I try to prove this equation using Fourier transform
If $$ K(y)= \left\{ \begin{array}{ll} 1-\mid y \mid &, \mid y \mid \leq 1 \\ 0 &, \mid y \mid \geq 1 \end{array} \right. $$ show that
$$k(u)=\int_{-\infty}^\infty e^{iuy}K(y) dy =\Bigg[\frac{\sin (u/2)}{(u/2)}\Bigg]^2$$
my attempt :
$$k(u)=\int_{-\infty}^0 e^{iuy}(1+y) dy +\int_{0}^{\infty} e^{iuy}(1-y) dy $$ How could I do ? Any attempt will be helpful
$$k(u)=\int_{-\infty}^\infty e^{iuy}K(y) dy =\int_{-1}^0 (1+y)e^{iuy} dy+\int_{0}^1 (1-y)e^{iuy} dy $$ $$=\int_{-1}^0 e^{iuy}dy+\int_{-1}^0ye^{iuy} dy+\int_{0}^1 e^{iuy}dy -\int_{0}^1ye^{iuy} dy$$ $$=\Bigg[ \frac{e^{iuy}}{iu} \Bigg]_{-1}^0 + \Bigg[ \frac{e^{iuy}}{u^2}(1-iuy) \Bigg]_{-1}^0 + \Bigg[ \frac{e^{iuy}}{iu} \Bigg]_{0}^1 - \Bigg[ \frac{e^{iuy}}{u^2}(1-iuy) \Bigg]_{0}^1$$ $$=\frac{1}{iu}-\frac{e^{-iu}}{iu}+\frac{1}{u^2}-\frac{e^{-iu}}{u^2}(1+iu)+\frac{e^{iu}}{iu}-\frac{1}{iu}-\frac{e^{iu}}{u^2}(1-iu)+\frac{1}{u^2}$$ $$=\frac{e^{iu}-e^{-iu}}{iu}+\frac{2}{u^2}-(\frac{e^{iu}+e^{-iu}}{u^2})-\frac{e^{iu}-e^{-iu}}{iu}$$ $$=\frac{2}{u^2}-(\frac{e^{iu}+e^{-iu}}{u^2})=\frac{2}{u^2}-\frac{2\cos u}{u^2}=\frac{2}{u^2}(1-\cos u)=\frac{4\sin^2 (u/2)}{u^2}$$ $$=\frac{\sin^2 (u/2)}{(u/2)^2}=\Bigg[\frac{\sin (u/2)}{(u/2)}\Bigg]^2$$