is a divergent integral, because for $0\lt \epsilon$,
$$\int_0^\theta \frac{d\phi}{\phi} \gt \int_\epsilon^\theta \frac{d\phi}{\phi} = \log(\theta) - \log(\epsilon) \to  \infty$$ 
as $\epsilon \to 0^{+}$.
Consequently, the positive part of the expectation does not exist.  It is immediate that the expectation of $X/W$ does not exist, either.
We left behind one exception to consider: when $X/Z$ is supported on a line passing through the origin, this argument breaks down (because then the density can equal zero--and in fact is zero for almost all $\theta$).  In this degenerate case, $X/Z$ reduces to a constant--equal to tangent of the slope of that line--and obviously that constant is its expectation.  This is the only such situation in which $X/Z$ has an expectation.
A: This is a follow-up to whuber's answer, and posted as a separate answer
because it is too long for a comment.
Lest people think that it is the bivariate normality 
of $X$ and $Y$ that is causing the problem, it is worth emphasizing that if $W$ is a continuous random variable whose density is nonzero on an open interval containing the origin, then $E\left[\frac 1W\right]$ does not exist. Since $\frac 1w$ diverges to $\pm\infty$ as $w$ approaches $0$, the integral for 
$E\left[\frac 1W\right]$, which is of the form
$$E\left[\frac 1W\right]=\int_{-\infty}^0 \frac 1w f_W(w)\,\mathrm dw
+ \int_0^{-\infty} \frac 1w f_W(w)\,\mathrm dw\tag{1}$$
is undefined because both integrals on the right side of
$(1)$ diverge and the
right side of $(1)$ is of the form $\infty-\infty$ (which is undefined).
