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Question Number 98589 by mathmax by abdo last updated on 14/Jun/20

calculate ∫_0 ^∞ ((sin(αx^2 ))/(x^2 +4))dx with α real

$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{sin}\left(\alpha\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dx}\:\:\mathrm{with}\:\alpha\:\mathrm{real} \\ $$

Answered by mathmax by abdo last updated on 15/Jun/20

let I =∫_0 ^∞ ((sin(αx^2 ))/(x^2 +4))dx ⇒2I =∫_(−∞) ^(+∞) ((sin(αx^2 ))/(x^2 +4))dx =Im(∫_(−∞) ^(+∞) (e^(iαx^2 ) /(x^2 +4))dx) let ϕ(z) = (e^(iαz^2 ) /(z^2 +4)) we can verify lim_(z→∞) ∣zϕ(z)∣=0 and ϕ(z) =(e^(iαz^2 ) /((z−2i)(z+2i))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2i) =2iπ×(e^(iα(2i)^2 ) /(4i)) =(π/2) e^(−4iα) =(π/2){cos(4α)−isin(4α)} ⇒ 2I =−(π/2)sin(4α) ⇒ I =−(π/4) sin(4α)

$$\mathrm{let}\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\alpha\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}\mathrm{dx}\:\Rightarrow\mathrm{2I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sin}\left(\alpha\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dx}\:=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{x}^{\mathrm{2}} } }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dx}\right) \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\:\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{z}^{\mathrm{2}} } }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}}\:\:\mathrm{we}\:\mathrm{can}\:\mathrm{verify}\:\mathrm{lim}_{\mathrm{z}\rightarrow\infty} \mid\mathrm{z}\varphi\left(\mathrm{z}\right)\mid=\mathrm{0}\:\:\mathrm{and} \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{e}^{\mathrm{i}\alpha\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{2i}\right)\left(\mathrm{z}+\mathrm{2i}\right)}\:\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{2i}\right)\:=\mathrm{2i}\pi×\frac{\mathrm{e}^{\mathrm{i}\alpha\left(\mathrm{2i}\right)^{\mathrm{2}} } }{\mathrm{4i}}\:=\frac{\pi}{\mathrm{2}}\:\mathrm{e}^{−\mathrm{4i}\alpha} \\ $$$$=\frac{\pi}{\mathrm{2}}\left\{\mathrm{cos}\left(\mathrm{4}\alpha\right)−\mathrm{isin}\left(\mathrm{4}\alpha\right)\right\}\:\Rightarrow\:\mathrm{2I}\:=−\frac{\pi}{\mathrm{2}}\mathrm{sin}\left(\mathrm{4}\alpha\right)\:\Rightarrow\:\mathrm{I}\:=−\frac{\pi}{\mathrm{4}}\:\mathrm{sin}\left(\mathrm{4}\alpha\right) \\ $$

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