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Question Number 128025 by Algoritm last updated on 03/Jan/21

Answered by Olaf last updated on 03/Jan/21

$$\mathrm{Let}{\mathrm{\Phi }}_n={\int }_0^{2\pi }{e}^{\cos \theta }\cos (n\theta -\sin \theta )d\theta$$$$\mathrm{Let}{\mathrm{\Omega }}_n={\int }_0^{2\pi }{e}^{\cos \theta }{e}^{i(n\theta -\sin \theta )}d\theta$$$${\mathrm{\Omega }}_n={\int }_0^{2\pi }{e}^{e^{-i\theta }}{e}^{in\theta }d\theta$$$${\mathrm{\Omega }}_n=i{\int }_0^{2\pi }(-{ie}^{-i\theta }{e}^{e^{-i\theta }})e^{i(n+1)\theta }d\theta$$$${\mathrm{\Omega }}_n=i{\left[e^{e^{-i\theta }}{e}^{i(n+1)\theta }\right]}_0^{2\pi }-i{\int }_0^{2\pi }{e}^{e^{-i\theta }}i(n+1)e^{i(n+1)\theta }d\theta$$$${\mathrm{\Omega }}_n=(n+1){\mathrm{\Omega }}_{n+1}$$$${\mathrm{\Omega }}_{n+1}=\frac{{\mathrm{\Omega }}_n}{n+1}\left(1\right)$$$${\mathrm{\Omega }}_0={\int }_0^{2\pi }{e}^{e^{-i\theta }}d\theta$$$${\mathrm{\Omega }}_0={\int }_{\mathcal{C}}{e}^{\stackrel{-}{z}}dz(\mathcal{C}:\mathrm{trigonometric}\mathrm{circle})$$$${\mathrm{\Omega }}_0={\int }_{\mathcal{C}}{e}^{z}dz=2\pi r(\mathrm{with}r=1)$$$${\mathrm{\Omega }}_0=2\pi \left(2\right)$$$$\left(1\right)\mathrm{and}\left(2\right):{\mathrm{\Omega }}_n=\frac{2\pi }{n!}\left(\mathrm{real}\mathrm{number}\right)$$$${\mathrm{\Phi }}_n=\mathrm{Re}\left({\mathrm{\Omega }}_n\right)={\mathrm{\Omega }}_n=\frac{2\pi }{n!}$$

Commented by mindispower last updated on 03/Jan/21

$$nicewaysir$$

Answered by mindispower last updated on 03/Jan/21

$$={\int }_0^{2\pi }Re\left(e^{in\theta +e^{-i\theta }}\right)d\theta$$$$=Re{\int }_0^{2\pi }{e}^{-in\theta +e^{i\theta }}d\theta ,z=e^{i\theta }\Rightarrow dz={ie}^{i\theta }d\theta$$$$=Re{\int }_C\frac{-{ie}^z}{z^{n+1}}dz$$$$=Re(.2i\pi Res(-i\frac{e^z}{z^{n+1}},0))$$$$=Re.2i\pi {\partial }_z^n.\frac{z^{n+1}}{n!}(-i\frac{e^z}{z^{n+1}},z=0)$$$$=Re(\frac{2\pi }{n!}.e^0)=\frac{2\pi }{n!}$$

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