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

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

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To get the best answers you should post some motivation and write down what work you have done. If it is homework put "homework" as a tag. I am downvoting this post. –  Nameless Dec 2 '12 at 10:36

2 Answers 2

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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 $+\infty.$

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$\frac{1}{1-a}$ doesn't go to $\pm\infty$ if a goes to 0. It should be 1. –  kram1032 Dec 2 '12 at 11:21
Thanks. I went too fast: problem was about $a \to 1^-$. I'll adjust! –  coffeemath Dec 2 '12 at 11:39

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?

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A nit: shouldn't $\epsilon$ appear in the limit here? –  coffeemath Dec 2 '12 at 11:42
Of course. Thanks. –  DonAntonio Dec 2 '12 at 12:35

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