So i have to prove the following equality;
$$\int_0^\infty\frac{\cos(\pi w/2)\cos(wx)}{1-w^2}dw=\left\{\begin{array}{ll}\frac{\pi}{2}\cos(x)&\mbox{ si }|x|\leq\frac{\pi}{2}\\0&\mbox{ si }|x|>\frac{\pi}{2}\end{array}\right.$$
So, what i have done is taking the function
$$f(x)=\left\{\begin{array}{ll}\frac{\pi}{2}\cos(x)&\mbox{ si }|x|\leq\frac{\pi}{2}\\0&\mbox{ si }|x|>\frac{\pi}{2}\end{array}\right.$$
By the general from of the fourier integral, we got
$$f(x)=\frac{1}{\pi}\int_0^\infty\left[A(\alpha)\cos(\alpha x)+B(\alpha)\sin(\alpha x)\right]d\alpha$$
where
$$A(\alpha)=\int_{-\infty}^\infty f(x)\cos(\alpha x)dx\quad\mbox{ and } \quad B(\alpha)=\int_{-\infty}^\infty f(x)\sin(\alpha x)dx.$$
So, being the function in parts, i should break the integral from $A(\alpha)$ in intervals so i get
$$\begin{array}{ll} A(\alpha)&=\int_{-\infty}^\frac{\pi}{2}f(x)\cos(\alpha x)dx+\int_\frac{\pi}{2}^\infty f(x)\cos(\alpha x)dx\\ &=\int_{-\infty}^\frac{\pi}{2}\frac{\pi}{2}\cos(x)\cos(\alpha x)dx+\int_\frac{\pi}{2}^\infty 0\cos(\alpha x)dx\\ &=\int_{-\infty}^\frac{\pi}{2}\frac{\pi}{2}\cos(x)\cos(\alpha x)dx\\ &=\frac{\pi}{2}\frac{\sin(x)\cos(\alpha x)-\alpha\cos(x)\sin(\alpha x)}{1-\alpha^2}\left|_{-\infty}^\frac{\pi}{2}\right. \end{array}$$
Is this correct? i'm having trouble getting the right answer, following this, i get to
$$f(x)=\frac{1}{2}\int_0^\infty\frac{\sin(x)}{1-\alpha^2}d\alpha$$
which is quite different from what i needed, so i think i may be doing something wrong, any help would be aprreciated, thanks.