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Question Number 122366 by rs4089 last updated on 16/Nov/20

Answered by mathmax by abdo last updated on 16/Nov/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\sin \left(\mathrm{sinx}\right)}{\mathrm{x}}{\mathrm{e}}^{\mathrm{cosx}}\mathrm{dx}\mathrm{the}\mathrm{function}\mathrm{under}\mathrm{integral}\mathrm{is}\mathrm{even}\Rightarrow$$$$2\mathrm{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sin \left(\mathrm{sinx}\right)}{\mathrm{x}}{\mathrm{e}}^{\mathrm{cosx}}\mathrm{dx}=\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\mathrm{e}}^{\mathrm{isinx}+\mathrm{cosx}}\frac{\mathrm{dx}}{\mathrm{x}}\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{ix}}}}{\mathrm{x}}\mathrm{dx}?\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{{\mathrm{e}}^{{\mathrm{e}}^{\mathrm{iz}}}}{\mathrm{z}}\mathrm{the}\mathrm{unique}\mathrm{pole}\mathrm{for}\mathrm{f}\mathrm{is}0$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{o})=2\mathrm{i}\pi \mathrm{e}\Rightarrow 2\mathrm{I}=2\pi \mathrm{e}\Rightarrow \mathrm{I}=\pi \mathrm{e}$$

Commented by mnjuly1970 last updated on 16/Nov/20

$$veryexcellentsirmax..$$

Commented by Bird last updated on 17/Nov/20

$$youarewelcome$$

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