I am trying to compute the Fourier transform of the triangle function \begin{equation} f(x) = \begin{cases} 1 -| x | & \text{if}~~ | x |\leq 1 \\ 0 & \text{if}~~ | x | > 1 \end{cases} \end{equation}
using the Fourier transform $\hat{f}$ = $\int_{-\infty}^{\infty} f$(x)e$^{-ikx}$dx. Plugging in the equation I get $\hat{f}$ =$\int_{-1}^{1} (1 - | x |)$ e$^{-ikx}$ dx which I expand to to $\hat{f}$ =$\int_{-1}^{0} e^{-ikx} + xe^{-ikx}$ dx + $\int_{0}^{1} e^{-ikx} - xe^{-ikx}$ dx. Evaluating the integrals then adding gives me $\frac{2}{k^2}$ - $\frac{2e^{-ik}(1-ik)}{k^2}$.
However, the result is apparently supposed to be $\frac{\sin^4y}{y^4}$, which I don't see how I can get to from what I have. Is there some different way to proceed to get this result?
$$\begin{aligned} \int_{-\infty}^\infty f(x) e^{-iyx} \, dx &= \int_{-1}^0 (1+x) e^{-iyx} \, dx + \int_0^1 (1-x) e^{-iyx} \, dx\\ &=\left[\frac{1}{y^2} - \frac{e^{iy}}{y^2} + \frac{i}{y} \right]+\left[\frac{1}{y^2} - \frac{e^{-iy}}{y^2} - \frac{i}{y} \right] \\&= \frac{2-2 \cos y}{y^2}. \end{aligned} $$
Using the trigonometric identity $$1 -\cos y = 2 \sin^2\frac{y}{2}$$ we can write the integral as
$$ \int_{-\infty}^\infty f(x) e^{-iyx} \, dx = \frac{4 \sin^2\frac{y}{2}}{y^2}$$ which will be found following the suggestion in the comments to use convolution.