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I have this integral, and the solution gives an indeterminate form for the value $\alpha = 1$, can you explain to me how to solve the indeterminate form?

$$\int_{\beta}^{+\infty} x^{-\alpha} dx = \frac{1}{1-\alpha} \left[ x^{-\alpha + 1}\right]_{\beta}^{+\infty} = \frac{1}{0} \left[ +\infty^{0} - 1\right] =?$$

Little side note, I know the solution of this integral I am really just curious od knowing how to tackle the indeterminate form.

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    $\begingroup$ do you know what is $ \int \frac{dx}{x} $ ? $\endgroup$
    – what'sup
    Oct 13, 2013 at 18:21
  • $\begingroup$ @what'sup - Yeah, log(x), so? $\endgroup$
    – Matteo
    Oct 13, 2013 at 18:22
  • $\begingroup$ @what'sup - Maybe I should specify in the question that I know what shoul be the result, I am interested in understanding how to solve the indeterminate form... $\endgroup$
    – Matteo
    Oct 13, 2013 at 18:24
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    $\begingroup$ @Matteo: the trick is to replace $\;\displaystyle\frac{1}{1-\alpha} \left[ x^{-\alpha + 1}\right]_{\beta}^{+\infty}\;$ by $\;\displaystyle\frac{1}{1-\alpha} \left[ x^{-\alpha + 1}-1\right]_{\beta}^{+\infty}\;$ to get a correct result at the limit (as shown in the two links proposed by what's up). $\endgroup$ Oct 13, 2013 at 19:18

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It's not really indeterminate. What you are trying to do is $$ \lim_{\alpha\to 1^+}\int_\beta^\infty x^{-\alpha}\,\, dx = \lim_{\alpha\to 1^+}\frac{x^{1-\alpha}}{1-\alpha}. $$ But this is of the form $\displaystyle \frac{1}{0}$ which is not indeterminate.

Your issue is that to reach that last formula in your post you have to assume the integral exists.

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  • $\begingroup$ but the integral does exist for $\alpha \to 1^{+}$ $\endgroup$
    – Matteo
    Oct 13, 2013 at 19:01
  • $\begingroup$ @Matteo: The integral exists for all $\alpha >1$. So, we can try to see if the limit exists. It makes no sense to say that the integral exists for $\alpha\rightarrow 1^+$. $\endgroup$
    – J126
    Oct 13, 2013 at 20:28

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