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Question Number 217383 by peter frank last updated on 12/Mar/25

$$\mathrm{show}\mathrm{that}\mathrm{the}\mathrm{intergral}$$$${\int }_1^{\mathrm{\infty }}\frac{1}{{\mathrm{x}}^{\mathrm{p}}}\mathrm{dx}\mathrm{converges}\mathrm{for}\mathrm{p}>1$$$$\mathrm{find}\mathrm{it}\mathrm{value}$$

Answered by mr W last updated on 12/Mar/25

$${\int }_1^{\mathrm{\infty }}\frac{dx}{x^p}={\left[\frac{1}{(p-1)x^{p-1}}\right]}_{\mathrm{\infty }}^1$$$$=\frac{1}{p-1}(1-\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x^{p-1}})=\frac{1}{p-1}$$

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