Why is it not true that $\int_{-\infty}^\infty{\rm} x\,dx=0 \, $ given that x is an odd function? I was given this to do at my own will and look at why it is divergent, but to be sincere I have no idea on how I can start approaching it. I just need explanation to why the answer is 0. I'm studying this integral:
$$\int_{-\infty}^\infty x\,dx=0. $$
Why is it divergent?
 A: Because of the way the double-sided improper integral is defined. The integral in question can be evaluated by choosing $a \in \mathbb{R}$ (for continuous functions, $a = 0$ is simple) and evaluating:
$$\int_{-\infty}^\infty x \, \mathrm{d}x = \lim_{M \to -\infty} \int_M^a x \,\mathrm{d}x + \lim_{N \to \infty} \int_a^Nf(x)\, \mathrm{d}x.$$
The two integrals on the right hand side diverge, so the integral does not exist.
Side: There is a concept called the Cauchy Principle Value that assigns these kinds of functions a finite value.
$$CPV\left(\int_{-\infty}^\infty f(x) \, \mathrm{d}x\right) = \lim_{M \to \infty} \int_{-M}^M f(x) \, \mathrm{d}x.$$
For odd functions, the CPV is 0.
A: The usual definition of that improper integral is
$$\lim_{a\to-\infty}\int_a^c x\,dx+\lim_{b\to\infty}\int_c^b x\,dx$$
where $c$ is some constant real number. (A theorem tells us that the choice of $c$ does not change this value.)
Both those integrals are divergent, therefore both limits do not exist, therefore the sum does not exist.
Therefore your integral is undefined.

That said, there is another definition for your integral, namely the Cauchy principal value, which is
$$\lim_{a\to\infty}\int_{-a}^a x\,dx$$
Under that definition, the integral inside the limit is always zero, so the total value is defined and is zero.
However, this is not the current standard definition of an improper integral, and if that is what you mean by your expression you need to say so.
A: Rory Daulton already perfectly answered your question, as Calculus textbooks usually introduce integrations in terms of the Riemann integral. This is constructed on compact intervals $[a,b]$ and can then be extended to intervals of the form $[a, \infty)$, $(-\infty, b]$, $(-\infty,+\infty)$ in the way described in Rory Daulton's answer.
On the other hand, integration can also be introduced in terms of the Lebesgue integral. In these cases, one first constructs the integral for positive (measurable) functions. An arbitrary (measurable) function $f$ is then said to be integrable if $ \int_{\mathbb R} \left\vert f(x) \right\vert dx < \infty$ and is then defined as:
$$ \int_{\mathbb R} f(x) dx = \int_{\mathbb R} f^+(x) dx - \int_{\mathbb R} f^-(x) dx $$
Here $f^+(x) = \max\{f(x),0\}$ and $f^-(x) = -\min\{f(x),0\}$.
In this case, it is easy to show that $\int_{\mathbb R} \left\vert x \right\vert dx = \infty$, so that $x$ is not integrable over $\mathbb R = (-\infty,+\infty)$, also when the integral is defined as the Lebesgue integral.
A: The improper Riemann integral $\int_{-\infty}^{\infty} f(x)\,dx$ is interpreted as 
$$\int_{-\infty}^{\infty}f(x)\,dx=\lim_{L_{-}\to \infty} \int_{-L_{-}}^0f(x)\,dx+\lim_{L_{+}\to \infty} \int_0^{L_{+}}f(x)\,dx \tag 1$$
Even for odd functions, neither integral in $(1)$ converges and neither does their sum.   However, if one interprets the improper integral in the sense of Cauchy Principal value, then $L_{-}=L_{+}=L$ and we have
$$\int_{-\infty}^{\infty}f(x)\,dx=\lim_{L\to \infty} \int_{-L}^{L}f(x)\,dx=0$$
where the odd symmetry of $f$ immediately gives the result.  Herein, $f(x)=x$ is an odd function and thus the aforementioned applies.
A: For improper integrals that exist according to the standard definition, you would need $\lim_{s \to \infty, t \to - \infty} \int_{t}^s x dx$ to exist. However, the only way this will happen is if e.g. you set $t = -s$. Otherwise you will get $\pm \infty$ for your answer for example if you set $t = -cs$ for some $c > 0$ such that $c \neq 1$. It could even happen that you take $s \to \infty$ in a non-linear fashion as a function of $t$, and then you may even get that the limit doesn't exist. The fact that you have to approach $\pm \infty$ at certain speeds, in order to get a unique answer for your integral, is why the integral is typically considered to be undefined. There are other improper integrals, considered to truly have defined answers, where you can break it up into pieces and then the answer for the overall integral you get is independent of how fast you approach each division point from either side.
