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Question Number 112366 by mathdave last updated on 07/Sep/20

Answered by mathmax by abdo last updated on 07/Sep/20

$$\mathrm{let}{\mathrm{S}}_{\mathrm{n}}=\mathrm{n}\sum _{\mathrm{k}=1}^{\mathrm{n}-1}\frac{\ln (\mathrm{k}+\mathrm{n})-\ln \left(\mathrm{n}\right)}{{\mathrm{n}}^2+{\mathrm{k}}^2}\Rightarrow$$$${\mathrm{S}}_{\mathrm{n}}=\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}-1}\frac{\ln \left(\frac{\mathrm{n}+\mathrm{k}}{\mathrm{n}}\right)}{1+\frac{{\mathrm{k}}^2}{{\mathrm{n}}^2}}=\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=1}^{\mathrm{n}-1}\frac{\ln (1+\frac{\mathrm{k}}{\mathrm{n}})}{1+{\left(\frac{\mathrm{k}}{\mathrm{n}}\right)}^2}\mathrm{so}{\mathrm{S}}_{\mathrm{n}}\mathrm{is}\mathrm{a}\mathrm{Rieman}\mathrm{sum}$$$$\mathrm{and}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}={\int }_0^1\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}\mathrm{changement}\mathrm{x}=\mathrm{tant}\mathrm{give}$$$${\int }_0^1\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}={\int }_0^{\frac{\pi }{4}}\frac{\ln (1+\mathrm{tant})}{1+{\tan }^2\mathrm{t}}(1+{\tan }^2\mathrm{t})\mathrm{dt}={\int }_0^{\frac{\pi }{4}}\ln (1+\tan \left(\mathrm{t}\right))\mathrm{dt}$$$${=}_{\mathrm{t}=\frac{\pi }{4}-\mathrm{u}}{\int }_0^{\frac{\pi }{4}}\ln (1+\tan (\frac{\pi }{4}-\mathrm{u}))\mathrm{du}={\int }_0^{\frac{\pi }{4}}\ln (1+\frac{1-\mathrm{tanu}}{1+\mathrm{tanu}})\mathrm{du}$$$$={\int }_0^{\frac{\pi }{4}}\ln \left(\frac{2}{1+\mathrm{tanu}}\right)\mathrm{du}=\frac{\pi }{4}\ln \left(2\right)-{\int }_0^{\frac{\pi }{4}}\ln (1+\mathrm{tanu})\mathrm{du}\Rightarrow$$$$2{\int }_0^{\frac{\pi }{4}}\ln (1+\mathrm{tanu})\mathrm{du}=\frac{\pi }{4}\ln \left(2\right)\Rightarrow {\int }_0^{\frac{\pi }{4}}\ln (1+\mathrm{tanu})\mathrm{du}=\frac{\pi }{8}\ln \left(2\right)$$$$={\int }_0^1\frac{\ln (1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{S}}_{\mathrm{n}}$$

Commented by mathdave last updated on 08/Sep/20

$$niceidea$$

Commented by mathmax by abdo last updated on 08/Sep/20

$$\mathrm{thanks}$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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