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Question Number 207652 by universe last updated on 22/May/24

$$\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}+{x}^{\mathrm{3}} \right){dx}\:\:=\:?{and}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\:\left(\mathrm{1}+{x}^{\mathrm{4}} \right){dx}\:=\:? \\$$$$\:\:\mathrm{and}\:\mathrm{if}\:\mathrm{possible}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{p} \\$$$$\:\mathrm{p}\:\:\:=\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{log}\left(\mathrm{1}+{x}^{{n}} \right){dx}\:=\:?\:\:\:\:\:\:{n}\in\mathbb{N} \\$$

Answered by Berbere last updated on 22/May/24

$${using}\:{x}^{{n}} +\mathrm{1}\:=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({x}−{e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{{n}}} \right)=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}{ln}\left({x}−{e}^{\frac{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}} \right) \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}−{e}^{{i}\left(\frac{\left.\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}\right)} \right){dx}={ln}\left({e}^{{i}\left(\pi+\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)} \left(\mathrm{1}−{xe}^{\frac{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{{n}}} \right)\right) \\$$$${using}\:{principal}\:{log}\:{representation} \\$$$${log}\left({z}\right)={ln}\mid{z}\mid+{iarg}\left({z}\right);{arg}\left({z}\right)\in\left[−\frac{\pi}{\mathrm{2}},\frac{\mathrm{3}\pi}{\mathrm{2}}\left[\right.\right. \\$$$${e}^{{i}\left(\pi+\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)} ={e}^{{i}\left(−\pi+\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)} ;{ln}\left({e}^{{i}\left(−\pi+\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)} \right)={i}\left(−\pi+\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right) \\$$$${p}=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{i}\left(−\pi+\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\pi\right)+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{xe}^{−{i}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\right)\pi} \right){dx} \\$$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}−{xe}^{−{i}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\right)\pi} \right){dx} \\$$$$\left.{l}\left.{n}\left(\mathrm{1}−{xa}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{{a}}\left(\left({ax}−\mathrm{1}\right){ln}\left(\mathrm{1}−{xa}\right)−{ax}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$=\frac{\mathrm{1}}{{a}}\left({a}−\mathrm{1}\right){ln}\left(\mathrm{1}−{a}\right)−\mathrm{1} \\$$$$=\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}{e}^{{i}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\right)\pi} \left[\left({e}^{−{i}\left(\frac{\mathrm{2}{k}+\mathrm{1}}{{n}}\right)\pi} −\mathrm{1}\right){ln}\left(\mathrm{1}−{e}^{−\frac{{i}\pi}{{n}}\left(\mathrm{2}{k}+\mathrm{1}\right)} \right)−\mathrm{1}\right] \\$$$$\\$$$$\\$$$$\\$$$$\\$$