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Question Number 100653 by Ar Brandon last updated on 28/Jun/20

Answered by mathmax by abdo last updated on 28/Jun/20

$$1)\mathrm{f}\left(\mathrm{x}\right)=\frac{\sin \left(\mathrm{nx}\right)}{\mathrm{sinx}}\mathrm{is}\mathrm{continue}\mathrm{on}]0,\frac{\pi }{2}]\Rightarrow \mathrm{integrable}\mathrm{at}\mathrm{V}\left(0\right)\mathrm{we}\mathrm{have}$$$$\mathrm{f}\left(\mathrm{x}\right)\sim \frac{\mathrm{nx}}{\mathrm{x}}=\mathrm{n}\mathrm{so}{\mathrm{I}}_{\mathrm{n}}\mathrm{exist}$$$$2){\mathrm{I}}_{\mathrm{n}}-{\mathrm{I}}_{\mathrm{n}-2}={\int }_0^{\frac{\pi }{2}}\frac{\sin \left(\mathrm{nx}\right)-\sin (\mathrm{n}-2)\mathrm{x}}{\mathrm{sinx}}\mathrm{dx}$$$$\mathrm{sinp}-\mathrm{sinq}=\mathrm{sinp}+\sin (-\mathrm{q})=\cos (\frac{\pi }{2}-\mathrm{p})+\cos (\frac{\pi }{2}+\mathrm{q})$$$$=2\cos \left(\frac{\pi -\mathrm{p}+\mathrm{q}}{2}\right)\cos \left(\frac{-\mathrm{p}-\mathrm{q}}{2}\right)=2\cos (\frac{\pi }{2}-\frac{\mathrm{p}-\mathrm{q}}{2})\cos \left(\frac{\mathrm{p}+\mathrm{q}}{2}\right)$$$$=2\cos \left(\frac{\mathrm{p}+\mathrm{q}}{2}\right)\sin \left(\frac{\mathrm{p}-\mathrm{q}}{2}\right)\Rightarrow \sin \left(\mathrm{nx}\right)-\sin (\mathrm{n}-2)\mathrm{x}=2\cos \left(\frac{\mathrm{nx}+\mathrm{nx}-2\mathrm{x}}{2}\right)\sin \left(\frac{\mathrm{nx}-\mathrm{nx}+2\mathrm{x}}{2}\right)$$$$=2\cos (\mathrm{n}-1)\mathrm{x}\mathrm{sinx}\Rightarrow {\mathrm{I}}_{\mathrm{n}}-{\mathrm{I}}_{\mathrm{n}-2}=2{\int }_0^{\frac{\pi }{2}}\cos (\mathrm{n}-1)\mathrm{x}\mathrm{dx}$$$$=2{\left[\frac{1}{\mathrm{n}-1}\sin (\mathrm{n}-1)\mathrm{x}\right]}_0^{\frac{\pi }{2}}=\frac{2}{\mathrm{n}-1}\sin (\mathrm{n}-1)\frac{\pi }{2}=\frac{2}{\mathrm{n}-1}\sin (\frac{\mathrm{n}\pi }{2}-\frac{\pi }{2})$$$$=-\frac{2}{\mathrm{n}-1}\cos \left(\frac{\mathrm{n}\pi }{2}\right)(\mathrm{n}⩾2)\Rightarrow {\mathrm{I}}_2-{\mathrm{I}}_0=-\frac{2}{1}\cos \left(\pi \right)=2\mathrm{but}{\mathrm{I}}_0=0\Rightarrow {\mathrm{I}}_2=2$$$$3)\mathrm{we}\mathrm{have}{\mathrm{I}}_{\mathrm{n}}-{\mathrm{I}}_{\mathrm{n}-2}=-\frac{2}{\mathrm{n}-1}\cos \left(\frac{\mathrm{n}\pi }{2}\right)\Rightarrow {\mathrm{I}}_{2\mathrm{n}}-{\mathrm{I}}_{2\mathrm{n}-2}=-\frac{2{(-1)}^{\mathrm{n}}}{\mathrm{n}-1}\Rightarrow$$$$\sum _{\mathrm{k}=2}^{\mathrm{n}}({\mathrm{I}}_{2\mathrm{k}}-{\mathrm{I}}_{2\mathrm{k}-2})=2\sum _{\mathrm{k}=2}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{\mathrm{k}-1}\Rightarrow$$$${\mathrm{I}}_4-{\mathrm{I}}_2+{\mathrm{I}}_6-{\mathrm{I}}_4+....{\mathrm{I}}_{2\mathrm{n}}-{\mathrm{I}}_{2\mathrm{n}-2}=2\sum _{\mathrm{k}=2}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{\mathrm{k}-1}\Rightarrow$$$${\mathrm{I}}_{2\mathrm{n}}={\mathrm{I}}_2+2\sum _{\mathrm{k}=2}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{\mathrm{k}-1}=2+2\sum _{\mathrm{k}=1}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{k}}}{\mathrm{k}}$$$${\mathrm{I}}_{2\mathrm{n}+1}-{\mathrm{I}}_{2\mathrm{n}-1}=-\frac{2}{2\mathrm{n}}\cos \left(\frac{(2\mathrm{n}+1)\pi }{2}\right)=-\frac{1}{\mathrm{n}}\cos (\mathrm{n}\pi +\frac{\pi }{2})=0\Rightarrow$$$${\mathrm{I}}_{2\mathrm{n}+1}={\mathrm{I}}_{2\mathrm{n}-1}={\mathrm{I}}_1=\frac{\pi }{2}$$

Commented by mathmax by abdo last updated on 28/Jun/20

$$\mathrm{error}\mathrm{at}{\mathrm{I}}_{2\mathrm{n}}\mathrm{we}\mathrm{have}{\mathrm{I}}_{\mathrm{n}}-{\mathrm{I}}_{\mathrm{n}-2}=\frac{-2{(-1)}^{\mathrm{n}}}{\mathrm{n}-1}\Rightarrow {\mathrm{I}}_{2\mathrm{n}}-{\mathrm{I}}_{2\mathrm{n}-2}=\frac{-2{(-1)}^{\mathrm{n}}}{2\mathrm{n}-1}$$$$\Rightarrow \sum _{\mathrm{k}=1}^{\mathrm{n}}({\mathrm{I}}_{2\mathrm{k}}-{\mathrm{I}}_{2\mathrm{k}-2})=2\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{2\mathrm{k}-1}\Rightarrow$$$${\mathrm{I}}_{2\mathrm{n}}-{\mathrm{I}}_0=2\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{2\mathrm{k}-1}\mathrm{but}{\mathrm{I}}_0=0\Rightarrow {\mathrm{I}}_{2\mathrm{n}}=2\sum _{\mathrm{k}=1}^{\mathrm{n}}\frac{{(-1)}^{\mathrm{k}-1}}{2\mathrm{k}-1}$$$$=2\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\frac{{(-1)}^{\mathrm{k}}}{2\mathrm{k}+1}\mathrm{and}\mathrm{we}\mathrm{see}\mathrm{thst}{\lim }_{\mathrm{n}\to +\mathrm{\infty }}{\mathrm{I}}_{2\mathrm{n}}=\frac{\pi }{2}$$$$$$

Commented by Ar Brandon last updated on 28/Jun/20

Thank you Sir ��

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