Here's the problem: Let $0\leq a <b$. Suppose $\phi :\mathbb{R} \rightarrow \mathbb{R}$ is even such that $$\phi (x) = \begin{cases} 1 & 0\leq x\leq a \\ \frac{x-b}{a-b} &a<x \leq b \\ 0 &x>b. \end{cases}$$ Show that $\widehat{\phi}(\omega)=\sqrt{\frac{2}{\pi}} \frac{2}{b-a} \frac{sin(\sigma \omega)sin(\tau \omega)}{\omega ^2} (\forall \omega \in \mathbb{R})$ where $\sigma=\frac{1}{2}(a+b)$ and $\tau =\frac{1}{2}(b-a)$. Note: Use the fact that $\phi$ is even to minimize computations.
Here's the start my solution: $\widehat{\phi}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(x)e^{-i\omega x}dx = \frac{1}{\sqrt{2\pi}}(\int_{0}^{a}e^{-i\omega x}dx + \int_{a}^{b}(\frac{x-b}{a-b})e^{-i\omega x}dx+\int_{b}^{\infty}0 dx)$
And I'm stuck here. Thanks a lot!