I have been having some trouble with the way to determine the convergence/divergence of an improper integral such as this: $$\int_0^a \frac{\arctan{(x)}}{1-x^3}dx,$$ where $a>1.$ It is evident that the integrand has an infinite discontinuity at $x=1,$ and the common course of action in order to establish the convergence/divergence of said integral would be to split the interval of integration at that point, and consider the two summands, both being improper integrals of their own, separately: $$\int_0^a \frac{\arctan{(x)}}{1-x^3}dx = \int_0^1 + \int_1^a.$$ However, both of the summands diverge, (for the integrand becomes infinite to the first order whatever side you approach $x=1$ from,) and they diverge into infinities of opposing signs. In short, this looks very much like an $\infty - \infty$ indeterminacy. Would it still be sound to say that the original integral diverges? This confusion is further facilitated by different textbooks. For instance, Courant and a bunch of others say we need to consider those „partial” improper integrals in complete separation, and if one diverges, I take it, the whole integral must diverge. Fichtenholtz, however, in his very influential „Course”, says the following:
…in case all the summands exist, except for when two of them become infinite of different signs.
It is my own translation from the Russian, as I am aware that this textbook has yet to receive a complete English translation, but I think I have related the idea right. So, how should one go about splitting the interval like that? When two integrals, like in the case I mentioned, become infinite of different signs, are we to state confidently the divergence of the integral, or should we try to find another way that could potentially clear the indeterminacy?