let-f-x-sin-2x-x-a-a-x-with-a-gt-0-calculate-the-fourier-trsnsform-of-f-

Question Number 35992 by abdo mathsup 649 cc last updated on 26/May/18

l e t f (x) = \frac{s i n (2 x)}{x} χ_{] - a, a [} (x) w i t h a > 0

c a l c u l a t e t h e f o u r i e r t r s n s f o r m o f f .

Commented by abdo mathsup 649 cc last updated on 27/May/18

F (f (x)) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{+ \infty} f (t) e^{- i x t} d t

= \frac{1}{\sqrt{2 π}} \int_{- a}^{a} \frac{s i n (2 t)}{t} e^{- i x t} d t

= \frac{1}{\sqrt{2 π}} \int_{- a}^{a} \frac{s i n (2 t)}{t} c o s (x t) d t

= \sqrt{\frac{2}{π}} \int_{0}^{a} \frac{s i n (2 t) c o s (x t)}{t} d t = \sqrt{\frac{2}{π}} w (x)

w (x) = \int_{0}^{a} \frac{s i n (2 t) c o s (x t)}{t} d t

w^{^{'}} (x) = - \int_{0}^{a} s i n (2 t) s i n (x t) d t b u t

s i n (2 t) s i n (x t) = \frac{1}{2} {c o s (x - 2) t - c o s (x + 2) t} \Rightarrow

w^{^{'}} (x) = - \frac{1}{2} {\int_{0}^{a} c o s (x - 2) t d t - \int_{0}^{a} c o s (x + 2) t d t}

= \frac{1}{2} {[\frac{1}{x + 2} s i n (x + 2) t]}_{0}^{a} - \frac{1}{2} {[\frac{1}{x - 2} s i n (x - 2) t]}_{0}^{a}

= \frac{1}{2 x + 4} s i n {(x + 2) a} - \frac{1}{2 x - 4} s i n {(x - 2) a} \Rightarrow

w (x) = \int_{0}^{x} \frac{s i n {(t + 2) a}}{2 t + 4} d t - \int_{0}^{x} \frac{s i n {(t - 2) a}}{2 t - 4} d t + c

(c = w (0) = \int_{0}^{a} \frac{s i n (2 t)}{t} d t) \dots b e c o n t i n u e d \dots