Flex Proof

In December 2020, as part of an Information Theory class, we encountered a question in a homework assignment prompting us to demonstrate that ${\int }_0^{\infty }(\frac{1-\cos x}{x}{)}^{2}dx=\pi /2$. Our teacher refrained from grading it upon realizing its impossibility to solve within the recommended time. However, I gained insight into a solution by utilizing Wolfram Alpha. Although I cannot recall the exact method employed to gain the insight (probably through the iterative examination of function graphs), the proof unfolded as follows:

The value of the integral ${\int }_0^{\infty }(\frac{1-\cos x}{x}{)}^{2}dx$ can be obtained by finding a function whose Fourier transform matches, up to a multiplicative constant, the function $x\mapsto \frac{1-\cos x}{x}$. The key insight lies in considering a piecewise-defined function $f:\mathbb{R}\to \mathbb{C}$, where $f=1$ over $(0,\frac{1}{2\pi }]$, $f=-1$ over $[-\frac{1}{2\pi },0)$, and $f=0$ elsewhere. Indeed: $\mathcal{F}f(y)=1/(2\pi ){\int }_{-1/(2\pi )}^0-e^{-iyx}dx+1/(2\pi ){\int }_0^{1/(2\pi )}{e}^{-iyx}dx=-i(1-\cos y)/(2\pi y)-i(1-\cos y)/(2\pi y)=-i/\pi (1-\cos y)/y$. (Skipping how the integral of $x\mapsto e^{-iyx}$ is computed for brevity here).

From this, we derive:

$$\begin{array}{rl}\forall x\in \mathbb{R},{\int }_{-\infty }^{\infty }(\frac{1-\cos x}{x}{)}^{2}dx& ={\int }_{-\infty }^{\infty }|-\mathcal{F}f(x)\frac{\pi }{i}{|}^{2}dx\\ & ={\pi }^2{\int }_{-\infty }^{\infty }|\mathcal{F}f(x){|}^{2}dx\\ & ={\pi }^2{\int }_{-\infty }^{\infty }|f(x){|}^{2}dx\text{ (by Plancherel)}\\ & ={\pi }^2{\int }_{-1/(2\pi )}^{1/(2\pi )}1dx\\ & =\pi .\end{array}$$

Therefore, by exploiting the evenness of the cosine function, we conclude that: ${\int }_0^{\infty }(\frac{1-\cos x}{x}{)}^{2}dx=\pi /2$. 💪