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Question Number 56310 by maxmathsup by imad last updated on 13/Mar/19
let f(x)=∫_(−∞) ^(+∞) cos(t^2 +xt +3)dt with x>0 1) find f(x) 2) calculate ∫_1 ^4 f(x)dx and ∫_1 ^(+∞) f(x)dx
$${let}\:{f}\left({x}\right)=\int_{−\infty} ^{+\infty} \:\:{cos}\left({t}^{\mathrm{2}} \:+{xt}\:+\mathrm{3}\right){dt}\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{1}} ^{\mathrm{4}} {f}\left({x}\right){dx}\:{and}\:\int_{\mathrm{1}} ^{+\infty} {f}\left({x}\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 14/Mar/19
1) f(x) =Re( ∫_(−∞) ^(+∞) e^(−i(t^2 +xt +3)) dt) but ∫_(−∞) ^(+∞) e^(−i(t^2 +xt +3)) dt =e^(−3i) ∫_(−∞) ^(+∞) e^(−i(t^2 +2(x/2)t +(x^2 /4) −(x^2 /4))) dt =e^(−3i) ∫_(−∞) ^(+∞) e^(−i(t+(x/2))^2 +i(x^2 /4)) dt =e^(i((x^2 /4)−3)) ∫_(−∞) ^(+∞) e^(−{(√i)(t +(x/2))}^2 ) dt changement (√i)(t+(x/2)) =u give f(x) =e^(i((x^2 /4)−3)) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/( (√i))) =(π/( (√i))) e^(i((x^2 /4)−3)) =π e^(−((iπ)/4)) e^(i((x^2 /4)−3)) =π e^(i((x^2 /4)−3−(π/4)) ) ⇒f(x) =π cos((x^2 /4)−3−(π/4)) .
$$\left.\mathrm{1}\right)\:{f}\left({x}\right)\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:{e}^{−{i}\left({t}^{\mathrm{2}} \:+{xt}\:+\mathrm{3}\right)} {dt}\right)\:\:{but} \\ $$$$\int_{−\infty} ^{+\infty} \:{e}^{−{i}\left({t}^{\mathrm{2}} \:+{xt}\:+\mathrm{3}\right)} {dt}\:={e}^{−\mathrm{3}{i}} \:\int_{−\infty} ^{+\infty} \:{e}^{−{i}\left({t}^{\mathrm{2}} \:+\mathrm{2}\frac{{x}}{\mathrm{2}}{t}\:\:+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\right)} {dt} \\ $$$$={e}^{−\mathrm{3}{i}} \:\int_{−\infty} ^{+\infty} \:\:{e}^{−{i}\left({t}+\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \:+{i}\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \:{dt}\:={e}^{{i}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}\right)} \:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left\{\sqrt{{i}}\left({t}\:+\frac{{x}}{\mathrm{2}}\right)\right\}^{\mathrm{2}} } {dt} \\ $$$${changement}\:\sqrt{{i}}\left({t}+\frac{{x}}{\mathrm{2}}\right)\:={u}\:{give} \\ $$$${f}\left({x}\right)\:={e}^{{i}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}\right)} \:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \:\:\frac{{du}}{\:\sqrt{{i}}}\:=\frac{\pi}{\:\sqrt{{i}}}\:{e}^{{i}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}\right)} \\ $$$$\left.=\pi\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} {e}^{{i}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}\right)} \:=\pi\:{e}^{{i}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}−\frac{\pi}{\mathrm{4}}\right.} \right)\:\Rightarrow{f}\left({x}\right)\:=\pi\:{cos}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}−\frac{\pi}{\mathrm{4}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 14/Mar/19
error from line 5 we have ∫_(−∞) ^(+∞) e^(−u^2 ) du =(√π) ⇒f(x)=Re(((√π)/( (√i))) e^(i((x^2 /4)−3)) ) ⇒ f(x) =(√π)cos((x^2 /4)−3−(π/4)) .
$${error}\:{from}\:{line}\:\mathrm{5}\:\:{we}\:{have}\:\int_{−\infty} ^{+\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:=\sqrt{\pi}\:\Rightarrow{f}\left({x}\right)={Re}\left(\frac{\sqrt{\pi}}{\:\sqrt{{i}}}\:{e}^{{i}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}\right)} \right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\sqrt{\pi}{cos}\left(\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}−\frac{\pi}{\mathrm{4}}\right)\:. \\ $$

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