This is a good question. One way to think about the issue is that the "convergence" of your integral is not just about whether the value is finite, but also about whether the finite value is well defined. This is somewhat related to indeterminant forms where $+\infty + (-\infty)$ is not a well-defined quantity. (You can't say they cancel out, since, heuristically, $\infty - \infty = (1+\infty)-\infty = 1+\infty -\infty = 1 + (\infty - \infty)$...)
So as long as one of the integrals on the right hand side diverges, the entire algebraic expression becomes indeterminate, and hence we say the integral diverges. (Diverges doesn't necessarily mean that the value must run-off to infinity; it can just mean that the value does not converge to a definite number/expression.)
(A similar issue also crops up when summing infinite series that doesn't converge absolutely. Riemann rearrangement theorem tells you that, depending on "how" you sum the series you can get the final number to be anything you want.)
Sometimes, however, it is advantageous to try to make sense of an integral which can be split into two divergent improper integrals, but also where one can argue that there should be some natural cancellation. For example, one may want to argue that $\int_{-a}^a \frac{1}{x^3} dx$ evaluates to 0 since it is the integral of an odd function. For this kind of situations, the notion of Cauchy principal value is useful. But notice that the definition is taken in the sense of a limit that rely on some cancellation, and so much in the same way of Riemann rearrangement theorem, "how" you take the limit can affect what value you get as the end result. (This is compatible with the notion that the integral diverges; as I said above, divergence should be taken to mean the lack of a well-defined, unique convergence.)
Edit. Let me add another example of an integral that remains finite but does not converge. What is the value of $\int_{0}^\infty \sin(x) dx$? For every fixed $a > 0$, $\int_0^a\sin(x) dx = 1 - \cos(a) $ is a number between 0 and 2. But the limit as $a\to \infty$ doesn't exist! If you pick a certain way to approach $\infty$, say choose a sequence $a_n = 2\pi n$, then you'll come to the conclusion that the "limit" is 0; but if you choose $a_n = (2n + 1)\pi$, then you get the conclusion that the limit is $2$. The idea here is roughly similar: you take a left limit and a right limit approaching the improper point, and depending on how you choose your representative points (by an algebraic relation between the speed at which the left and right limits approach the improper point, say), you can get different answers.
If the integral involves $|x|$, we split the interval of integration by $0$, and unwind the definition of $|x|$ on each part: $|x|=x$ on the positive part, $|x|=-x$ on the negative. Then integrate and add the results. If only the convergence of an improper integral is of interest, we don't calculate the values, but only check whether each part converges.
More generally: if the integral involves $|g(x)|$ where $g$ is some function of $x$, the interval of integration is split by the points where $g(x)$ changes sign.
Best Answer
HINT:
Note that $ x^t e^{-x}\ge e^{-1} x^t$ for $x\in [0,1]$.