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Question Number 42260 by math khazana by abdo last updated on 21/Aug/18

let f(a) = ∫_(−∞) ^(+∞) cos(ax^2 )dx with a>0 1) calculate f(a) interms of a ) calculate ∫_(−∞) ^(+∞) cos(2x^2 )dx 3) find the value of ∫_(−∞) ^(+∞) cos(x^2 +x+1)dx .

$${let}\:{f}\left({a}\right)\:=\:\int_{−\infty} ^{+\infty} \:{cos}\left({ax}^{\mathrm{2}} \right){dx}\:{with}\:{a}>\mathrm{0} \\ $$ $$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right)\:{interms}\:{of}\:{a} \\ $$ $$\left.\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\:\:{cos}\left(\mathrm{2}{x}^{\mathrm{2}} \right){dx} \\ $$ $$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:{cos}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right){dx}\:. \\ $$

Commented bymaxmathsup by imad last updated on 21/Aug/18

3) ∫_(−∞) ^(+∞) cos(x^2 +x+1)dx =∫_(−∞) ^(+∞) cos((x+(1/2))^2 +(3/4))dx =_(x+(1/2)=((√3)/2)t) ∫_(−∞) ^(+∞) cos((3/4)(t^2 +1))((√3)/2) dt =((√3)/2) ∫_(−∞) ^(+∞) cos((3/4)t^2 +(3/4))dt =((√3)/2) ∫_(−∞) ^(+∞) {cos((3/4)t^2 )cos((3/4))−sin((3/4)t^2 )sin((3/4))}dt =((√3)/2) cos((3/4)) ∫_(−∞) ^(+∞) cos((3/4)t^2 )dt −((√3)/2) sin((3/4))∫_(−∞) ^(+∞) sin((3/4)t^2 )dt =((√3)/2) cos((3/4))((√(2π))/(2(√(3/4)))) −((√3)/2) sin((3/4))((√(2π))/(2(√(3/4)))) =((√(2π))/2) cos((3/4)) −((√(2π))/2) sin((3/4)) ⇒ ∫_(−∞) ^(+∞) cos(x^2 +x +1)dx =(√(π/2)){cos((3/4))−sin((3/4))} .

$$\left.\mathrm{3}\right)\:\int_{−\infty} ^{+\infty} \:{cos}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right){dx}\:=\int_{−\infty} ^{+\infty} \:{cos}\left(\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right){dx} \\ $$ $$=_{{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\:\:\:\:\:\int_{−\infty} ^{+\infty} \:\:{cos}\left(\frac{\mathrm{3}}{\mathrm{4}}\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{dt} \\ $$ $$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:{cos}\left(\frac{\mathrm{3}}{\mathrm{4}}{t}^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\right){dt} \\ $$ $$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\int_{−\infty} ^{+\infty} \:\:\left\{{cos}\left(\frac{\mathrm{3}}{\mathrm{4}}{t}^{\mathrm{2}} \right){cos}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−{sin}\left(\frac{\mathrm{3}}{\mathrm{4}}{t}^{\mathrm{2}} \right){sin}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right\}{dt} \\ $$ $$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{cos}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\:\int_{−\infty} ^{+\infty} \:{cos}\left(\frac{\mathrm{3}}{\mathrm{4}}{t}^{\mathrm{2}} \right){dt}\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{sin}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\int_{−\infty} ^{+\infty} \:{sin}\left(\frac{\mathrm{3}}{\mathrm{4}}{t}^{\mathrm{2}} \right){dt} \\ $$ $$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{cos}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}}\:\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{sin}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}\sqrt{\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$ $$=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}}\:{cos}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\:\:−\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}}\:{sin}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\:\Rightarrow \\ $$ $$\int_{−\infty} ^{+\infty} \:{cos}\left({x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\right){dx}\:=\sqrt{\frac{\pi}{\mathrm{2}}}\left\{{cos}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)−{sin}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right\}\:. \\ $$ $$ \\ $$

Commented bymaxmathsup by imad last updated on 21/Aug/18

we have ∫_(−∞) ^(+∞) cos(ax^2 )dx−i ∫_(−∞) ^(+∞) sin(ax^2 )dx = ∫_(−∞) ^(+∞) e^(−iax^2 ) dx =∫_(−∞) ^(+∞) e^(−((√(ia))x)^2 ) dx =_((√(ia))x =t) ∫_(−∞) ^(+∞) e^(−t^2 ) (dt/(√(ia))) = (1/((√a)e^((iπ)/4) )) (√π)=((√π)/(√a)) e^(−((iπ)/4)) =((√π)/(√a)) {cos((π/4))−isin((π/4))} ⇒ f(a) =((√π)/(√a)) ((√2)/2) =((√(2π))/(2(√a))) . also we have ∫_(−∞) ^(+∞) sin(ax^2 )dx =((√(2π))/(2(√a))) 2) ∫_(−∞) ^(+∞) cos(2x^2 )dx =f(2) = ((√(2π))/(2(√2))) =((√π)/2) .

$${we}\:{have}\:\int_{−\infty} ^{+\infty} \:{cos}\left({ax}^{\mathrm{2}} \right){dx}−{i}\:\int_{−\infty} ^{+\infty} \:{sin}\left({ax}^{\mathrm{2}} \right){dx} \\ $$ $$=\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{iax}^{\mathrm{2}} } {dx}\:\:=\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\sqrt{{ia}}{x}\right)^{\mathrm{2}} } {dx}\:=_{\sqrt{{ia}}{x}\:={t}} \:\:\:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\frac{{dt}}{\sqrt{{ia}}} \\ $$ $$=\:\frac{\mathrm{1}}{\sqrt{{a}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:\sqrt{\pi}=\frac{\sqrt{\pi}}{\sqrt{{a}}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:=\frac{\sqrt{\pi}}{\sqrt{{a}}}\:\left\{{cos}\left(\frac{\pi}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}\right)\right\}\:\Rightarrow \\ $$ $${f}\left({a}\right)\:=\frac{\sqrt{\pi}}{\sqrt{{a}}}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\:=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}\sqrt{{a}}}\:\:.\:{also}\:{we}\:{have}\:\int_{−\infty} ^{+\infty} \:\:{sin}\left({ax}^{\mathrm{2}} \right){dx}\:=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}\sqrt{{a}}} \\ $$ $$\left.\mathrm{2}\right)\:\int_{−\infty} ^{+\infty} \:\:{cos}\left(\mathrm{2}{x}^{\mathrm{2}} \right){dx}\:={f}\left(\mathrm{2}\right)\:=\:\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\:. \\ $$

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