Why is $\int^\infty_{-\infty} \frac{x}{x^2+1} dx$ not zero? We break up $\int^\infty_{-\infty} \dfrac{x}{x^2+1} dx$ into:
$$\lim_{t\to -\infty} \int^0_t \dfrac{x}{x^2+1} dx + \lim_{t\to \infty} \int^t_0 \dfrac{x}{x^2+1} dx$$
So, evaluated, this gives;
$$\lim_{t \to \infty} \left(\frac{1}{2} \ln (1+t^2)\right) - \lim_{t \to -\infty} \left(\frac{1}{2} \ln (1+t^2)\right)$$
But those two terms are essentially identical! It should be zero!  Plus, the integrand is an odd function, so why is this undefined?  Is it just an unspoken rule that once you encounter $\infty$ in a mathematical expression you should stop evaluating immediately?
 A: This integral does not converge absolutely since $\frac{|x|}{1+x^2}\sim\frac 1{|x|}$ for $|x|\to\infty$ and $\int\limits_{x=1}^\infty \frac{1}{|x|} = \infty$. Though to tackle problems like yours you may consider Cauchy principal value which in your case exists and indeed $0$ since the function is odd.
A: Several years ago I added this example to Wikipedia's article titled improper integral:
$$
\lim_{a\rightarrow\infty}\int_{-a}^a\frac{2x\,dx}{x^2+1}=0\text{ and }\lim_{a\rightarrow\infty}\int_{-2a}^a\frac{2x\,dx}{x^2+1}=-\ln 4.
$$
In both cases the upper and lower bounds of integration both approach infinities.  But the two values we get are different.  This can happen only if the positive and negative parts are both infinite, i.e. $\displaystyle \int_{-\infty}^0 \dfrac{2x\;dx}{x^2+1}=-\infty$ and $\displaystyle \int_0^\infty \dfrac{2x\;dx}{x^2+1}=\infty$.
A: If you "interpret" the integral $ \displaystyle \int_{-\infty}^{\infty} \frac{x}{x^2 + 1} dx$ as $\displaystyle \lim_{R \rightarrow \infty} \int_{-R}^{R} \frac{x}{x^2 + 1} dx$, then it is indeed zero. (This is called the Cauchy principal value).
However, note that the integral $ \displaystyle \int_{-\infty}^{\infty} \frac{x}{x^2 + 1} dx$ can also be interpreted as, for instance, $$\displaystyle \lim_{R \rightarrow \infty} \int_{-R}^{kR} \frac{x}{x^2 + 1} dx = \lim_{R \rightarrow \infty}  \frac12 \log \left( \frac{1 + k^2R^2}{1 + R^2} \right) = \log(k) \text{ where }k > 0.$$ You could also take other functions of $R$ such that the lower limit tends to negative infinity and upper limit tends to infinity as $R \rightarrow \infty$ to get different answers.
Hence, $ \displaystyle \int_{-\infty}^{\infty} \frac{x}{x^2 + 1} dx$ is not zero and in fact cannot be assigned any value unless you know how the lower limit and upper limit approach $\infty$.
A: We should agree that, whatever it is we want 
$$\int_{-\infty}^{\infty}f(x)\,dx$$
to be, the expression should still respect the usual rules of integration. In particular, for any real number $c$, we "should" have
$$\int_{-\infty}^{\infty}f(x)\,dx = \int_{-\infty}^cf(x)\,dx + \int_c^{\infty}f(x)\,dx.$$
Otherwise, the value of the integral may depend on how we choose to evaluate the integral, and that is no good.
This in turn means that in order for
$$\int_{-\infty}^{\infty}f(x)\,dx$$
to make sense, we need both
$$\int_{-\infty}^cf(x)\,dx\quad\text{and}\quad\int_{c}^{\infty}f(x)\,dx$$
to make sense separately and independently, and this to happen for every real number $c$. 
So in order for
$$\int_{-\infty}^{\infty}\frac{x}{1+x^2}\,dx$$
to make sense as a number, we need
$$\textbf{both}\quad\int_{-\infty}^c\frac{x}{1+x^2}\,dx\quad\textbf{and}\quad \int_{c}^{\infty}\frac{x}{1+x^2}\,dx\quad\textbf{to each make sense for every }c.$$
However,
$$\int_c^{\infty}\frac{x}{1+x^2}\,dx$$
does not exist for any value of $c$, so we cannot make sense of
$$\int_{-\infty}^{\infty}\frac{x}{1+x^2}\,dx$$
as a number. 
Now, it is tempting to say that since the function is odd and the interval is symmetric about the origin, the integral "should" be equal to $0$. Unfortunately, that runs into serious trouble pretty soon. Consider for example trying to argue that way with
$$\int_{-\infty}^{\infty}\sin x\,dx.$$
Okay, that "should" be $0$ because $\sin x$ is an odd function. However, I claim that in fact, the integral "should" be $2$. Why? Well,
$$\begin{align*}
\int_{-\infty}^{\infty}&\sin x\,dx\\
 &= \cdots + \int_{-4\pi}^{-2\pi}\sin x\,dx +\int_{-2\pi}^0\sin x\,dx + \int_{0}^{\pi}\sin x\,dx +\int_{\pi}^{3\pi}\sin x\,dx + \int_{3\pi}^{5\pi}\sin x\,dx + \cdots
\end{align*}$$
now, every integral except for $\int_{0}^{\pi}\sin x\,dx$ is equal to $0$; and 
$$\int_0^{\pi}\sin x\,dx = 2.$$
So, "clearly", the whole integral, $\int_{-\infty}^{\infty}\sin x\,dx$ "should" equal $2$, not $0$. And by choosing other ways of breaking up $(-\infty,\infty)$, I could give you good reasons why the integral "should be" any particular number you want between $-2$ and $2$. 
This just won't do; and so we solve the problem by reaching the only conclusion possible: the original integral simply does not exist. Just because our function is odd is not enough reason to conclude the integral "should" be $0$. 
A: Such an integral is defined ONLY if each of the two parts is defined:
$\displaystyle\lim\limits_{t\to -\infty} \int^0_t \dfrac{x}{x^2+1} dx$
and
$\displaystyle\lim\limits_{t\to \infty} \int^t_0 \dfrac{x}{x^2+1} dx$
Then, if BOTH of these limits exist separately, the full integral is defined as the sum of the two.  But, as your work shows, each of these has a limit of $\infty$, i.e., neither exists.  So, the full integral, $\displaystyle\int^\infty_{-\infty} \frac{x}{x^2+1} dx$, is undefined.
