Question Number 132928 by metamorfose last updated on 17/Feb/21
$$\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{k}}\right) \\ $$
Commented by Olaf last updated on 17/Feb/21
$${sorry}\:{sir},\:{I}\:{deleted}\:{my}\:{answer}. \\ $$$${it}\:{was}\:{wrong}. \\ $$
Answered by mnjuly1970 last updated on 17/Feb/21
$$\:\:{S}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{k}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{{x}+{k}} \\ $$$$\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{{x}+{k}}{dx} \\ $$$$\:\underset{{convergent}} {\overset{{conditional}} {=}}\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \Sigma\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}+\frac{{k}}{\mathrm{2}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\underset{{k}=} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}+{k}}\:−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+{k}}\right\}{dx} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{−\gamma+\Sigma\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}+{k}}−\left(−\gamma+\Sigma\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{\frac{{x}}{\mathrm{2}}+{k}}\right)\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\boldsymbol{\psi}\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\boldsymbol{\psi}\left(\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)\right\}{dx} \\ $$$${note}:\:\boldsymbol{\psi}\left({z}+\mathrm{1}\right)=−\boldsymbol{\gamma}+\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+{z}}\right) \\ $$$$\therefore\:{S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{ln}\left(\boldsymbol{\Gamma}\left(\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right)−\mathrm{2}{ln}\left(\boldsymbol{\Gamma}\left(\frac{{x}}{\mathrm{2}}+\mathrm{1}\right)\right)_{\mathrm{0}} ^{\mathrm{1}} \right. \\ $$$$=−{ln}\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−{ln}\left(\boldsymbol{\Gamma}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right)= \\ $$$$=−{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\pi}\:.\sqrt{\pi}\:\right)={ln}\left(\frac{\mathrm{2}}{\pi}\right)…\checkmark\checkmark\checkmark \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}−{x}}{\left(\mathrm{1}+{x}\right){ln}\left({x}\right)}{dx}\:\:\:\:\:.{m}.{n}.{july}.\mathrm{1970}. \\ $$$$\:\: \\ $$
Commented by metamorfose last updated on 17/Feb/21
$${thanks}\:,\:{it}'{s}\:{an}\:{intresting}\:{solution} \\ $$
Commented by mnjuly1970 last updated on 17/Feb/21
$$\:{grateful}\:{sir}\:….. \\ $$$$ \\ $$