let-f-t-0-cos-t-1-x-2-1-x-2-dx-with-t-0-find-a-explicit-form-of-f-t-

Question Number 56829 by maxmathsup by imad last updated on 24/Mar/19

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

f i n d a e x p l i c i t f o r m o f f (t)

Answered by Smail last updated on 25/Mar/19

f (t) = I m (\int_{0}^{\infty} \frac{e^{- i t (1 + x^{2})}}{1 + x^{2}} d x)

z (t) = \int_{0}^{\infty} \frac{e^{- i t (1 + x^{2})}}{1 + x^{2}} d x

z^{'} (t) = - i \int_{0}^{\infty} e^{- i t (1 + x^{2})} d x = - {i e}^{- i t} \int_{0}^{\infty} e^{- {i x}^{2} t} d x

l e t u = \sqrt{i t} x \Rightarrow d x = \frac{d u}{\sqrt{i t}}

z^{'} (t) = - \frac{\sqrt{i} e^{- i t}}{\sqrt{t}} \int_{0}^{\infty} e^{- u^{2}} d u

= - \frac{\sqrt{i} e^{- i t}}{\sqrt{t}} \times \frac{\sqrt{π}}{2}

z (t) - z (0) = - \frac{\sqrt{i π}}{2} \int_{0}^{t} \frac{e^{- i u}}{\sqrt{u}} d u

z (0) = \int_{0}^{\infty} \frac{d x}{1 + x^{2}} = \frac{π}{2}

l e t θ = \sqrt{i u} \Rightarrow d θ = \frac{\sqrt{i} d u}{2 \sqrt{u}}

z (t) = - \sqrt{π} \int_{0}^{\sqrt{i t}} e^{- θ^{2}} d θ + \frac{π}{2}

= - \sqrt{π} \int_{0}^{\sqrt{i t}} \underset{n = 0}{\sum^{\infty}} \frac{{(- θ^{2})}^{n}}{n!} d θ + \frac{π}{2}

= - \sqrt{π} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} {(\sqrt{i t})}^{2 n + 1}}{n! (2 n + 1)} + \frac{π}{2}

= \frac{π}{2} - \sqrt{π} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} {(i t)}^{n} . \sqrt{i t}}{n! (2 n + 1)}

= \frac{π}{2} - \sqrt{π t} \underset{n = 0}{\sum^{\infty}} \frac{{(- t)}^{n} e^{i (\frac{π}{2}) n} \times e^{i \frac{π}{4}}}{n! (2 n + 1)}

= \frac{π}{2} - \sqrt{π t} \underset{n = 0}{\sum^{\infty}} \frac{{(- t)}^{n} e^{i (\frac{n π}{2} + \frac{π}{4})}}{n! (2 n + 1)}

T h u s

f (t) = \frac{π}{2} - \sqrt{π t} \underset{n = 0}{\sum^{\infty}} \frac{{(- t)}^{n} c o s (\frac{2 n + 1}{4} π)}{n! (2 n + 1)}

Commented by maxmathsup by imad last updated on 26/Mar/19

t h a n k s s i r s m a i l .