Improper Riemann integrals

Question

Calculate the improper Riemann integral $$I_a := \int\limits_0^1\frac{dx}{x^a}$$

Does $I_a$ have a limit as $a \to 1^-$?

Answer 1

The integrand is infinite at $0$ so as an improper integral one calculates $$\lim_ {t \to 0^+} \int_t^1 \frac{dx}{x^a}=\lim _{t \to 0^+} \frac{1}{1-a}-\frac{t^{1-a}}{1-a}=\frac{1}{1-a}.$$ This, as $a \to 1^-$ , goes to $+\infty.$

Answer 2

Hint: it clearly has to be $\,a\neq 1$ , so:

$$I_a=\lim_{\epsilon\to 0}\int_\epsilon^1\frac{dx}{x^a}=\frac{1}{1-a}\lim_{\epsilon\to 0}\left(1-\epsilon^{1-a}\right)$$

When does the above limit exist finitely?