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Here is a calculus fact. If $a>0$ then $\int\limits_{a}^{\infty}\frac{1}{x^p}dx$ is convergent if $p>1$ and divergent if $p\leq 1$ .My question is that is the fact rue.If so how can I prove it?

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    $\begingroup$ Fix $p$ and calculate $\lim\limits_{b\rightarrow\infty}\int_a^b 1/x^p\,dx$. $\endgroup$ Mar 14, 2015 at 18:14

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We may represent the integral as

$$ \int_{a}^{\infty} x^{-p} dx $$

Integrating using the power rule:

$$ \frac{x^{-p + 1}}{-p + 1} \bigg|_a^\infty $$

If $p>1$ then $-p+1<0$; $\infty$ raised to a negative power will go to zero. Therefore the integral will converge.

Note that this is a general argument, ignoring the fact that $\int x^{-1} = \ln x$

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  • $\begingroup$ the lower bound is $a$ not $0$. (+1) $\endgroup$
    – Math-fun
    Mar 14, 2015 at 18:25
  • $\begingroup$ Corrected, thanks. $\endgroup$
    – baum
    Mar 14, 2015 at 18:26
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Notice ( $ p \neq 1 $ )

$$ \int\limits_a^{\infty} x^{-p } dx = \lim_{\alpha \to \infty} \int\limits_a^{\alpha} x^{-p} dx = \lim_{\alpha \to \infty} \bigg( \frac{x^{1-p}}{1-p} \bigg)_{x=a}^{x=\alpha}=$$

$$ = \lim_{\alpha \to \infty} \bigg( \frac{ \alpha^{1-p}}{1-p} - \frac{ a^{1-p}}{1-p} \bigg)$$

Notice, if $1-p > 0 $ which means $1 > p $, then the limit goes to $\infty$ since $n^k \to \infty$ if $n > 1$ and similarly if $p > 1 $, then limits goes to $ - \frac{ a^{1-p}}{1-p} $.

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