Differentiation Questions

Question Number 165194 by mnjuly1970 last updated on 27/Jan/22

$$\\$$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {ln}\:\left(\:\mathrm{1}+\:{cos}\:\left({x}\right)\right).{cos}\:\left({nx}\:\right){dx}=? \\$$

Answered by mindispower last updated on 27/Jan/22

$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){cos}\left({nx}\right){dx} \\$$$${x}=\mathrm{2}{y} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left({y}\right)\right){cos}\left(\mathrm{2}{ny}\right){dy} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left({y}\right)\right){cos}\left(\mathrm{2}{ny}\right){dy}+\mathrm{2}\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} {ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left({y}\right)\right){cos}\left(\mathrm{2}{ny}\right){dy} \\$$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}{ny}\right){ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left({y}\right)\right){dy} \\$$$$=\mathrm{4}{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}{ny}\right)+\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}{ny}\right){ln}\left({cos}\left({y}\right)\right){dy} \\$$$$\left.=−\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}{cos}\left(\mathrm{2}{ky}\right){cos}\left(\mathrm{2}{ny}\right){dy}−\mathrm{8}{ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\mathrm{2}{ny}\right){dy} \\$$$$=−\mathrm{4}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}\left({k}+{n}\right){x}\right)+{cos}\left(\mathrm{2}\left({k}−{n}\right){x}\right){dx} \\$$$${n}\in\mathbb{N},{k}+{n}\neq\mathrm{0};\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}\left({k}+{n}\right){x}\right){dx}=\mathrm{0} \\$$$$=−\mathrm{4}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}\left(\mathrm{2}\left({k}−{n}\right){x}\right){dx}=−\mathrm{4}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\$$$$=−\mathrm{2}\pi\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}=\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\pi,\forall{n}\geqslant\mathrm{1} \\$$$${n}=\mathrm{0} \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} {ln}\left(+{cos}\left({x}\right)\right){dx}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {ln}\left(\mathrm{2}\right)+{ln}\left({cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\$$$$=\mathrm{2}\pi{ln}\left(\mathrm{2}\right)+\int_{\mathrm{0}} ^{\pi} {ln}\left({cos}^{\mathrm{2}} \left({x}\right)\right)\mathrm{2}{dx} \\$$$$=\mathrm{2}\pi{ln}\left(\mathrm{2}\right)+\mathrm{8}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({x}\right)\right){dx} \\$$$$=−\mathrm{2}\pi{ln}\left(\mathrm{2}\right) \\$$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} {ln}\left(\mathrm{1}+{cos}\left({x}\right)\right){cos}\left({x}\right){dx}=\begin{cases}{−\mathrm{2}\pi{ln}\left(\mathrm{2}\right)\:,{n}=\mathrm{0}}\\{\mathrm{2}\pi.\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}};{n}\geqslant\mathrm{1}}\end{cases} \\$$$$\\$$

Commented by mnjuly1970 last updated on 27/Jan/22

$${very}\:{nice}\:\:..\:{really}\:{very}\:{nice}\:{solution} \\$$$${sir}\:{power}\:\:{thanks}\:{alot} \\$$

Commented by mindispower last updated on 27/Jan/22

$${thanxSir}\:{withe}\:{Pleasur} \\$$$${Have}\:{a}\:{nice}\:{Day} \\$$

Answered by aleks041103 last updated on 27/Jan/22

$${IBP}: \\$$$${I}_{{n}} =\left[\frac{\mathrm{1}}{{n}}{ln}\left(\mathrm{1}+{cos}\left({x}\right)\right){sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\mathrm{2}\pi} +\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left({x}\right){sin}\left({nx}\right)}{\mathrm{1}+{cos}\left({x}\right)}{dx} \\$$$$\Rightarrow{I}_{{n}} =\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left({x}\right){sin}\left({nx}\right)}{\mathrm{1}+{cos}\left({x}\right)}{dx} \\$$$${let}\:{z}={e}^{{ix}} \\$$$$\Rightarrow{dz}={ie}^{{ix}} {dx}={izdx}\Rightarrow{dx}=\frac{{dz}}{{iz}} \\$$$${sin}\left({x}\right)=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}=\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{iz}} \\$$$${sin}\left({nx}\right)=\frac{{z}^{\mathrm{2}{n}} −\mathrm{1}}{\mathrm{2}{iz}^{{n}} } \\$$$${cos}\left({x}\right)=\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}}=\frac{{z}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{z}} \\$$$$\Rightarrow\frac{{sin}\left({x}\right){sin}\left({nx}\right)}{\mathrm{1}+{cos}\left({x}\right)}{dx}=\frac{\left({z}^{\mathrm{2}} −\mathrm{1}\right)\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)}{−\mathrm{4}{iz}^{{n}+\mathrm{2}} \left(\mathrm{1}+\frac{{z}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{z}}\right)}{dz}= \\$$$$={i}\frac{\left({z}−\mathrm{1}\right)\left({z}+\mathrm{1}\right)\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\mathrm{2}{z}^{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)^{\mathrm{2}} }{dz}=\frac{{i}}{\mathrm{2}}\:\frac{\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)}{{z}^{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)}{dz} \\$$$${Also}\:{when}\:{x}\in\left[\mathrm{0},\mathrm{2}\pi\right],\:{z}={e}^{{ix}} \in\Gamma:\mid{z}\mid=\mathrm{1}. \\$$$$\Rightarrow{I}_{{n}} =\frac{{i}}{\mathrm{2}{n}}\underset{\Gamma} {\oint}\frac{\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)}{{z}^{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)}{dz}=\frac{{i}}{\mathrm{2}{n}}\underset{\Gamma} {\oint}{f}\left({z}\right){dz} \\$$$${This}\:{can}\:{be}\:{evaluated}\:{using}\:{the} \\$$$${Residue}\:{theorem}. \\$$$${Poles}\:{of}\:{f}\left({z}\right): \\$$$$\left.\mathrm{1}\right){z}=−\mathrm{1} \\$$$$\underset{{z}\rightarrow−\mathrm{1}} {{lim}}\frac{\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)}{{z}^{{n}+\mathrm{1}} \left({z}+\mathrm{1}\right)}=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \underset{{z}\rightarrow−\mathrm{1}} {{lim}}\frac{{z}^{\mathrm{2}{n}} −\mathrm{1}}{{z}+\mathrm{1}}= \\$$$$\overset{{l}'{H}} {=}\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \underset{{z}\rightarrow−\mathrm{1}} {{lim}}\frac{\mathrm{2}{nz}^{\mathrm{2}{n}−\mathrm{1}} }{\mathrm{1}}=\mathrm{4}{n}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \nrightarrow\pm\infty \\$$$$\Rightarrow{z}=−\mathrm{1}\:{is}\:{not}\:{a}\:{pole}! \\$$$$\left.\mathrm{2}\right)\:{z}=\mathrm{0}\:−\:{obv}.\:{a}\:{pole}\:{of}\:\left({n}+\mathrm{1}\right)−{th}\:{order}. \\$$$$\Rightarrow\underset{\Gamma} {\oint}{f}\left({z}\right){dz}=\mathrm{2}\pi{iRes}\left({f},\mathrm{0}\right) \\$$$$\Rightarrow{I}_{{n}} =−\frac{\pi}{{n}}{Res}\left({f},\mathrm{0}\right) \\$$$$\\$$$${Res}\left({f},\mathrm{0}\right)=\frac{\mathrm{1}}{{n}!}\underset{{z}\rightarrow\mathrm{0}} {{lim}}\left[\frac{{d}^{{n}} }{{dz}^{{n}} }\left({z}^{{n}+\mathrm{1}} {f}\left({z}\right)\right)\right]= \\$$$$=\frac{\mathrm{1}}{{n}!}\underset{{z}\rightarrow\mathrm{0}} {{lim}}\left[\frac{{d}^{{n}} }{{dz}^{{n}} }\left(\frac{\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\left({z}+\mathrm{1}\right)}\right)\right] \\$$$$\frac{\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\left({z}+\mathrm{1}\right)}=\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}= \\$$$$=\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)\left(\mathrm{1}−\frac{\mathrm{2}}{{z}+\mathrm{1}}\right)= \\$$$$={z}^{\mathrm{2}{n}} +\frac{\mathrm{2}}{{z}+\mathrm{1}}−\mathrm{1}−\frac{\mathrm{2}{z}^{\mathrm{2}{n}} }{{z}+\mathrm{1}} \\$$$$\Rightarrow\frac{{d}^{{n}} }{{dz}^{{n}} }\left(\frac{\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}{n}} −\mathrm{1}\right)}{\left({z}+\mathrm{1}\right)}\right)= \\$$$$=\frac{\left(\mathrm{2}{n}\right)!}{{n}!}{z}^{{n}} +\mathrm{2}\left(−\mathrm{1}\right)^{{n}} {n}!\frac{\mathrm{1}}{\left({z}+\mathrm{1}\right)^{{n}+\mathrm{1}} }−\mathrm{2}\frac{{d}^{{n}} }{{dz}^{{n}} }\left(\frac{{z}^{\mathrm{2}{n}} }{{z}+\mathrm{1}}\right) \\$$$$\Rightarrow{Res}\left(\mathrm{0},{f}\right)=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} −\frac{\mathrm{2}}{{n}!}\:\frac{{d}^{{n}} }{{dz}^{{n}} }\left(\frac{{z}^{\mathrm{2}{n}} }{{z}+\mathrm{1}}\right)_{{z}=\mathrm{0}} \\$$$${Since}\:{we}\:{want}\:\:\frac{{d}^{{n}} }{{dz}^{{n}} }\left(\frac{{z}^{\mathrm{2}{n}} }{{z}+\mathrm{1}}\right)\:{at}\:{z}=\mathrm{0}\:\left(\mid{z}\mid=\mathrm{0}<\mathrm{1}\right)\:{we}\:{can} \\$$$${use}\:\frac{\mathrm{1}}{\mathrm{1}+{z}}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{z}\right)^{{i}} \\$$$$\Rightarrow\:\frac{{d}^{{n}} }{{dz}^{{n}} }\left(\frac{{z}^{\mathrm{2}{n}} }{{z}+\mathrm{1}}\right)_{{z}=\mathrm{0}} =\:\frac{{d}^{{n}} }{{dz}^{{n}} }\left(\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−{z}\right)^{{i}+\mathrm{2}{n}} \right)_{{z}=\mathrm{0}} = \\$$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{n}+{i}\right)!}{\left({n}+{i}\right)!}\left(\left(−{z}\right)^{{i}+{n}} \right)_{{z}=\mathrm{0}} \\$$$${since}\:{for}\:{i}\geqslant\mathrm{0}\:{and}\:{n}\geqslant\mathrm{1},\:{i}+{n}\geqslant\mathrm{1}\:{then} \\$$$$\left(\left(−{z}\right)^{{i}+{n}} \right)_{{z}=\mathrm{0}} =\mathrm{0} \\$$$$\Rightarrow\frac{{d}^{{n}} }{{dz}^{{n}} }\left(\frac{{z}^{\mathrm{2}{n}} }{{z}+\mathrm{1}}\right)_{{z}=\mathrm{0}} =\:\mathrm{0} \\$$$$\Rightarrow{Res}\left(\mathrm{0},{f}\right)=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \\$$$$\Rightarrow{I}_{{n}} =\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \pi}{{n}} \\$$$$\\$$

Commented by mindispower last updated on 28/Jan/22

$${nice}\:{Solution}\:{Sir} \\$$