Why is it that the integral only exists when $r-\alpha -1<1$ ? Please check the question link,Thanks in advance!
Doubt: It says here the integral exists only if the power to x is less than one, how so?
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integration
improper-integrals
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1Have you computed the antiderivative of $x^{r-a-1}$? – 2017-01-18
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0for $r-a-1>1$ integral is diverge. – 2017-01-18
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0So when it diverges, it does not exist? – 2017-01-18
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0Is $\mu'_r=\infty$ meaningful here? – 2017-01-18
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0I think it is meaningful, it is also mentioned that $\alpha >0$. – 2017-01-18
1 Answers
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The indefinite integral is given by \begin{align*} \alpha \int_1^{\infty} x^{r - \alpha - 1} dx = \alpha \frac{x^{r - \alpha}}{r - \alpha} \Big\rvert^{\infty}_1 = \frac{\alpha}{r- \alpha} (\lim_{x \to \infty} x^{r-\alpha} - 1) \end{align*}where $\lim_{x \to \infty}x^{r-\alpha}$ converges only if $r-\alpha < 0$, i.e. $r < \alpha$.
The $r-\alpha - 1 < 1$ bit seems unnecessary. It can be simplified to say $r-\alpha < 2$, but if $r-\alpha < 0$, then it is certainly less than 2 as well. So the second condition suffices.