= Solution
With the stated <Fourier transform> convention, the <inverse Fourier transform> and the deterministic <autocorrelation of a square-integrable function> are
$$
\boxed{f(x)=\frac1{2\pi}\int_{\mathbb R}\widetilde f(k)e^{ikx}dk,\qquad h(x)=\int_{\mathbb R}\overline{f(y)}f(x+y)dy.}
$$
Assume initially sufficient decay and smoothness, for example a <Schwartz function>, to justify interchanging integrations and <Fourier inversion>. With $s=x+y$, the <cross-correlation> identity is
$$
\begin{aligned}
\widetilde h(k)&=\int_{\mathbb R}\int_{\mathbb R}e^{-ikx}\overline{f(y)}f(x+y)dy\,dx\\
&=\left[\int_{\mathbb R}e^{iky}\overline{f(y)}dy\right]\left[\int_{\mathbb R}e^{-iks}f(s)ds\right]=|\widetilde f(k)|^2.
\end{aligned}
$$
Evaluating $h$ directly and by its inverse transform at zero proves the <Parseval identity>:
$$
\boxed{\int_{\mathbb R}|f(x)|^2dx=h(0)=\frac1{2\pi}\int_{\mathbb R}|\widetilde f(k)|^2dk.}
$$
Approximation extends the identity to <L2 space>, as the <Plancherel theorem>.
For the compact cosine profile, the <Fourier transform of a truncated cosine> follows from $\cos x=(e^{ix}+e^{-ix})/2$:
$$
\begin{aligned}
\widetilde f(k)&=\frac{\sin[(k-1)\pi/2]}{k-1}+\frac{\sin[(k+1)\pi/2]}{k+1}\\
&=\boxed{\frac{2\cos(k\pi/2)}{1-k^2}}.
\end{aligned}
$$
At $k=\pm1$ the apparent singularities are removable, with value $\pi/2$. The direct squared norm is $\int_{-\pi/2}^{\pi/2}\cos^2x\,dx=\pi/2$. Hence <Parseval identity> and evenness give
$$
\int_0^\infty\frac{\cos^2(k\pi/2)}{(1-k^2)^2}dk=\frac{\pi^2}{8}.
$$
Putting $t=\pi k/2$ makes the left side $\pi^3/8$ times the requested integral. Therefore
$$
\boxed{\int_0^\infty\frac{\cos^2t}{(\pi^2/4-t^2)^2}dt=\frac1\pi.}
$$
At $t=\pi/2$ the integrand's continuous extension is $1/\pi^2$, so no principal-value prescription is required.
Back to article page
