Question Number 216722 by issac last updated on 17/Feb/25
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Answered by MrGaster last updated on 17/Feb/25
![∫_0 ^∞ ((1/ℓ)−(1/p))dℓ=∫_0 ^∞ (((p−ℓ)/(ℓp)))dℓ ∫_0 ^∞ ((1/ℓ)−(1/(ℓ+1))+(1/(ℓ+2))+(1/(ℓ+3))+…)dℓ=lim_(N→∞) [ln(ℓ+1)ln(ℓ+2)+ln(ℓ+3)+…)]_0 ^N lim_(N→∞) [ln(N)−ln(N+1)(N+2)(N+3)…)]=lim_(N→∞) [ln(N)−ln(N!)] lim_(N→∞) [ln(N)−(N ln(N)−N+(1/2)ln(2πN)+O((1/N)))]=∞ diverges](Q216725.png)
$$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\ell}−\frac{\mathrm{1}}{{p}}\right){d}\ell=\int_{\mathrm{0}} ^{\infty} \left(\frac{{p}−\ell}{\ell{p}}\right){d}\ell \\ $$$$\left.\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\ell}−\frac{\mathrm{1}}{\ell+\mathrm{1}}+\frac{\mathrm{1}}{\ell+\mathrm{2}}+\frac{\mathrm{1}}{\ell+\mathrm{3}}+\ldots\right){d}\ell=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{ln}\left(\ell+\mathrm{1}\right)\mathrm{ln}\left(\ell+\mathrm{2}\right)+\mathrm{ln}\left(\ell+\mathrm{3}\right)+\ldots\right)\right]_{\mathrm{0}} ^{{N}} \\ $$$$\left.\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{ln}\left({N}\right)−\mathrm{ln}\left({N}+\mathrm{1}\right)\left({N}+\mathrm{2}\right)\left({N}+\mathrm{3}\right)\ldots\right)\right]=\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{ln}\left({N}\right)−\mathrm{ln}\left({N}!\right)\right] \\ $$$$\underset{{N}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{ln}\left({N}\right)−\left({N}\:\mathrm{ln}\left({N}\right)−{N}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\pi{N}\right)+{O}\left(\frac{\mathrm{1}}{{N}}\right)\right)\right]=\infty \\ $$$$\mathrm{diverges} \\ $$