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Question Number 39787 by abdo mathsup 649 cc last updated on 10/Jul/18

calculste I_λ = ∫_(−∞) ^(+∞) ((cos(λx^n ))/(1+x^2 )) dx with λ from R and n integr natural 2) find the vslue of ∫_(−∞) ^(+∞) ((cos(3 x^9 ))/(1+x^2 )) dx .

$${calculste}\:\:{I}_{\lambda} \:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\lambda{x}^{{n}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\:{with} \\ $$$$\lambda\:{from}\:{R}\:{and}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{vslue}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{3}\:{x}^{\mathrm{9}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:. \\ $$

Commented by abdo mathsup 649 cc last updated on 11/Jul/18

1) we have I_λ = Re( ∫_(−∞) ^(+∞) (e^(iλx^n ) /(1+x^2 ))dx) let ϕ(z) = (e^(iλz^n ) /(1+z^2 )) ϕ(z)= (e^(iλz^n ) /((z−i)(z+i))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (z−i)ϕ(z) =lim_(z→i) (e^(iλz^n ) /(z+i)) = (e^(λ i^(n+1) ) /(2i)) =(e^(λ{cos((((n+1)π)/2) +isin((((n+1)π)/2))}) /(2i)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ e^(λcos((((n+1)π)/2))) ((cos)(λsin((((n+1)π)/2))) +isin(λsin((((n+1)π)/2))))/(2i)) ⇒I_λ = π e^(λ cos((((n+1)π)/2))) cos{λsin((((n+1)π)/2))} .

$$\left.\mathrm{1}\right)\:\:{we}\:{have}\:{I}_{\lambda} =\:{Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\lambda{x}^{{n}} } }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\right)\:\:{let} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\lambda{z}^{{n}} } }{\mathrm{1}+{z}^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)=\:\frac{{e}^{{i}\lambda{z}^{{n}} } }{\left({z}−{i}\right)\left({z}+{i}\right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)\:={lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right) \\ $$$$={lim}_{{z}\rightarrow{i}} \:\frac{{e}^{{i}\lambda{z}^{{n}} } }{{z}+{i}}\:=\:\frac{{e}^{\lambda\:{i}^{{n}+\mathrm{1}} } }{\mathrm{2}{i}}\:=\frac{{e}^{\lambda\left\{{cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\:+{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\right\}\right.} }{\mathrm{2}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{e}^{\lambda{cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)} \frac{\left.{cos}\right)\left(\lambda{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\right)\:\:+{isin}\left(\lambda{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\right)}{\mathrm{2}{i}} \\ $$$$\Rightarrow{I}_{\lambda} =\:\pi\:\:{e}^{\lambda\:{cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)} \:{cos}\left\{\lambda{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\right\}\:. \\ $$

Commented by math khazana by abdo last updated on 12/Jul/18

2) let take λ=3 and n=9 we get ∫_(−∞) ^(+∞) ((cos(3x^9 ))/(1+x^2 ))dx =πe^(3cos(5π)) cos{3 sin(5π)} = π e^(−3) =(π/e^3 ) .

$$\left.\mathrm{2}\right)\:{let}\:{take}\:\lambda=\mathrm{3}\:{and}\:{n}=\mathrm{9}\:{we}\:{get} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{3}{x}^{\mathrm{9}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\pi{e}^{\mathrm{3}{cos}\left(\mathrm{5}\pi\right)} \:{cos}\left\{\mathrm{3}\:{sin}\left(\mathrm{5}\pi\right)\right\} \\ $$$$=\:\pi\:{e}^{−\mathrm{3}} \:\:=\frac{\pi}{{e}^{\mathrm{3}} }\:\:. \\ $$$$ \\ $$

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