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Question Number 45063 by rahul 19 last updated on 08/Oct/18

$${\int }_0^{2\pi }{e}^{\cos \theta }\cos \left(\sin \theta \right)d\theta =?$$

Commented by maxmathsup by imad last updated on 08/Oct/18

$$I=Re\left({\int }_0^{2\pi }{e}^{cos\theta +isin\theta }d\theta \right)but{e}^{cos\theta +isin\theta }=e^{e^{i\theta }}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(e^{i\theta }\right)}^n}{n!}$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{e^{in\theta }}{n!}=\sum _{n=0}^{\mathrm{\infty }}\frac{cos\left(n\theta \right)}{n!}+i\sum _{n=0}^{\mathrm{\infty }}\frac{sin\left(n\theta \right)}{n!}\Rightarrow$$$${\int }_0^{2\pi }{e}^{e^{i\theta }}d\theta =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{\int }_0^{2\pi }cos\left(n\theta \right)d\theta +i\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}{\int }_0^{2\pi }sin\left(n\theta \right)d\theta$$$$=2\pi +\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{2\pi }cos\left(n\theta \right)d\theta +i\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n!}{\int }_0^{2\pi }sin\left(n\theta \right)d\theta but$$$${\int }_0^{2\pi }cos\left(n\theta \right)d\theta ={\left[\frac{1}{n}sin\left(n\theta \right)\right]}_0^{2\pi }=0(n⩾1)also$$$${\int }_0^{2\pi }sin\left(n\theta \right)d\theta ={[-\frac{1}{n}cos\left(n\theta \right)]}_0^{2\pi }=0(n⩾1)\Rightarrow {\int }_0^{2\pi }{e}^{e^{i\theta }}d\theta =2\pi \Rightarrow$$$${\int }_0^{2\pi }{e}^{cos\theta }cos\left(sin\theta \right)d\theta =2\pi .$$

Commented by tanmay.chaudhury50@gmail.com last updated on 08/Oct/18

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