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Question Number 65922 by mathmax by abdo last updated on 05/Aug/19
calculate ∫_0 ^∞   ((sin(3x^2 ))/x^2 )dx
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{3}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\: \\ $$
Commented by mathmax by abdo last updated on 08/Aug/19
let A =∫_0 ^∞   ((sin(3x^2 ))/x^2 )dx ⇒A =_((√3)x =t)    3∫_0 ^∞  ((sin(t^2 ))/t^2 )(dt/( (√3)))  =(1/( (√3)))∫_0 ^∞   ((sin(t^2 ))/t^2 )dt  let ϕ(x) =∫_0 ^∞   ((sin(t^2 )e^(−xt^2 ) )/t^2 )dt  with x≥0  ϕ^′ (x) =−∫_0 ^∞   e^(−xt^2 ) sin(t^2 )dt =−Im(∫_0 ^∞  e^(−xt^2 +it^2 ) dt) and  ∫_0 ^∞  e^((−x+i)t^2 ) dt =_((√(−x+i))t =u)    ∫_0 ^∞  e^(−u^2 ) (du/( (√(−x+i))))  =((√π)/(2(√(−x+i)))) ⇒ϕ(x) =((√π)/2)∫ (dx/( (√(−x+i)))) +c  −x+i =(√(1+x^2 )) e^(iarctan(−(1/x)))  =(√(1+x^2 ))e^(−i arctan((1/x)))   =(√(1+x^2 )) e^(−i((π/2)−arctanx))  =−i(√(1+x^2 ))e^(iatctanx)  ⇒  (√(−x+i)) =(√(−i))(1+x^2 )^(1/4)  e^((i/2)arctan(x))  =e^(−((iπ)/4))  (1+x^2 )^(1/4)  e^((i/2)arctanx)  ⇒  ϕ(x) =((√π)/2) e^((iπ)/4)   ∫   (1+x^2 )^(−(1/4)) e^(−(i/2)arctanx) dx +c  ....be continued....
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{3}{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\:\Rightarrow{A}\:=_{\sqrt{\mathrm{3}}{x}\:={t}} \:\:\:\mathrm{3}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }\frac{{dt}}{\:\sqrt{\mathrm{3}}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} }{dt}\:\:{let}\:\varphi\left({x}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right){e}^{−{xt}^{\mathrm{2}} } }{{t}^{\mathrm{2}} }{dt}\:\:{with}\:{x}\geqslant\mathrm{0} \\ $$$$\varphi^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{xt}^{\mathrm{2}} } {sin}\left({t}^{\mathrm{2}} \right){dt}\:=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}^{\mathrm{2}} +{it}^{\mathrm{2}} } {dt}\right)\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−{x}+{i}\right){t}^{\mathrm{2}} } {dt}\:=_{\sqrt{−{x}+{i}}{t}\:={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } \frac{{du}}{\:\sqrt{−{x}+{i}}} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}\sqrt{−{x}+{i}}}\:\Rightarrow\varphi\left({x}\right)\:=\frac{\sqrt{\pi}}{\mathrm{2}}\int\:\frac{{dx}}{\:\sqrt{−{x}+{i}}}\:+{c} \\ $$$$−{x}+{i}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{{iarctan}\left(−\frac{\mathrm{1}}{{x}}\right)} \:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$$=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{−{i}\left(\frac{\pi}{\mathrm{2}}−{arctanx}\right)} \:=−{i}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{e}^{{iatctanx}} \:\Rightarrow \\ $$$$\sqrt{−{x}+{i}}\:=\sqrt{−{i}}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctan}\left({x}\right)} \:={e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:{e}^{\frac{{i}}{\mathrm{2}}{arctanx}} \:\Rightarrow \\ $$$$\varphi\left({x}\right)\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\int\:\:\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{4}}} {e}^{−\frac{{i}}{\mathrm{2}}{arctanx}} {dx}\:+{c} \\ $$$$….{be}\:{continued}…. \\ $$