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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

I =∫_0 ^∞ ((sin(sinx))/x)e^(cosx) dx the function under integral is even ⇒ 2I=∫_(−∞) ^(+∞) ((sin(sinx))/x)e^(cosx) dx =Im(∫_(−∞) ^(+∞) e^(isinx+cosx) (dx/x)) ∫_(−∞) ^(+∞) (e^e^(ix) /x)dx ? let ϕ(z) =(e^e^(iz) /z) the unique pole for f is0 ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,o) =2iπ e ⇒2I =2πe ⇒I =π e

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{sin}\left(\mathrm{sinx}\right)}{\mathrm{x}}\mathrm{e}^{\mathrm{cosx}} \mathrm{dx}\:\:\mathrm{the}\:\mathrm{function}\:\mathrm{under}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{even}\:\Rightarrow \\ $$$$\mathrm{2I}=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sin}\left(\mathrm{sinx}\right)}{\mathrm{x}}\mathrm{e}^{\mathrm{cosx}} \mathrm{dx}\:=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\mathrm{e}^{\mathrm{isinx}+\mathrm{cosx}} \frac{\mathrm{dx}}{\mathrm{x}}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{e}^{\mathrm{ix}} } }{\mathrm{x}}\mathrm{dx}\:\:?\:\:\:\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{e}^{\mathrm{iz}} } }{\mathrm{z}}\:\:\mathrm{the}\:\mathrm{unique}\:\mathrm{pole}\:\mathrm{for}\:\mathrm{f}\:\mathrm{is0} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{o}\right)\:\:=\mathrm{2i}\pi\:\mathrm{e}\:\Rightarrow\mathrm{2I}\:=\mathrm{2}\pi\mathrm{e}\:\Rightarrow\mathrm{I}\:=\pi\:\mathrm{e} \\ $$

Commented by mnjuly1970 last updated on 16/Nov/20

very excellent sir max..

$${very}\:{excellent}\:{sir}\:{max}.. \\ $$

Commented by Bird last updated on 17/Nov/20

you are welcome

$${you}\:{are}\:{welcome} \\ $$

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