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Question-122366




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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