Improper Integral - If there is an asymptote, does it necessarily have to diverge? Context: After learning that the harmonic series diverges, I have begun to doubt just merely looking at graphs 
Here's an improper integral with two vertical asymptotes: $x=0$ and $x=0.5$.  
$$\int_{0}^1\frac{1}{2x^2-x}dx$$ 
It diverges.
My line of reasoning: The function is not bounded at $x=0$ and $x=0.5$, and is not integrable, despite being continuous almost everywhere from $0$ to $1$.  
Another possible line of reasoning:
The integral can be split up into: $\int_{0^+}^{0.5^-}\frac{1}{2x^2-x}dx$ + $\int_{0.5^+}^{1}\frac{1}{2x^2-x}dx$
Each one of these integrals, when evaluated, diverge. So the improper integral diverges.  
Which line of reasoning is more logical?
Is there a function that has vertical asymptote $x=k$, but when evaluated with limits at $x=k^-$ and $x=k^+$, converges?
(If this is true, then the second line of reasoning might be correct.)  

I assume by "diverges" you mean is not absolutely (and so Lebesgue) integrable? Because in this case you can just compute the integral of its absolute value
$$ \int_0^\frac{1}{2} \frac{1}{x-2x^2} + \int_\frac{1}{2}^2 \frac{1}{2x^2-x} $$
and notice that, for instance, the second integrand is $+\infty$. Because the integral of its negative part is also $+\infty$, it's Lebesgue integral simply makes no sense.
Do not split your integral into two different parts before knowing it's absolute value is integrable (or the argument does not change sign), because that operation makes no sense.
