let-f-x-0-1-cos-xt-2-t-2-e-xt-2-dt-with-x-gt-0-1-find-a-simple-form-of-f-x-2-calculate-0-1-cos-2t-2-t-2-e-3t-2-dt-

Question Number 38454 by maxmathsup by imad last updated on 25/Jun/18

l e t f (x) = \int_{0}^{\infty} \frac{1 - c o s ({x t}^{2})}{t^{2}} e^{- {x t}^{2}} d t w i t h x > 0

1) f i n d a s i m p l e f o r m o f f (x)

2) c a l c u l a t e \int_{0}^{\infty} \frac{1 - c o s (2 t^{2})}{t^{2}} e^{- 3 t^{2}} d t .

Commented by math khazana by abdo last updated on 26/Jun/18

t h e Q i s f (x) = \int_{0}^{\infty} \frac{1 - c o s ({a t}^{2})}{t^{2}} e^{- {x t}^{2}} d t

Commented by abdo.msup.com last updated on 27/Jun/18

w e h a v e f (x) = \int_{0}^{\infty} \frac{1 - c o s ({a t}^{2})}{t^{2}} e^{- {x t}^{2}} d t \Rightarrow

f^{^{'}} (x) = - \int_{0}^{\infty} (1 - c o s ({a t}^{2}) e^{- {x t}^{2}} d t

= \int_{0}^{\infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t - \int_{0}^{\infty} e^{- {x t}^{2}} d t b u t

\int_{0}^{\infty} e^{- {x t}^{2}} d t =_{\sqrt{x} t = u} \int_{0}^{\infty} e^{- u^{2}} \frac{d u}{\sqrt{x}}

= \frac{1}{\sqrt{x}} \int_{0}^{\infty} e^{- u^{2}} d u = \frac{π}{2 \sqrt{x}} a n d

\int_{0}^{\infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t = \frac{1}{2} \int_{- \infty}^{+ \infty} c o s ({a t}^{2}) e^{- {x t}^{2}} d t

= \frac{1}{2} R e (\int_{- \infty}^{\infty} e^{{i a t}^{2} - {x t}^{2}} d t) b u t

\int_{- \infty}^{+ \infty} e^{(- x + i a) t^{2}} d t =_{\sqrt{- x + i a} t = u} \int_{- \infty}^{+ \infty} e^{- u^{2}} \frac{d u}{\sqrt{- x + i a}}

= \frac{\sqrt{π}}{\sqrt{- x + i a}} b u t

- x + i a = \sqrt{x^{2} + a^{2}} {\frac{- x}{\sqrt{x^{2} + a^{2}}} + \frac{i a}{\sqrt{x^{2} + a^{2}}}}

= r e^{i θ} \Rightarrow r = \sqrt{x^{2} + a^{2}} a n d t a n θ = - \frac{a}{x} \Rightarrow

θ = - a r c t a n (\frac{a}{x}) \Rightarrow - x + i a = r e^{- i a r c t a n (\frac{a}{x})}

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

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

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

f (x) = \frac{\sqrt{π}}{2} \int_{.}^{x} {(t^{2} + a^{2})}^{- \frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{a}{t})) d t

+ c

Commented by abdo.msup.com last updated on 27/Jun/18

e r r o r a t t h e f i n a l l i n e s

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

- \frac{π}{2 \sqrt{x}} \Rightarrow

f (x) = \frac{\sqrt{π}}{2} \int_{.}^{x} {(t^{2} + a^{2})}^{- \frac{1}{4}} c o s (\frac{1}{2} a r c t a n (\frac{a}{t})) d t

- π \sqrt{x} + c