For which $a$, the integral $\int_{0}^{1}\frac{1-x^a}{1-x}dx$ converges

Question

Find $\forall a \in R$ for which the integral:

$$ \int_{0}^{1}\frac{1-x^a}{1-x}dx $$

Converges.

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My Question

Lets define $f(x)$

$$ f(x) = \frac{1-x^a}{1-x} $$

Why cant we say that $f(x)$ is continuous in the interval except for $x = 1$.

Therefore, we can define:

$$ \forall 1 \neq 0 \in R: g(x) = f(x) $$

And:

$$ x = 1: g(x) = 0 $$

Therefore, both integrals of $g(x)$ and $f(x)$ are the same as the functions are the same except for 1 point.

Now, bby using LHopital

$$ \lim_{x \to 1}g(x) = 1 $$

Therefore, $g(x)$ is continuous therefore integrable in the closed inteval: $[0,1]$ and we can conclude:

$$ \int_{0}^{1}g(x)dx $$

Converges

Therefore our original integral converges.

Why not that way?

Answer 1

You can integrate without worrying about the end-point, as long as the function converges. You need to treat both ends of the interval, especially when $a$ is not positive.