Taking inverse Fourier transform of $\frac{\sin^2(\pi s)}{(\pi s)^2}$ How do I show that
$$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds = \begin{cases} 1+x & \text{if }-1 \le x \le 0 \\ 1-x & \text{if }0 \le x \le 1 \\ 0 & \text{otherwise} \end{cases}$$
I know that $\sin^2(\pi s)=\frac{1-\cos(2\pi s)}{2}=\frac{1-(e^{2\pi i s}-e^{-2\pi i s}))/2}{2}$, so
$$\int_{-\infty}^\infty \frac{\sin^2(\pi s)}{(\pi s)^2} e^{2\pi isx} \, ds=2\int_0^\infty \frac{2e^{2\pi isx}-(e^{2\pi is(1+x)}+e^{2\pi i s(-1+x)})}{4\pi^2s^2} \, ds$$
I am also allowed to use the known identity
$$\int_{-\infty}^\infty \frac{1-\cos(a \pi x)}{(\pi x)^2} \, dx = |a|$$
for some real number $a$.

Note that we can write
$$\begin{align}
\int_{-\infty}^\infty\frac{\sin^2(\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds&=\frac12\int_{-\infty}^\infty\frac{1-\cos(2\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds\\\\
&=\int_0^\infty \frac{1-\cos(2\pi s)}{(\pi s)^2}\,\cos(2\pi sx)\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-\frac12\left(\cos(2\pi s(x+1))+\cos(2\pi s(x-1))\right)}{(\pi s)^2}\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-1}{(\pi s)^2}\,ds\\\\
&+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x+1))}{(\pi s)^2}\,ds\\\\
&+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x-1))}{(\pi s)^2}\,ds\\\\
&=-|x|+\frac12|x+1|+\frac12|x-1|
\end{align}$$
