Why does $\int_{-\infty}^\infty x \ dx$ not converge? I understand that $\displaystyle\int_{a}^\infty x \ dx$ or $\displaystyle\int_{-\infty}^a x \ dx$ does not converge, since $f(x)=x$ is unbounded. 
However why does $\displaystyle\int_{-\infty}^\infty x \ dx$ not converge? 
Intuition tells that, as an odd function, $f(x)=x$ should allow

$$\int_{-\infty}^\infty x \ dx =\int_0^\infty x \ dx + \int_{0}^\infty -x \ dx=0.$$

 A: Your intuition is true in some sense. You are looking for something called Cauchy principal value. 

However, since the function $f(x)=x$ is not absolutely integrable over $\mathbb{R}$, namely, $\int_{\mathbb{R}}|x|\ dx=\infty$, one needs to be careful if one really wants to define the improper integral
$$
\int_{-\infty}^{\infty}x\ dx.
$$
More importantly, one should keep in mind that $\infty$ is not a real number. It can be viewed as an extended real number though, one needs to know what kind of arithmetic operations are (and are not allowed). For instance, $\infty-\infty$ is not defined (it is so called indeterminate form).

[Added:] The symbol $\int_{-\infty}^{+\infty}$  can have at least 3 meanings: 


*

*1.$\int_{-\infty}^{+\infty}=\int_{-\infty}^a+\int_{a}^{+\infty}$;   

*2.$\int_{-\infty}^{+\infty}=\lim_{a\to\infty}\lim_{b\to\infty}\int_a^b$;   

*3.$\int_{-\infty}^{+\infty}=\lim_{b\to\infty}\int_{-b}^b$ which is called the principal value.


For improper Riemann integrals such as $\int_{-\infty}^\infty f(x)\ dx$, there is no ambiguity if the integrand is either non-negative or absolutely convergent (as then one can use the Lebesgue theory to show that all the different interpretations mentioned above agree), but otherwise if one wants to be fully rigorous, one has to specify the precise interpretation of the integral. 
A: Your intuition says $$\infty -\infty=0$$ 
But this is not true!
Why?
For example, we can write $$\int_{-\infty} ^\infty xdx=\lim_{n\to\infty}\int_{-n}^{n+1}xdx$$
Now you see why we can't say anything?
A: Try this going to infinity in a non-symmetric way, it makes it easier to see

$$\lim_{t\to\infty}\int_{-t^2}^t x \,dx = {1\over 2}\left(t^2-t^4\right)=-\infty$$
$$\lim_{t\to\infty}\int_{-\log t}^t x \,dx = {1\over 2}\left(t^2-\log^2t\right)=\infty$$

If you're feeling saucy, you can make it finite too,

$$\lim_{t\to\infty}\int_{-t}^{t+\epsilon(t)} x \,dx = {1\over 2}\left((t+\epsilon(t))^2-t^2\right)=t\epsilon(t) + {1\over 2}\epsilon(t)^2$$

if I use the quadratic formula and say I want this to be $1$, for example,
$$\epsilon(t)^2+2t\epsilon(t) -2= 0\implies \epsilon(t) = -t\pm\sqrt{t^2+2}$$
Since this is positive, we know that $\epsilon(t)>0$ so the plus value holds, and with that function we get

$$\int_{-t}^{\sqrt{t^2+2}}x\,dx = 1$$

for all $t$ in particular it holds in the limit as $t\to\infty$
By letting things go at different rates you can make this whatever you want.
A: We say that $$\int_{-\infty}^{+\infty}f(t)dt$$
is convergent
$$\iff \lim_{(x,y)\to(-\infty,+\infty)}\int_x^yf(t)dt \in \mathbb R.$$
$\int_{-\infty}^{+\infty}xdx$ is divergent since
if $x=-\sqrt{2t}$ and $y=\sqrt{2t+2A}$,
we find that the limit when $t\to+\infty$is $A$.
