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).


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Question Number 45771 by maxmathsup by imad last updated on 16/Oct/18

$f i n d f (x) = \int_{0}^{\infty} c o s (x + t^{2}) d t a n d g (x) = \int_{0}^{\infty} s i n (x + t^{2}) d t$ $2) f i n d t h e v a l u e o f f^{^{'}} (x) a n d g^{^{'}} (x) .$
	Answered by maxmathsup by imad last updated on 18/Oct/18
	$1)we have f(x)−ig(x)=∫_0 ^∞ e^(−i(x+t^2 )) dt = e^(−ix) ∫_0 ^∞ e^(−it^2 ) dt =e^(−ix) ∫_0 ^∞ e^(−((√i)t)^2 ) dt =_(t(√i)=u) e^(−ix) ∫_0 ^∞ e^(−u^2 ) (du/(√i)) =e^(−ix) e^(−i(π/4)) ∫_0 ^∞ e^(−u^2 ) du =((√π)/2) e^(−i(x+(π/4))) =((√π)/2){ cos(x+(π/4))−isin(x+(π/4))} ⇒ f(x)=((√π)/2) cos(x+(π/4)) and g(x)=((√π)/2)sin(x+(π/4)) .$
	$1) w e h a v e f (x) - i g (x) = \int_{0}^{\infty} e^{- i (x + t^{2})} d t = e^{- i x} \int_{0}^{\infty} e^{- {i t}^{2}} d t$ $= e^{- i x} \int_{0}^{\infty} e^{- {(\sqrt{i} t)}^{2}} d t =_{t \sqrt{i} = u} e^{- i x} \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{i}}$ $= e^{- i x} e^{- i \frac{π}{4}} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{\sqrt{π}}{2} e^{- i (x + \frac{π}{4})} = \frac{\sqrt{π}}{2} {c o s (x + \frac{π}{4}) - i s i n (x + \frac{π}{4})} \Rightarrow$ $f (x) = \frac{\sqrt{π}}{2} c o s (x + \frac{π}{4}) a n d g (x) = \frac{\sqrt{π}}{2} s i n (x + \frac{π}{4}) .$
	Commented by maxmathsup by imad last updated on 18/Oct/18

	$2) w e h a v e f^{^{'}} (x) = - \frac{\sqrt{π}}{2} s i n (x + \frac{π}{4}) a n d g^{^{'}} (x) = \frac{\sqrt{π}}{2} c o s (x + \frac{π}{4}) .$
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