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## Main Question or Discussion Point

I had a question regarding convergent and divergent integrals. I want to know the "exact" definition of an improper integral that converges. Wikipedia states that

If you solve [tex]\int_0^2 \frac{dx}{1-x}[/tex], it turns out that it diverges twice, on both integrals (0 to 1, and 1 to 2). My textbook, which has never failed me so far, states that it diverges. Its solution solves only one of the integrals and it of course goes to infinity. Therefore, it diverges.

The problem lies in the fact that this integral "diverges" twice, at the same rate but in opposite directions. When evaluating this integral, this logically would lead to 0 because they're opposites of eacher.

What's your guys' opinions on whether this converges or diverges?

For a while, I took that as a valid answer and claimed that any integral that has a finite answer must be convergent. However, I came up with a problem.An improper integral converges if the limit defining it exists.

If you solve [tex]\int_0^2 \frac{dx}{1-x}[/tex], it turns out that it diverges twice, on both integrals (0 to 1, and 1 to 2). My textbook, which has never failed me so far, states that it diverges. Its solution solves only one of the integrals and it of course goes to infinity. Therefore, it diverges.

**However**, if you solve the integral algebraically, you're left with [tex]ln|\dfrac{1-b}{1-b}|[/tex], which ends up being 0 as [tex]\lim_{b\rightarrow 1}[/tex]. Clearly, this integral has a limit that exists.The problem lies in the fact that this integral "diverges" twice, at the same rate but in opposite directions. When evaluating this integral, this logically would lead to 0 because they're opposites of eacher.

What's your guys' opinions on whether this converges or diverges?