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Question Number 148285 by Khalmohmmad last updated on 26/Jul/21

$${\int }_1^{\mathrm{\infty }}\frac{1}{x^2\ln x}dx=?$$

Commented by ahmedNasser last updated on 26/Jul/21

$$letu=lnx\therefore du=\frac{1}{x}dxx=e^u$$$${\int }_0^{\mathrm{\infty }}\frac{1}{e^{2u}\cdot u}\cdot e^{u}du={\int }_0^{\mathrm{\infty }}{u}^{-1}{e}^{-u}du=\mathrm{\Gamma }\left(0\right)=\mathrm{\infty }$$

Answered by puissant last updated on 26/Jul/21

$$diverge..$$

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