Question Number 130135 by mathmax by abdo last updated on 22/Jan/21
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{6}} }\mathrm{dx} \\ $$
Answered by Dwaipayan Shikari last updated on 22/Jan/21
$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}+\mathrm{1}−\mathrm{1}} \:}{\left(\mathrm{1}+{x}\right)^{\mathrm{6}+{a}+\mathrm{1}−{a}−\mathrm{1}} }{dx}\:\Rightarrow{I}\left({a}\right)=\frac{\Gamma\left({a}+\mathrm{1}\right)\Gamma\left(\mathrm{5}−{a}\right)}{\Gamma\left(\mathrm{6}\right)} \\ $$$${I}'\left({a}\right)=\frac{\mathrm{1}}{\mathrm{120}}\left(\Gamma'\left({a}+\mathrm{1}\right)\Gamma\left(\mathrm{5}−{a}\right)−\Gamma'\left(\mathrm{5}−{a}\right)\Gamma\left({a}+\mathrm{1}\right)\right) \\ $$$${I}'\left(\mathrm{0}\right)=−\frac{\gamma}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{120}}\left(−\mathrm{24}\gamma+\mathrm{24}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$$$=−\frac{\mathrm{5}}{\mathrm{12}} \\ $$$${Means}\:{in}\:\:{General}\:\int_{\mathrm{0}} ^{\infty} \frac{{logx}}{\left(\mathrm{1}+{x}\right)^{{n}} }{dx}=−\frac{\mathrm{1}}{{n}}\underset{{n}=\mathrm{1}} {\overset{{n}−\mathrm{2}} {\sum}}\frac{\mathrm{1}}{{n}}\: \\ $$