Zadanie

Zadanie Trivial: Mam pytanie odno�nie transformaty Fouriera...
Czy da si� policzy� do jako� �atwiej?
$f(x) = \begin{cases} x+1 & -1 \leq x < 0 \\ 1-x^2 & 0 \leq x \leq 1 \\ 0 & |x| > 1 \end{cases} \\\\ F(u) \overset{\textup{df}}= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}f(x) e^{-iux} \mbox dx = \frac{1}{\sqrt{2\pi}} \left( \int_{-1}^{0}(x+1) e^{-iux} \mbox dx + \int_{0}^{1}(1-x^2) e^{-iux} \mbox dx \right ) = \frac{1}{\sqrt{2\pi}} \left( \left[ \left(\frac{x+1}{-iu} + \frac{1}{u^2} \right ) e^{-iux}\right ]_{-1}^{0} + \left[ \left( \frac{1-x^2}{-iu} - \frac{2x}{u^2} - \frac{2}{iu^3} \right ) e^{-iux}\right ]_{0}^{1} \right) = ... = \frac{1}{\sqrt{2\pi}u^2}\left ( \frac{2-2e^{-iu}}{iu} + 1-e^{iu}-2e^{-iu} \right ).$