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Question Number 45771 by maxmathsup by imad last updated on 16/Oct/18
find f(x)=∫_0 ^∞ cos(x+t^2 )dtand g(x)=∫_0 ^∞ sin(x+t^2 )dt 2) find the value of f^′ (x) and g^′ (x).
$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{cos}\left({x}+{t}^{\mathrm{2}} \right){dtand}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{sin}\left({x}+{t}^{\mathrm{2}} \right){dt} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{f}^{'} \left({x}\right)\:{and}\:{g}^{'} \left({x}\right). \\ $$
Answered by maxmathsup by imad last updated on 18/Oct/18
1)we have f(x)−ig(x)=∫_0 ^∞ e^(−i(x+t^2 )) dt = e^(−ix) ∫_0 ^∞ e^(−it^2 ) dt =e^(−ix) ∫_0 ^∞ e^(−((√i)t)^2 ) dt =_(t(√i)=u) e^(−ix) ∫_0 ^∞ e^(−u^2 ) (du/( (√i))) =e^(−ix) e^(−i(π/4)) ∫_0 ^∞ e^(−u^2 ) du =((√π)/2) e^(−i(x+(π/4))) =((√π)/2){ cos(x+(π/4))−isin(x+(π/4))} ⇒ f(x)=((√π)/2) cos(x+(π/4)) and g(x)=((√π)/2)sin(x+(π/4)) .
$$\left.\mathrm{1}\right){we}\:{have}\:{f}\left({x}\right)−{ig}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{i}\left({x}+{t}^{\mathrm{2}} \right)} {dt}\:=\:{e}^{−{ix}} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{it}^{\mathrm{2}} } {dt} \\ $$$$={e}^{−{ix}} \:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\sqrt{{i}}{t}\right)^{\mathrm{2}} } {dt}\:=_{{t}\sqrt{{i}}={u}} \:{e}^{−{ix}} \:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}^{\mathrm{2}} } \:\frac{{du}}{\:\sqrt{{i}}} \\ $$$$={e}^{−{ix}} \:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{u}^{\mathrm{2}} } {du}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:{e}^{−{i}\left({x}+\frac{\pi}{\mathrm{4}}\right)} =\frac{\sqrt{\pi}}{\mathrm{2}}\left\{\:{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)−{isin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right\}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\:{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right)\:{and}\:{g}\left({x}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 18/Oct/18
2) we have f^′ (x)=−((√π)/2)sin(x+(π/4)) and g^′ (x)=((√π)/2) cos(x+(π/4)).
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)=−\frac{\sqrt{\pi}}{\mathrm{2}}{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\:{and}\:{g}^{'} \left({x}\right)=\frac{\sqrt{\pi}}{\mathrm{2}}\:{cos}\left({x}+\frac{\pi}{\mathrm{4}}\right). \\ $$

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