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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) 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}) .