Why is it that the integral only exists when $r-\alpha -1<1$ ? Please check the question link,Thanks in advance!
2026-04-12 13:31:39.1776000699
Doubt: It says here the integral exists only if the power to x is less than one, how so?
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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.