anonymous 4 years ago find the fourier transform of the following function: $f(x)=x , -1\leq x \leq 1, f(x) = 0 \abs{x} \geq 1$

1. anonymous

ah, that \abs is supposed to be the absolute value encompassing x

2. nikvist

$F(\omega)=\int\limits_{-\infty}^{+\infty}f(x)e^{-i\omega x}dx=\int\limits_{-1}^{1}xe^{-i\omega x}dx=$$=\int\limits_{-1}^{1}x(\cos{\omega x}-i\sin{\omega x})dx=$$=\underbrace{\int\limits_{-1}^{1}x\cos{\omega x}dx}_{=0}-i\int\limits_{-1}^{1}x\sin{\omega x}dx=-2i\int\limits_{0}^{1}x\sin{\omega x}dx=$$=-2i\frac{\sin{\omega}-\omega\cos{\omega}}{\omega^2}$