Studying the divergence or otherwise of an improper integral

Question

I'm supposed to study the following improper integral : $$\int^{1/2}_{-\infty}\frac{1}{1-x^{1/3}} dx$$

This could be an exercise out of 5 in our final exam paper, so I reckon there's a fast way to do it, but I just couldn't manage to determine what that might be. I don't see a way to apply any of the convergence criteria and finding the primitive of this integral took me way too long .

How can we avoid computing the antiderivative in this particular case and determine the nature of this integral ?

Answer 1

Perhaps the most useful way is to compare the integrand to a certain series at infinity.
Since $\forall x \in (-\infty, 1/2]$ the function is well defined, so the only problematic point is just $-\infty$. When $x\to -\infty$, we have $$\frac{1}{1-x^{1/3}}=-\frac{1}{x^{1/3}}+o(\frac{1}{x^{1/3}})$$ now that $|\frac{1}{3}|<1$, the integral must diverge.
A more detailed explanation may be like this: For your integral, its divergence or convergence completely completely depends on whether the integrands "major part" $-\frac{1}{x^{1/3}}$ diverges or converges under such an improper integral. Undoubtedly, $\int_{-\infty}^{1/2}-\frac{1}{x^{1/3}}=+\infty$, you can derive it by finding its anti-derivative. Therefore, your integral diverges.

Answer 2

Near the origin, the function is nice and bounded so there are no integrability problems. You just need to look at what happens at the tail, where $x\to-\infty$, so you're essentially looking at the integral $$\int_{-\infty}^{0}x^{-1/3}dx$$

The nature of which I am sure you can determine.