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Question Number 39373 by maxmathsup by imad last updated on 05/Jul/18
find the values of integrals  A = ∫_(−∞) ^(+∞)  cos(x^2  +x+1)dx   and B = ∫_(−∞) ^(+∞)  sin(x^2  +x+1)dx
$${find}\:{the}\:{values}\:{of}\:{integrals} \\ $$$${A}\:=\:\int_{−\infty} ^{+\infty} \:{cos}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right){dx}\:\:\:{and}\:{B}\:=\:\int_{−\infty} ^{+\infty} \:{sin}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right){dx} \\ $$
Commented by math khazana by abdo last updated on 06/Jul/18
we have A−iB = ∫_(−∞) ^(+∞)  e^(−i(x^2  +x+1)) dx  =e^(−i)   ∫_(−∞) ^(+∞)   e^(−i(x^2  +2 (x/2) +(1/4) −(1/4))) dx  =e^(−i )  ∫_(−∞) ^(+∞)   e^(−i(x+(1/2))^2  +(i/4))  dx  = e^(−(3/4)i)   ∫_(−∞) ^(+∞ )  e^(−((√i)(x+(1/2))^2 )) dx  =_((√i)(x+(1/2)) =t)   e^(−((3i)/4))    ∫_(−∞) ^(+∞)   e^(−t^2 )   (dt/( (√i)))  = (e^(−(3/4)i) /e^(i(π/4)) ) (√π)= e^(−i(((3+π)/4)))  (√π)  =(√π){ cos((π/4) +(3/4))−isin((π/4)+(3/4))} ⇒  A=(√π)cos( (π/4) +(3/4)) and  B =(√π)sin((π/4) +(3/4)).
$${we}\:{have}\:{A}−{iB}\:=\:\int_{−\infty} ^{+\infty} \:{e}^{−{i}\left({x}^{\mathrm{2}} \:+{x}+\mathrm{1}\right)} {dx} \\ $$$$={e}^{−{i}} \:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{i}\left({x}^{\mathrm{2}} \:+\mathrm{2}\:\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{4}}\right)} {dx} \\ $$$$={e}^{−{i}\:} \:\int_{−\infty} ^{+\infty} \:\:{e}^{−{i}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{{i}}{\mathrm{4}}} \:{dx} \\ $$$$=\:{e}^{−\frac{\mathrm{3}}{\mathrm{4}}{i}} \:\:\int_{−\infty} ^{+\infty\:} \:{e}^{−\left(\sqrt{{i}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \right)} {dx} \\ $$$$=_{\sqrt{{i}}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\:={t}} \:\:{e}^{−\frac{\mathrm{3}{i}}{\mathrm{4}}} \:\:\:\int_{−\infty} ^{+\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:\:\frac{{dt}}{\:\sqrt{{i}}} \\ $$$$=\:\frac{{e}^{−\frac{\mathrm{3}}{\mathrm{4}}{i}} }{{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:\sqrt{\pi}=\:{e}^{−{i}\left(\frac{\mathrm{3}+\pi}{\mathrm{4}}\right)} \:\sqrt{\pi} \\ $$$$=\sqrt{\pi}\left\{\:{cos}\left(\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{4}}\right)−{isin}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\right)\right\}\:\Rightarrow \\ $$$${A}=\sqrt{\pi}{cos}\left(\:\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{4}}\right)\:{and} \\ $$$${B}\:=\sqrt{\pi}{sin}\left(\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{3}}{\mathrm{4}}\right). \\ $$

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