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Question Number 87881 by Rio Michael last updated on 06/Apr/20

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{x}dx=$$

Commented by mathmax by abdo last updated on 06/Apr/20

$$thefunctionx\to \frac{1}{x}isodd\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x}=0$$$$alsowecanwrite{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dx}{x}={lim}_{\xi \to +\mathrm{\infty }}{\int }_{-\xi }^{\xi }\frac{dx}{x}$$$$={lim}_{\xi \to +\mathrm{\infty }}{[ln\mid x\mid ]}_{-\xi }^{\xi }=0$$

Answered by mind is power last updated on 07/Apr/20

$$PV\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{x}dx\right)=0$$$$cauchyprincipalvalue$$$$but{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dx}{x}notexistRiemann$$

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