Expected root of quadratic random polynomial Suppose $A,B,C$ are i.i.d. random variables with uniform distribution on $[-1,1]$. I'm interested in the expected roots of the polynomial $Ax^2 + Bx + C$, which are complex random variables given by
$$Z_1 = \frac{-B+\sqrt{B^2-4AC}}{2A}$$
and
$$Z_2 = \frac{-B-\sqrt{B^2-4AC}}{2A}.$$
Making simulations, I computed 
$$E[Z_1] \approx 0.3559 + 0.0005i$$ 
and 
$$E[Z_2] \approx -0.6421 - 0.0005i.$$
To confirm this resuts, I need to calculate this values mathematically. For $E[Z_1]$ for instance, this means to calculate the integral 
$$\frac{1}{8}\int_{-1}^1 \int_{-1}^1 \int_{-1}^1  \frac{-b+\sqrt{b^2-4ac}}{2a}\ da\ db\ dc.$$
Unfortunately, looks like this integral has different values when we change the order of integration. I tried to compute with Wolframalpha. It gives me zero or can't compute depending on the order. Probably this is because the term $\frac{1}{2a}$ goes to infinity in the interval of integration, so we can't use Fubini's Theorem. I'm not sure if Wolframalpha just failed to compute some integrals or $E[Z_1]$ is really not defined. This second scenario means $Z_1$ has no expected value, so the random polynomial $Ax^2 + Bx + C$ has no expected root. I think this is a strange scenario, therefore I really need to confirm whether this is the case or not.   
 A: Your $Z_1$ and $Z_2$ are not well defined until you have made a choice of which complex root to take.  That choice could affect their distributions.   (It actually does not, by virtue of the symmetries of $A$, $B$, and $C$ around $0$.)
Regardless, since $Z_1+Z_2=-B/A$ is well-defined, suppose you have made such a choice and that the $Z_i$ have finite expectations.  From the independence of $A$ and $B$ and the fact that the density of $A$ does not approach zero near $A=0$, it follows from I've heard that ratios or inverses of random variables often are problematic, in not having expectations. Why is that? that $-B/A$ has no expectation.  But since $E[-B/A]=E[Z_1+Z_2]$, that creates a contradiction demonstrating at least one of $Z_1$ and $Z_2$ cannot have an expectation.

You can also argue from the symmetry of this problem that the expectation of $\sqrt{B^2-4AC}/(2A)$, if it exists, must be zero.  (The distribution of $(A,B,C)$ and the distribution of $(-A,B,-C)$ are the same, but the corresponding distributions of $\sqrt{B^2-4AC}/(2A)$ are negatives of each other.  Ergo, their expectations must be negatives of each other, too.)  Therefore the expectation of each $Z_i$ is just $E[-B/(2A)]$.  This has a simpler expression as an integral:
$$E[-B/2A] = \frac{1}{4}\int_{-1}^1 \int_{-1}^1 -\frac{b}{2a} da db$$
We might try to evaluate it as an iterated integral (according to Fubini's Theorem).  However, the inner integral (with respect to $a$) diverges at $0$:
$$\lim_{t\to 0^{+}} \int_t^1 \frac{-da}{a} = \lim_{t\to 0^{+}}\log(t) \to -\infty$$
while
$$\lim_{t\to 0^{-}} \int_{-1}^t \frac{-da}{a} = \lim_{t\to 0^{-}}(-\log(-t)) \to \infty,$$
demonstrating it is undefined.  That is why it is invalid to change the order of integration--Fubini's Theorem does not apply--to obtain $0$ for the integral over $b$ and thus get the (wrong) value of $0$ for the expectation.
In either analysis, the source of the difficulty is clear: $A$ has a non-negligible density in any neighborhood of zero.
