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.
Best Answer
This is a good question! It is all a matter of how to define an improper integral. The traditional way is to write
$$\int_a^b f(t)\, dt = \lim_{x \to a} \lim_{y \to b} \int_x^y f(t)\, dt$$
whenever one of the bounds is worrisome. In particular, we can write
$$\int_a^b f(t) \,dt = \lim_{x \to a} \lim_{y \to b} \left(\int_x^1 f(t) \, dt + \int_1^y f(t)\, dt \right)$$ or $$\lim_{x \to a} \lim_{y \to b} \left(\int_x^1 f(t) \,dt \right) + \lim_{x \to a} \lim_{y \to b} \left( \int_1^y f(t) \,dt \right) $$
so that
$$\int_a^b f(t)\, dt = \lim_{x \to a} \left(\int_x^1 f(t) \,dt \right) + \lim_{y \to b} \left(\int_1^y f(t)\, dt \right).$$
Now convergence means that the integral is finite and the only way for the sum of two numbers to be finite is if both are finite. Hence your professor's comment. All of the above assumes $a < 1 < b$ and that there is no trouble with $f$ near $t=1$.
But you are right that you can get cancelation if you take the limits differently. Instead of taking the limit to a and then the limit to b sequentially, if you took them at the same time--that is to say you combined them--you could possibly get cancellation.
Here's an example. If define instead the improper integral as
$$\int_{-\pi/2}^{\pi/2} \tan t\, dt = \lim_{x \to \pi/2} \int_{-x}^x \tan t\,dt$$
then
$$\int_{-\pi/2}^{\pi/2} \tan t\, dt = \lim_{x \to \pi/2} \left(- \ln \cos(x) + \ln \cos(-x) \right) = 0.$$
What you are thinking of is a useful concept and it often comes up with the term Principal Value in more advanced calculus courses.