let-f-t-0-sin-x-2-e-tx-2-x-2-dx-with-t-gt-0-find-a-simple-form-of-f-t-

Question Number 33978 by abdo imad last updated on 28/Apr/18

l e t f (t) = \int_{0}^{\infty} \frac{s i n (x^{2}) e^{- {t x}^{2}}}{x^{2}} d x w i t h t > 0

f i n d a s i m p l e f o r m o f f^{^{'}} (t) .

Commented by math khazana by abdo last updated on 01/May/18

a f t e r v e r i f y i n g c o n d i t i o n o f d e r i v a b i t y f o r f w e h a v e

f^{^{'}} (t) = \int_{0}^{\infty} - x^{2} \frac{s i n (x^{2}) e^{- {t x}^{2}}}{x^{2}} d x

= - \int_{0}^{\infty} s i n (x^{2}) e^{- {t x}^{2}} d x \Rightarrow 2 f^{^{'}} (x) = - \int_{- \infty}^{+ \infty} s i n (x^{2}) e^{- {t x}^{2}} d x

= I m (\int_{- \infty}^{+ \infty} e^{- {i x}^{2}} e^{- {t x}^{2}} d x) b u t

\int_{- \infty}^{+ \infty} e^{- {i x}^{2}} e^{- {t x}^{2}} d x = \int_{- \infty}^{+ \infty} e^{- (t + i) x^{2}} d x

=_{\sqrt{t + i} x = u} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{t + i}} = \frac{\sqrt{π}}{\sqrt{t + i}} b u t

∣ t + i ∣= \sqrt{1 + t^{2}} \Rightarrow t + i = \sqrt{1 + t^{2}} (\frac{t}{\sqrt{1 + t^{2}}} + \frac{i}{\sqrt{1 + t^{2}}})

= r e^{i θ} \Rightarrow r = \sqrt{1 + t^{2}} a n d c o s θ = \frac{t}{\sqrt{1 + t^{2}}} a n d

s i n θ = \frac{1}{\sqrt{1 + t^{2}}} \Rightarrow t a n θ = \frac{1}{t} \Rightarrow θ = a r c t a n (\frac{1}{t})

t + i = {(1 + t^{2})}^{\frac{1}{2}} e^{i (\frac{π}{2} - a r c t a n t)} \Rightarrow

\sqrt{t + i} = {(1 + t^{2})}^{\frac{1}{4}} e^{i (\frac{π}{4} - \frac{1}{2} a r c t a n t)}

\frac{1}{\sqrt{t + i}} = {(1 + t^{2})}^{- \frac{1}{4}} e^{i (\frac{1}{2} a r c t a n t - \frac{π}{4})} \Rightarrow

2 f^{^{'}} (x) = {(1 + t^{2})}^{- \frac{1}{4}} s i n (\frac{1}{2} a r c t a n t - \frac{π}{4}) \Rightarrow

f^{^{'}} (x) = \frac{1}{2} {(1 + t^{2})}^{- \frac{1}{4}} s i n (\frac{1}{2} a r c t a n t - \frac{π}{4}) .

Commented by math khazana by abdo last updated on 01/May/18

f^{^{'}} (t) = \frac{1}{2} {(1 + t^{2})}^{- \frac{1}{4}} s i n (\frac{1}{2} a r c t a n t - \frac{π}{4}) .

Answered by candre last updated on 30/Apr/18

f (t) = \underset{0}{\int^{\infty}} \frac{\sin x^{2} e^{- {t x}^{2}}}{x^{2}} d x

l e t s u s e t h e f a c t t h a t

\underset{- \infty}{\int^{+ \infty}} e^{- {a x}^{2}} d x = \sqrt{\frac{π}{a}}

\sin x = \frac{e^{i x} - e^{- i x}}{2 i}

w e g e t

\frac{d f}{d t} = \underset{0}{\int^{\infty}} \frac{\partial}{\partial t} \frac{\sin x^{2} e^{- {t x}^{2}}}{x^{2}} d x

= - \underset{0}{\int^{\infty}} \sin x^{2} e^{- {t x}^{2}} d x

= - \underset{0}{\int^{\infty}} \frac{e^{{i x}^{2}} - e^{- {i x}^{2}}}{2 i} e^{- {t x}^{2}} d x

= - \frac{1}{2 i} [\underset{0}{\int^{\infty}} e^{(i - t) x^{2}} d x - \underset{0}{\int^{\infty}} e^{- (i + t) x^{2}} d x]

= \frac{\sqrt{π}}{4 i} (\sqrt{\frac{1}{t + i}} - \sqrt{\frac{1}{t - i}})

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