Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 114526 by bobhans last updated on 19/Sep/20

$$\frac{1}{\pi }\underset{0}{\stackrel{\pi }{\int }}{e}^{2\cos \theta }d\theta ?$$

Commented by JDamian last updated on 19/Sep/20

https://youtu.be/-UhFu0g9740

Commented by mathmax by abdo last updated on 19/Sep/20

$$\mathrm{the}\mathrm{problem}\mathrm{how}\mathrm{to}\mathrm{calculate}\mathrm{\Sigma }\frac{1}{{(\mathrm{n}!)}^2}...!$$

Answered by mathmax by abdo last updated on 19/Sep/20

$$\mathrm{I}=\frac{1}{\pi }{\int }_0^{\pi }{\mathrm{e}}^{2\cos \theta }\mathrm{d}\theta \Rightarrow \pi \mathrm{I}={\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{2\cos \theta }\mathrm{d}\theta +{\int }_{\frac{\pi }{2}}^{\pi }{\mathrm{e}}^{2\cos \theta }\mathrm{d}\theta (\to \theta =\frac{\pi }{2}+\mathrm{u})$$$$={\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{2\cos \theta }\mathrm{d}\theta +{\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{-2\mathrm{sinu}}\mathrm{du}={\int }_0^{\frac{\pi }{2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\left(2\cos \theta \right)}^{\mathrm{n}}}{\mathrm{n}!}\mathrm{d}\theta$$$$+{\int }_0^{\frac{\pi }{2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-2\sin \theta )}^{\mathrm{n}}}{\mathrm{n}!}\mathrm{d}\theta =\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{2^{\mathrm{n}}}{\mathrm{n}!}{\int }_0^{\frac{\pi }{2}}{\cos }^{\mathrm{n}}\theta \mathrm{d}\theta$$$$+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-2)}^{\mathrm{n}}}{\mathrm{n}!}{\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}}\theta \mathrm{d}\theta \mathrm{changement}\theta =\frac{\pi }{2}-\mathrm{t}\mathrm{give}$$$${\int }_0^{\frac{\pi }{2}}{\sin }^{\mathrm{n}}\theta \mathrm{d}\theta ={\int }_0^{\frac{\pi }{2}}{\cos }^{\mathrm{n}}\mathrm{t}\mathrm{dt}\Rightarrow$$$$\pi \mathrm{I}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\{\frac{2^{\mathrm{n}}}{\mathrm{n}!}+\frac{{(-2)}^{\mathrm{n}}}{\mathrm{n}!}\}{\int }_0^{\frac{\pi }{2}}{\cos }^{\mathrm{n}}\theta \mathrm{d}\theta \mathrm{let}{\mathrm{w}}_{\mathrm{n}}={\int }_0^{\frac{\pi }{2}}{\cos }^{\mathrm{n}}\theta \mathrm{d}\theta \left(\mathrm{wallis}\mathrm{integral}\right)$$$$\pi \mathrm{I}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{2^{\mathrm{n}}}{\mathrm{n}!}(1+{(-1)}^{\mathrm{n}}){\mathrm{W}}_{\mathrm{n}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{2^{2\mathrm{n}}}{\left(2\mathrm{n}\right)!}\left(2\right){\mathrm{W}}_{2\mathrm{n}}\Rightarrow$$$$\mathrm{I}=\frac{1}{\pi }\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{2^{2\mathrm{n}+1}}{\left(2\mathrm{n}\right)!}{\mathrm{W}}_{2\mathrm{n}}\left({\mathrm{w}}_{2\mathrm{n}}\mathrm{is}\mathrm{known}\mathrm{by}\mathrm{recurrence}\right)$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com