Question Number 116554 by Bird last updated on 04/Oct/20
$${study}\:{the}\:{integral}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}^{{n}} {x}}{{x}^{{n}} }{dx} \\ $$$${n}\:{integr}\:{and}\:{n}\geqslant\mathrm{2} \\ $$
Answered by Olaf last updated on 05/Oct/20
$$\mathrm{I}_{\mathrm{2}{n}} \:=\:\frac{{n}\pi}{\left(\mathrm{2}{n}\right)!}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{\mathrm{2}{n}} ^{{n}−{k}} \left(−\mathrm{1}\right)^{{n}−{k}} {k}^{\mathrm{2}{n}−\mathrm{1}} \\ $$$$\mathrm{I}_{\mathrm{2}{n}+\mathrm{1}} \:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{2}{n}\right)!}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{\mathrm{2}{n}+\mathrm{1}} ^{{n}−{k}} \left(−\mathrm{1}\right)^{{n}−{k}} \left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}{n}} \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{demonstration}\:\mathrm{is}\:\mathrm{very}\:\mathrm{long}\:\mathrm{and} \\ $$$$\mathrm{not}\:\mathrm{so}\:\mathrm{easy}\:\mathrm{for}\:\mathrm{this}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{apps}. \\ $$