Question Number 191049 by Mingma last updated on 16/Apr/23
Commented by Mingma last updated on 17/Apr/23
Nice solution, sir!
Answered by aleks041103 last updated on 17/Apr/23
$${cos}\left({sin}\left({x}\right)\right)={Re}\left({exp}\left({i}\:{sin}\left({x}\right)\right)\right) \\ $$$$\Rightarrow{cos}\left({sin}\left({x}\right)\right){exp}\left({cos}\left({x}\right)\right)= \\ $$$$={Re}\left({exp}\left({cos}\left({x}\right)+{i}\:{sin}\left({x}\right)\right)\right)= \\ $$$$={Re}\left({e}^{{e}^{{ix}} } \right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {cos}\left({sin}\left({x}\right)\right){exp}\left({cos}\left({x}\right)\right){dx}= \\ $$$$={Re}\left(\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {e}^{{e}^{{ix}} } {dx}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {e}^{{e}^{{ix}} } {dx}=? \\ $$$${z}={e}^{{ix}} \Rightarrow{dz}={i}\:{z}\:{dx}\:\Rightarrow\:{dx}=−{i}\frac{{dz}}{{z}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {e}^{{e}^{{ix}} } {dx}=−{i}\oint_{\Gamma} \frac{{e}^{{z}} }{{z}}{dz} \\ $$$${where}\:\Gamma:\mid{z}\mid=\mathrm{1} \\ $$$${By}\:{residue}\:{theorem}: \\ $$$$\oint_{\Gamma} \frac{{e}^{{z}} }{{z}}{dz}=\mathrm{2}\pi{i}\:{Res}\left({z}=\mathrm{0}\right) \\ $$$${Res}\left({z}=\mathrm{0}\right)=\underset{{z}\rightarrow\mathrm{0}} {{lim}}\frac{{e}^{{z}} }{{z}}\left({z}−\mathrm{0}\right)=\mathrm{1}\Rightarrow \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {e}^{{e}^{{ix}} } {dx}=−{i}\left(\mathrm{2}\pi{i}\right)=\mathrm{2}\pi \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {cos}\left({sin}\left({x}\right)\right){exp}\left({cos}\left({x}\right)\right){dx}=\mathrm{2}\pi \\ $$