Question Number 217383 by peter frank last updated on 12/Mar/25
$$\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{intergral} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{\mathrm{x}^{\mathrm{p}} }\mathrm{dx}\:\:\mathrm{converges}\:\mathrm{for}\:\mathrm{p}>\mathrm{1} \\ $$$$\:\mathrm{find}\:\mathrm{it}\:\mathrm{value} \\ $$
Answered by mr W last updated on 12/Mar/25
![∫_1 ^∞ (dx/x^p )=[(1/((p−1)x^(p−1) ))]_∞ ^1 =(1/(p−1))(1−lim_(x→∞) (1/x^(p−1) ))=(1/(p−1))](https://www.tinkutara.com/question/Q217396.png)
$$\int_{\mathrm{1}} ^{\infty} \frac{{dx}}{{x}^{{p}} }=\left[\frac{\mathrm{1}}{\left({p}−\mathrm{1}\right){x}^{{p}−\mathrm{1}} }\right]_{\infty} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{p}−\mathrm{1}}\left(\mathrm{1}−\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{x}^{{p}−\mathrm{1}} }\right)=\frac{\mathrm{1}}{{p}−\mathrm{1}} \\ $$