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Question Number 192388 by sudipmoi last updated on 16/May/23

Answered by aleks041103 last updated on 22/May/23

$$p>0$$$${\int }_1^{\mathrm{\infty }}\frac{sin\left(x\right)dx}{x^p}={[-\frac{cos\left(x\right)}{x^p}]}_1^{\mathrm{\infty }}-{\int }_1^{\mathrm{\infty }}(-cos\left(x\right))d\left(\frac{1}{x^p}\right)=$$$$=cos\left(1\right)-p{\int }_1^{\mathrm{\infty }}\frac{cos\left(x\right)}{x^{p+1}}dx$$$${\int }_1^{\mathrm{\infty }}\mid \frac{cos\left(x\right)}{x^{p+1}}\mid dx<{\int }_1^{\mathrm{\infty }}\frac{dx}{x^{p+1}}=-{\left[\frac{1}{{px}^p}\right]}_1^{\mathrm{\infty }}=\frac{1}{p}$$$$\Rightarrow theintegralisabsolutelyconvergent$$$$\Rightarrow {\int }_1^{\mathrm{\infty }}\frac{cos\left(x\right)}{x^{p+1}}dxconverges$$$$\Rightarrow {\int }_1^{\mathrm{\infty }}\frac{sin\left(x\right)dx}{x^p}convergesforp>0$$

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