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Question Number 113004 by malwan last updated on 10/Sep/20
prove that _0 ∫^( ∞) cos(x^2 )dx = _0 ∫^( ∞) sin(x^2 )dx =((√π)/(2(√2)))
$${prove}\:{that} \\ $$$$\:_{\mathrm{0}} \int^{\:\infty} \:{cos}\left({x}^{\mathrm{2}} \right){dx}\:=\:\:_{\mathrm{0}} \int^{\:\infty} {sin}\left({x}^{\mathrm{2}} \right){dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Answered by mathdave last updated on 10/Sep/20
solution let y=x^2 ,x=y^(1/2) and dx=(1/2)y^((1/2)−1) I=(1/2)∫_0 ^∞ ((siny)/y^(1/2) )dy recall to Maz identity which state dat ∫_0 ^∞ ((sin(ax))/x^n )dx=((πa^(n−1) )/(2Γ(n)sin(((πn)/2)))) (a=1,n=(1/2),Γ((1/2))=(√π)) I=(1/2)∫_0 ^∞ ((siny)/y^(1/2) )dy=(1/2)•(π/(2Γ((1/2))sin((π/4)))) I=(1/4)•(π/( (√π)×(1/( (√2)))))=(((√π)×(√2))/4)=((√π)/(2(√2))) ∫_0 ^∞ sin(x^2 )dx=((√π)/(2(√2))) Q.E.D by mathdave(11/09/2020)
$${solution}\: \\ $$$${let}\:{y}={x}^{\mathrm{2}} \:,{x}={y}^{\frac{\mathrm{1}}{\mathrm{2}}} \:{and}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}{y}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{y}}{{y}^{\frac{\mathrm{1}}{\mathrm{2}}} }{dy} \\ $$$${recall}\:{to}\:{Maz}\:{identity}\:{which}\:{state}\:{dat} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({ax}\right)}{{x}^{{n}} }{dx}=\frac{\pi{a}^{{n}−\mathrm{1}} }{\mathrm{2}\Gamma\left({n}\right)\mathrm{sin}\left(\frac{\pi{n}}{\mathrm{2}}\right)}\:\:\:\:\:\left({a}=\mathrm{1},{n}=\frac{\mathrm{1}}{\mathrm{2}},\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\sqrt{\pi}\right) \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}{y}}{{y}^{\frac{\mathrm{1}}{\mathrm{2}}} }{dy}=\frac{\mathrm{1}}{\mathrm{2}}\bullet\frac{\pi}{\mathrm{2}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\bullet\frac{\pi}{\:\sqrt{\pi}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}=\frac{\sqrt{\pi}×\sqrt{\mathrm{2}}}{\mathrm{4}}=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \mathrm{sin}\left({x}^{\mathrm{2}} \right){dx}=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:\:\:\:\:\:{Q}.{E}.{D} \\ $$$${by}\:{mathdave}\left(\mathrm{11}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by malwan last updated on 11/Sep/20
thank you so much
$${thank}\:{you}\:{so}\:{much} \\ $$
Commented by Tawa11 last updated on 06/Sep/21
great sir
$$\mathrm{great}\:\mathrm{sir} \\ $$
Answered by mathmax by abdo last updated on 10/Sep/20
let I =∫_0 ^∞ cos(x^2 )dx and J =∫_0 ^∞ sin(x^2 )dx ⇒ I−iJ =∫_0 ^∞ e^(−ix^2 ) dx =∫_0 ^∞ e^(−((√i)x)^2 ) dx =_(x(√i)=t) ∫_0 ^∞ e^(−t^2 ) (dt/( (√i))) =(1/( (√i)))∫_0 ^∞ e^(−t^2 ) dt =(1/e^((iπ)/4) )×((√π)/2)=e^(−((iπ)/4)) ×(√π)=((√π)/2){cos((π/4))−isin((π/4))} =((√π)/2){(1/( (√2)))−(i/( (√2)))} =((√π)/(2(√2))) −i((√π)/(2(√2))) ⇒I =J =((√π)/(2(√2)))
$$\mathrm{let}\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:\mathrm{and}\:\mathrm{J}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{I}−\mathrm{iJ}\:=\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{e}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx}\:\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\left(\sqrt{\mathrm{i}}\mathrm{x}\right)^{\mathrm{2}} } \mathrm{dx}\:=_{\mathrm{x}\sqrt{\mathrm{i}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \:\frac{\mathrm{dt}}{\:\sqrt{\mathrm{i}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{i}}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{e}^{−\mathrm{t}^{\mathrm{2}} } \mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} }×\frac{\sqrt{\pi}}{\mathrm{2}}=\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:×\sqrt{\pi}=\frac{\sqrt{\pi}}{\mathrm{2}}\left\{\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right\} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right\}\:=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:−\mathrm{i}\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow\mathrm{I}\:=\mathrm{J}\:=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$
Commented by malwan last updated on 11/Sep/20
great Sir thank you
$${great}\:{Sir} \\ $$$${thank}\:{you} \\ $$

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