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Question Number 49232 by Abdo msup. last updated on 04/Dec/18
let f(a)= ∫_(−∞) ^(+∞) cos(x^2 +ax +1)dx 1)calculate f(a) and f^′ (a) 2) find f^((n)) (a)
$${let}\:{f}\left({a}\right)=\:\int_{−\infty} ^{+\infty} {cos}\left({x}^{\mathrm{2}} \:+{ax}\:+\mathrm{1}\right){dx} \\ $$$$\left.\mathrm{1}\right){calculate}\:{f}\left({a}\right)\:{and}\:{f}^{'} \left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:{f}^{\left({n}\right)} \left({a}\right)\: \\ $$
Commented by Abdo msup. last updated on 06/Dec/18
letg(a)=∫_(−∞) ^(+∞) sin(x^2 +ax+1)dx we hsve f(a)−ig(a) =∫_(−∞) ^(+∞) e^(−i(x^2 +ax+1)) dx =∫_(−∞) ^(+∞) e^(−i(x^(2 ) +2(a/2)x +(a^2 /4) +1−(a^2 /4))) dx =e^(−i(1−(a^2 /4))) ∫_(−∞) ^(+∞) e^(−((√i)(x+(a/2)))^2 ) dx =_((√i)(x+(a/2)) =u) e^(−i(1−(a^2 /4))) ∫_(−∞) ^(+∞) e^(−u^2 ) (du/( (√i))) =(√π) e^(−i(1−(a^2 /4))) e^(−((iπ)/4)) =(√π) e^(−i(1+(π/4)−(a^2 /2))) (√π){cos(1+(π/4)−(a^2 /4))−isin(1+(π/4)−(a^2 /4))} ⇒ f(a) =(√π)cos(1+(π/4)−(a^2 /4)) also wehave g(a) =(√π)sin(1+(π/4) −(a^2 /4)).
$${letg}\left({a}\right)=\int_{−\infty} ^{+\infty} \:{sin}\left({x}^{\mathrm{2}} \:+{ax}+\mathrm{1}\right){dx}\:{we}\:{hsve} \\ $$$${f}\left({a}\right)−{ig}\left({a}\right)\:=\int_{−\infty} ^{+\infty} \:{e}^{−{i}\left({x}^{\mathrm{2}} +{ax}+\mathrm{1}\right)} {dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\:{e}^{−{i}\left({x}^{\mathrm{2}\:} \:+\mathrm{2}\frac{{a}}{\mathrm{2}}{x}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\:+\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)} {dx} \\ $$$$={e}^{−{i}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)} \:\int_{−\infty} ^{+\infty} \:\:{e}^{−\left(\sqrt{{i}}\left({x}+\frac{{a}}{\mathrm{2}}\right)\right)^{\mathrm{2}} } {dx} \\ $$$$=_{\sqrt{{i}}\left({x}+\frac{{a}}{\mathrm{2}}\right)\:={u}} \:\:\:\:{e}^{−{i}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)} \:\int_{−\infty} ^{+\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \frac{{du}}{\:\sqrt{{i}}} \\ $$$$=\sqrt{\pi}\:\:{e}^{−{i}\left(\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)} \:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:=\sqrt{\pi}\:{e}^{−{i}\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$$\sqrt{\pi}\left\{{cos}\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)−{isin}\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)\right\}\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\sqrt{\pi}{cos}\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)\:{also}\:{wehave} \\ $$$${g}\left({a}\right)\:=\sqrt{\pi}{sin}\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}\:−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right). \\ $$
Commented by Abdo msup. last updated on 06/Dec/18
f^′ (a) =−(√π)(−(a/2))sin(1+(π/4) −(a^2 /4)) =((a(√π))/2) sin(1+(π/4)−(a^2 /4)) .
$${f}^{'} \left({a}\right)\:=−\sqrt{\pi}\left(−\frac{{a}}{\mathrm{2}}\right){sin}\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}\:−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right) \\ $$$$=\frac{{a}\sqrt{\pi}}{\mathrm{2}}\:{sin}\left(\mathrm{1}+\frac{\pi}{\mathrm{4}}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}\right)\:. \\ $$

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