Question Number 130203 by mnjuly1970 last updated on 23/Jan/21
Answered by Dwaipayan Shikari last updated on 23/Jan/21
$${I}\left({a}\right)−{I}\left(\beta\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} }{\left(\mathrm{1}+{x}\right){logx}}−\frac{{x}^{\beta} }{\left(\mathrm{1}+{x}\right){log}\left({x}\right)}{dx} \\ $$$${I}'\left({a}\right)−{I}'\left(\beta\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{a}+{n}+\mathrm{1}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{\beta+{n}+\mathrm{1}} \\ $$$${I}\left({a}\right)−{I}\left(\beta\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \left({log}\left({a}+{n}+\mathrm{1}\right)−{log}\left(\beta+{n}+\mathrm{1}\right)\right)+{C} \\ $$$${I}\left({a}\right)−{I}\left(\beta\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {log}\left(\frac{{a}+{n}+\mathrm{1}}{\beta+{n}+\mathrm{1}}\right)\:\:\:\:\left({C}=\mathrm{0}\:\:{when}\:{a}=\mathrm{0}\:,\:\beta=\mathrm{0}\right) \\ $$$${I}\left(\mathrm{0}\right)−{I}\left(\mathrm{1}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} {log}\left(\frac{{n}+\mathrm{1}}{{n}+\mathrm{2}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {log}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}+\mathrm{1}}\right) \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{a}} }{\left(\mathrm{1}+{x}\right){log}\left({x}\right)}{dx}\:\Rightarrow{I}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{x}^{{a}} }{\left(\mathrm{1}+{x}\right)}{dx}=−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} −{x}^{{a}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}+\mathrm{1}} }{\mathrm{1}−{x}^{\mathrm{2}} }−\frac{{x}^{{a}} }{\mathrm{1}−{x}^{\mathrm{2}} }{dx}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{u}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−{u}}−\frac{\mathrm{1}−{u}^{\frac{{a}}{\mathrm{2}}} }{\mathrm{1}−{u}}{du}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{a}+\mathrm{2}}{\mathrm{2}}\right)\right) \\ $$$${I}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{a}+\mathrm{2}}{\mathrm{2}}\right)\:{da}\:={log}\left(\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)}\right)+{C} \\ $$$${a}=\mathrm{0}\:\:\:\:{C}=−\frac{\mathrm{1}}{\mathrm{2}}{log}\left(\pi\right) \\ $$$${I}\left({a}\right)={log}\left(\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)\sqrt{\pi}}\right)\Rightarrow{I}\left(\mathrm{1}\right)={log}\left(\frac{\mathrm{2}}{\pi}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 23/Jan/21
$${tayeb}\:{allah}\:{mr}\:{dwaypayan} \\ $$$$\:{nice}\:{way}... \\ $$
Commented by Dwaipayan Shikari last updated on 23/Jan/21
$${With}\:{pleasure} \\ $$
Answered by Dwaipayan Shikari last updated on 23/Jan/21
$${Another}\:{way} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {log}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)=−{log}\left(\frac{\mathrm{2}}{\mathrm{1}}.\frac{\mathrm{2}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{2}}.\frac{\mathrm{3}}{\mathrm{4}}...\right)={log}\left(\frac{\mathrm{2}}{\pi}\right) \\ $$