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Question Number 33170 by prof Abdo imad last updated on 11/Apr/18

$$provethat{\int }_0^{\mathrm{\infty }}\frac{\mid sinx\mid }{x}dxisdivergent.$$

Commented by prof Abdo imad last updated on 13/Apr/18

$${\int }_0^{\mathrm{\infty }}\frac{\mid sinx\mid }{x}dx={lim}_{n\to +\mathrm{\infty }}{A}_{n}with$$$$A_n={\int }_0^{n\pi }\frac{\mid sinx\mid }{x}dxbut$$$$A_n=\sum _{k=0}^n{\int }_{k\pi }^{(k+1)\pi }\frac{\mid sinx\mid }{x}dx$$$${=}_{x=k\pi +t}\sum _{k=0}^n{\int }_0^{\pi }\frac{sint}{k\pi +t}dtbut$$$$0⩽t⩽\pi \Rightarrow k\pi ⩽k\pi +t⩽(k+1)\pi \Rightarrow$$$$\frac{1}{(k+1)\pi }⩽\frac{1}{k\pi +t}⩽\frac{1}{k\pi }\Rightarrow \frac{sint}{k\pi +t}⩾\frac{sint}{(k+1)\pi }\mathrm{\forall }t\in [0,\pi ]$$$$\Rightarrow {\int }_0^{\pi }\frac{sint}{k\pi +t}dt⩾\frac{1}{(k+1)\pi }{\int }_0^{\pi }sintdt=\frac{2}{(k+1)\pi }\Rightarrow$$$$A_n⩾\frac{2}{\pi }\sum _{k=0}^n\frac{1}{k+1}\Rightarrow A_n⩾\frac{2}{\pi }\sum _{k=1}^{n+1}\frac{1}{k}\Rightarrow$$$$A_n⩾\frac{2}{\pi }{H}_{n+1}{}_{n\to +\mathrm{\infty }}\to +\mathrm{\infty }so{lim}_{n\to +\mathrm{\infty }}{A}_n=+\mathrm{\infty }$$$$andtheintegralisdivergent.$$

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