Necessary condition for the convergence of an improper integral. My calculus professor mentioned the other day that whenever we separate an improper integral into smaller integrals, the improper integral is convergent iff the two parts of the integral are convergent. For example:
$$ \int_0^{+\infty} \frac{\log t}{t} \mathrm{d}t = \underbrace{\int_0^1 \frac{\log t}{t} \mathrm{d}t}_{\text{Diverges to }-\infty} + \underbrace{\int_1^{+\infty} \frac{\log t}{t} \mathrm{d}t}_{\text{Diverges to }+\infty} 
$$
So the integral would not converge, because one of the parts of the integral is divergent (or both in this case).
However, I don't see why the part that diverges to $-\infty$ and the part that diverges to $+\infty$ cannot  cancel out and make it converge, how ever counterintuitive it may seem. It's what happens with the Dirichlet Integral to some extent, although the areas are bounded. 
It could be that the problem arises when things tend to $\pm\infty$ but if I recall correctly this is not a problem for the sum of an infinite series, e.g.:
$$\sum_{n=1}^{\infty}A_n+B_n=
\underbrace{\sum_{n=1}^{\infty}A_n}_{\text{Diverges to }+\infty} + 
\underbrace{\sum_{n=1}^{\infty}B_n}_{\text{Diverges to }-\infty} 
\nRightarrow \nexists \sum_{n=1}^{\infty}A_n+B_n \vee \sum_{n=1}^{\infty}A_n+B_n= \pm \infty $$
Where does the problem arise?
Also, if my use of symbology is incorrect (which I suspect it is) please tell me so. I'm trying to writing more formally and efficiently.

$$
\int_{-\infty}^{+\infty} x\,dx = \text{what?}
$$
$$
\int_{-\infty}^{+\infty} x\,dx = \underbrace{\int_{-\infty}^0 x\,dx}_{\text{Diverges to }-\infty} + \underbrace{\int_0^\infty x\,dx}_{\text{Diverges to }+\infty}
$$
But
$$
\lim_{a\to\infty} \int_{-a}^a x\,dx = 0.
$$
Here's a more involved example:
$$
\lim_{a\to\infty}\int_{-a}^a \frac{x\,dx}{1+x^2} = 0 \tag 1
$$
but
$$
\lim_{a\to\infty}\int_{-a}^{2a} \frac{x\,dx}{1+x^2} = \lim_{a\to\infty} \frac 1 2 \log_e \frac{1+4a^2}{1+a^2} = \log_e 2.
$$
Re-arranging a sum or an integral can result in a different value only if positive and negative parts both diverge to infinity.
The result in $(1)$ is the "principal value" of this improper integral.
