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Question Number 216722 by issac last updated on 17/Feb/25

Answered by MrGaster last updated on 17/Feb/25

$${\int }_0^{\mathrm{\infty }}(\frac{1}{\ell }-\frac{1}{p})d\ell ={\int }_0^{\mathrm{\infty }}\left(\frac{p-\ell }{\ell p}\right)d\ell$$$${{\int }_0^{\mathrm{\infty }}(\frac{1}{\ell }-\frac{1}{\ell +1}+\frac{1}{\ell +2}+\frac{1}{\ell +3}+\dots )d\ell =\underset{N\to \mathrm{\infty }}{\lim }[\ln (\ell +1)\ln (\ell +2)+\ln (\ell +3)+\dots )]}_0^N$$$$\underset{N\to \mathrm{\infty }}{\lim }[\ln \left(N\right)-\ln (N+1)(N+2)(N+3)\dots )]=\underset{N\to \mathrm{\infty }}{\lim }[\ln \left(N\right)-\ln (N!)]$$$$\underset{N\to \mathrm{\infty }}{\lim }[\ln \left(N\right)-(N\ln \left(N\right)-N+\frac{1}{2}\ln \left(2\pi N\right)+O\left(\frac{1}{N}\right))]=\mathrm{\infty }$$$$\mathrm{diverges}$$

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