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Question Number 48043 by maxmathsup by imad last updated on 18/Nov/18

let f(α) =∫_(−∞) ^(+∞) ((cos(αx^3 ))/(x^2 +8)) dx 1)calculate f(α) 2) calculate ∫_(−∞) ^(+∞) ((cos(2x^3 ))/(x^2 +8))dx .

$${let}\:{f}\left(\alpha\right)\:=\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\alpha{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}\:{dx} \\ $$$$\left.\mathrm{1}\right){calculate}\:{f}\left(\alpha\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left(\mathrm{2}{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx}\:. \\ $$

Commented by maxmathsup by imad last updated on 20/Nov/18

1) we have f(α)=Re( ∫_(−∞) ^(+∞) (e^(iαx^3 ) /(x^2 +8))dx) let consider the complex function ϕ(z) = (e^(iαz^3 ) /(z^2 +8)) ⇒ϕ(z) =(e^(iαz^3 ) /(z^2 −(2(√2)i)^2 )) =(e^(iαz^3 ) /((z−2(√2)i)(z+2(√2)i))) so the poles of ϕ are +^− 2(√2)i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2i(√2)) but Res(ϕ,2i(√2)) =lim_(z→2i(√2)) (z−2i(√2))ϕ(z) =(e^(iα(2i(√2))^3 ) /(4i(√2))) =(e^(−iα(16i(√2))) /(4i(√2))) = (e^(16α(√2)) /(4i(√2))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ .(e^(16α(√2)) /(4i(√2))) =((π e^(16α(√2)) )/(2(√2))) ⇒ f(α) = ((π(√2))/4) e^(16α(√2)) . 2)∫_(−∞) ^(+∞) ((cos(2x^3 ))/(x^2 +8))dx =f(2) =((π(√2))/4) e^(32(√2)) .

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left(\alpha\right)={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\frac{{e}^{{i}\alpha{x}^{\mathrm{3}} } }{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx}\right)\:{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{i}\alpha{z}^{\mathrm{3}} } }{{z}^{\mathrm{2}} +\mathrm{8}}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{e}^{{i}\alpha{z}^{\mathrm{3}} } }{{z}^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{2}}{i}\right)^{\mathrm{2}} }\:=\frac{{e}^{{i}\alpha{z}^{\mathrm{3}} } }{\left({z}−\mathrm{2}\sqrt{\mathrm{2}}{i}\right)\left({z}+\mathrm{2}\sqrt{\mathrm{2}}{i}\right)} \\ $$$${so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\overset{−} {+}\mathrm{2}\sqrt{\mathrm{2}}{i}\:\:{residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\:{but} \\ $$$${Res}\left(\varphi,\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{i}\sqrt{\mathrm{2}}} \:\:\:\:\left({z}−\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\varphi\left({z}\right)\:=\frac{{e}^{{i}\alpha\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } }{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:=\frac{{e}^{−{i}\alpha\left(\mathrm{16}{i}\sqrt{\mathrm{2}}\right)} }{\mathrm{4}{i}\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{{e}^{\mathrm{16}\alpha\sqrt{\mathrm{2}}} }{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:.\frac{{e}^{\mathrm{16}\alpha\sqrt{\mathrm{2}}} }{\mathrm{4}{i}\sqrt{\mathrm{2}}}\:=\frac{\pi\:{e}^{\mathrm{16}\alpha\sqrt{\mathrm{2}}} }{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$${f}\left(\alpha\right)\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:{e}^{\mathrm{16}\alpha\sqrt{\mathrm{2}}} . \\ $$$$\left.\mathrm{2}\right)\int_{−\infty} ^{+\infty} \:\:\:\frac{{cos}\left(\mathrm{2}{x}^{\mathrm{3}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{8}}{dx}\:={f}\left(\mathrm{2}\right)\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}\:{e}^{\mathrm{32}\sqrt{\mathrm{2}}} . \\ $$

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