Integration Questions

Question Number 152141 by Ar Brandon last updated on 26/Aug/21

$$\int_{\mathrm{0}} ^{+\infty} \frac{\left(\mathrm{sin}{x}\right)^{\mathrm{2}{n}+\mathrm{1}} }{{x}}{dx}=\frac{\pi}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix} \\$$

Answered by Olaf_Thorendsen last updated on 26/Aug/21

$$\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{sin}^{\mathrm{2}{n}+\mathrm{1}} {x}}{{x}}\:{dx} \\$$$$\mathrm{Let}\:{f}\left({x}\right)\:=\:\mathrm{sin}^{\mathrm{2}{n}} {x} \\$$$$\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{sin}{x}}{{x}}{f}\left({x}\right)\:{dx} \\$$$$\\$$$${f}\:\mathrm{is}\:\mathrm{a}\:\pi−\mathrm{periodic}\:\mathrm{function}.\:\mathrm{We}\:\mathrm{can} \\$$$$\mathrm{apply}\:\mathrm{the}\:\mathrm{Lobachevsky}−\mathrm{Dirchlet} \\$$$$\mathrm{integral}\:\mathrm{formula}\:: \\$$$$\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{sin}{x}}{{x}}{f}\left({x}\right)\:{dx}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {f}\left({x}\right){dx} \\$$$$\Rightarrow\:\mathrm{I}_{{n}} \:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{sin}^{\mathrm{2}{n}} {x}\:{dx}\:=\:{W}_{\mathrm{2}{n}} \:=\:\frac{\pi}{\mathrm{2}}.\frac{\left(\mathrm{2}{n}\right)!}{\left(\mathrm{2}^{{n}} {n}!\right)^{\mathrm{2}} } \\$$$$\left(\mathrm{W}_{\mathrm{2}{n}} \::\:\mathrm{integral}\:\mathrm{of}\:\mathrm{Wallis}\right) \\$$$$\mathrm{I}_{{n}} \:=\:\frac{\pi}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }.\frac{\left(\mathrm{2}{n}\right)!}{{n}!{n}!}\:=\:\frac{\pi}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\mathrm{C}_{{n}} ^{\mathrm{2}{n}} \\$$

Commented by Ar Brandon last updated on 26/Aug/21

$$\mathrm{Gra} ext{\c{c}} \mathrm{cias}\:\mathrm{se}\overset{ } {\mathrm{n}or}\:! \\$$

Commented by puissant last updated on 26/Aug/21

$${hum} \\$$

Answered by Kamel last updated on 26/Aug/21

$${sin}^{\mathrm{2}{n}} \left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{2}{n}−{k}} \left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{k}} {e}^{{ikx}} {e}^{−{i}\left(\mathrm{2}{n}−{k}\right){ix}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{n}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{k}} }{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{k}} {e}^{{ikx}} {e}^{−\left(\mathrm{2}{n}−{k}\right){ix}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{k}} }{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{k}} {e}^{{i}\left(\mathrm{2}{n}−{k}\right){x}} {e}^{−{ikx}} \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{n}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{k}} }{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{k}} \left({e}^{−\left(\mathrm{2}{n}−\mathrm{2}{k}\right){ix}} +{e}^{{i}\left(\mathrm{2}{n}−\mathrm{2}{k}\right){x}} \right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{n}} +\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{k}} }{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }{C}_{\mathrm{2}{n}} ^{{k}} {cos}\left(\mathrm{2}{n}−\mathrm{2}{k}\right) \\$$$$\:\therefore\:\int_{\mathrm{0}} ^{+\infty} \frac{{sin}^{\mathrm{2}{n}+\mathrm{1}} \left({x}\right)}{{x}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{n}} \int_{\mathrm{0}} ^{+\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}+\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+{k}} }{\mathrm{2}^{\mathrm{2}{n}} }{C}_{\mathrm{2}{n}} ^{{k}} \int_{\mathrm{0}} ^{+\infty} \frac{{sin}\left(\left(\mathrm{2}{n}−\mathrm{2}{k}+\mathrm{1}\right){x}\right)−{sin}\left(\left(\mathrm{2}{n}−\mathrm{2}{k}−\mathrm{1}\right){x}\right)}{{x}}{dx} \\$$$$\:{We}\:{have}:\:\:\:\:\:\mathrm{2}{n}−\mathrm{2}{k}+\mathrm{1}>\mathrm{0},\:\mathrm{2}{n}−\mathrm{2}{k}−\mathrm{1}>\mathrm{0}\:\forall{k}=\overline {\mathrm{0},{n}−\mathrm{1}},\:{n}\geqslant\mathrm{1} \\$$$$\:\:\:\therefore\:\int_{\mathrm{0}} ^{+\infty} \frac{{sin}\left({ax}\right)}{{x}}{dx}\overset{{a}>\mathrm{0}} {=}\frac{\pi}{\mathrm{2}} \\$$$${So}\:\::\:\:\int_{\mathrm{0}} ^{+\infty} \frac{{sin}^{\mathrm{2}{n}+\mathrm{1}} \left({x}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} }\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix} \\$$

Commented by Ar Brandon last updated on 26/Aug/21

$$\mathrm{Oh}\:\mathrm{my}!\:\mathrm{Thanks}\:\mathrm{Sir} \\$$