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 $a$.
 A: Note that we can write
$$\begin{align}
\int_{-\infty}^\infty\frac{\sin^2(\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds&=\frac12\int_{-\infty}^\infty\frac{1-\cos(2\pi s)}{(\pi s)^2}e^{i2\pi sx}\,ds\\\\
&=\int_0^\infty \frac{1-\cos(2\pi s)}{(\pi s)^2}\,\cos(2\pi sx)\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-\frac12\left(\cos(2\pi s(x+1))+\cos(2\pi s(x-1))\right)}{(\pi s)^2}\,ds\\\\
&=\int_0^\infty \frac{\cos(2\pi sx)-1}{(\pi s)^2}\,ds\\\\
&+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x+1))}{(\pi s)^2}\,ds\\\\
&+\frac12\int_0^\infty \frac{1-\cos(2\pi s(x-1))}{(\pi s)^2}\,ds\\\\
&=-|x|+\frac12|x+1|+\frac12|x-1|
\end{align}$$
A: We can prove this formally without using the given identity; just using the various Fourier transform identities.
First consider, \begin{align*} \mathcal F^{-1}(\sin(\pi s)) &= \frac{1}{2i} \int_{\mathbb R} (e^{i\pi s} - e^{-i\pi s})e^{2\pi i xs} ds \\ &= \frac{1}{2i} \int_{\mathbb R} e^{2i\pi s\left(x + \tfrac 1 2 \right)} +  e^{2i\pi s\left(x - \tfrac 1 2 \right)} \\ &= \frac{1}{2i}\left[\delta\left(x + \tfrac 1 2 \right) - \delta\left(x-\tfrac 1 2 \right) \right].
\end{align*} Then \begin{align*}\mathcal F^{-1}(\sin (\pi s)) &= \mathcal \pi F^{-1}\left(s \frac 1 {\pi s} \sin(\pi s) \right) \\
&= -\frac{1}{2 i} \frac{d}{dx} \mathcal F^{-1}\left(\frac{\sin(\pi s)}{\pi s}\right).
\end{align*} Then \begin{align*}F^{-1}\left(\frac{\sin(\pi s)}{\pi s} \right) &= -2i \int^x_{-\infty}[F^{-1}(\sin (\pi s))](y) dy \\ 
&=\int^x_{-\infty} \left[\delta\left( y - \tfrac 1 2 \right) - \delta\left(y+\tfrac 1 2 \right) \right] dy = \left\{ \begin{matrix} 1, & x \in \left(-\tfrac 1 2, \tfrac 1 2\right) \\ 0, & \text{otherwise}\end{matrix} \right.
\end{align*} Finally, \begin{align*}\mathcal F^{-1}\left( \frac{\sin^2(\pi s)}{(\pi s)^2} \right) &= F^{-1}\left(\frac{\sin(\pi s)}{\pi s} \right) * F^{-1}\left(\frac{\sin(\pi s)}{\pi s} \right) \\ 
&= \int_{\mathbb R} \chi_{(-1/2,1/2)}(y) \chi_{(-1/2,1/2)}(x-y) dy \\
&= \int^{1/2}_{-1/2} \chi_{(x-1/2, x+1/2)}(y) dy.
\end{align*} where $\chi_A$ denotes the indicator function on the set $A$. If $\lvert x \rvert > 1$, then the function is zero over the range of integration, so the integral is zero. For $x \in (-1,0)$, we have $$\int^{1/2}_{-1/2} \chi_{(x-1/2, x+1/2)}(y) dy = \int^{1/2 + x}_{-1/2} dy = 1+x$$ and for $x \in (0,1)$, we get $$\int^{1/2}_{-1/2} \chi_{(x-1/2, x+1/2)}(y) dy = \int^{1/2}_{-1/2+x} dx = 1-x.$$ Thus $$\mathcal F^{-1}\left( \frac{\sin^2(\pi s)}{(\pi s)^2} \right) = \left \{\begin{matrix} 
0, & \lvert x \rvert > 1 \\
1+x, & x \in (-1,0), \\
1-x, & x \in (0,1).
\end{matrix}\right.$$
