Question Number 113714 by mathdave last updated on 14/Sep/20 | ||
Answered by Olaf last updated on 15/Sep/20 | ||
$$\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}} \mathrm{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx} \\ $$$$\mathrm{I}_{\mathrm{0}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}\right){dx}}{\mathrm{1}+{x}} \\ $$$$\mathrm{I}_{\mathrm{0}} \:=\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}^{\mathrm{2}} \mathrm{2} \\ $$$$\mathrm{I}_{{n}+\mathrm{1}} +\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{n}+\mathrm{1}} +{x}^{{n}} }{\mathrm{1}+{x}}\mathrm{ln}\left(\mathrm{1}+{x}\right){dx} \\ $$$$\mathrm{I}_{{n}+\mathrm{1}} +\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \mathrm{ln}\left(\mathrm{1}+{x}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} {x}^{{k}} \\ $$$$\mathrm{ln}\left(\mathrm{1}+{x}\right)\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}},\:\mid{x}\mid<\mathrm{1} \\ $$$$\mathrm{I}_{{n}+\mathrm{1}} +\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}{dx} \\ $$$$\mathrm{I}_{{n}+\mathrm{1}} +\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{{n}+{k}+\mathrm{1}} }{{k}+\mathrm{1}}{dx} \\ $$$$\mathrm{I}_{{n}+\mathrm{1}} +\mathrm{I}_{{n}} \:=\:\left[\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} {x}^{{n}+{k}+\mathrm{1}} }{\left({k}+\mathrm{1}\right)\left({n}+{k}+\mathrm{1}\right)}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\mathrm{I}_{{n}+\mathrm{1}} +\mathrm{I}_{{n}} \:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{\left({k}+\mathrm{1}\right)\left({n}+{k}+\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)\left({n}+{k}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{{n}}\left(\frac{\mathrm{1}}{{k}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+{k}+\mathrm{1}}\right) \\ $$$$\mathrm{I}_{{n}+\mathrm{1}} +\mathrm{I}_{{n}} \:=\:\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \left(\frac{\mathrm{1}}{{k}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+{k}+\mathrm{1}}\right) \\ $$$$... \\ $$ | ||
Commented by mathdave last updated on 15/Sep/20 | ||
$${well}\:{done}\:{but}\:{wait}\:{ooo}\:{if}\:{integrate}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{k}+{n}+\mathrm{1}} {dx}\:{this}\:{supposed}\:{to}\:{gave} \\ $$$$\frac{{x}^{{k}+{n}+\mathrm{2}} }{{k}+{n}+\mathrm{2}} \\ $$ | ||
Commented by Tawa11 last updated on 06/Sep/21 | ||
$$\mathrm{great}\:\mathrm{sir} \\ $$ | ||