Prove that $\sum\limits_{n=1}^\infty\frac{\text{Si}^2(\pi n)}{n^2}=\frac{\pi^2}2$ I was doing numerical calculations and found
$$\sum _{n=1}^{\infty } \left(\frac{\text{Si}(\pi  n)}{\pi n}\right)^2\overset{?}{=}\frac{1}{2},$$
where $\text{Si}(x)$ means the sine integral. Interestingly, it seems that only when the terms are squared can the result be a rational number.
I attempted to evaluate the following series expansion about $x=0$:
$$\frac{\text{Si}(\pi  x)}{\pi  x}=\sum_{n=1}^\infty\frac{(-1)^n(\pi x)^{2 n}}{(2 n+1)^2 (2 n)!}=1-\frac{\pi ^2 n^2}{9}+\frac{13 \pi ^4 n^4}{2025}-\frac{8 \pi ^6 n^6}{33075}+O\left(n^8\right).$$
But it didn't give me any clue. I still have no idea about how to evaluate
$$\sum_{n=1}^\infty\ \displaystyle\left(\frac{\text{Si}(\pi  n)}{\pi n}\right)^2.$$
 A: $$\frac{\text{Si}(\pi n)}{\pi n}=\int_{0}^{1}\frac{\sin(\pi n x)}{\pi n x}\,dx \tag{1}$$
hence for any $z\in(0,1)$
$$ f(z)\stackrel{\text{def}}{=}\sum_{n\geq 1}\frac{\text{Si}(\pi n)}{\pi n}\,\cos(\pi n z)=\int_{0}^{1}\frac{W_z(x)}{x}\,dx \tag{2}$$
where $W_z(x)$ is a piecewise-linear function over $(0,1)$, with a jump discontinuity at $z$, $W_z(0^+)=W_z(1^-)=0$ and derivative equal to $-\frac{1}{2}$ at any point of $(0,1)$ which differs from $z$, by the Fourier series of the sawtooth wave. It follows that
$$ f(z) = \int_{0}^{z}-\frac{dx}{2}+\int_{z}^{1}\frac{1-x}{2x}\,dx=-\frac{1+\log z}{2}\tag{3}$$
and by Parseval's theorem
$$ \sum_{n\geq 1}\left(\frac{\text{Si}(\pi n)}{\pi n}\right)^2 = 2\int_{0}^{1}f(z)^2\,dz=\frac{1}{2}\int_{0}^{1}(1+\log z)^2\,dz\stackrel{\color{green}{\checkmark}}{=}\frac{1}{2}\tag{4}$$
as conjectured.
A: I combined both of my answers into one answer.

If $-\pi \le a \le \pi$, and if we define the value of $\left(\frac{\operatorname{Si}(az)}{z} \right)^{2}$ at $z=0$ to to be its limiting value, then we can use the Abel-Plana formula under the weakened conditions mentioned in achille hui's answer to a previous question to get that
$$\sum_{n=0}^{\infty} \left(\frac{\operatorname{Si}(an)}{n} \right)^{2} = a^{2} + \sum_{n=1}^{\infty} \left(\frac{\operatorname{Si}(an)}{n} \right)^{2} =\int_{0}^{\infty} \left(\frac{\operatorname{Si}(ax)}{x} \right)^{2} \, dx + \frac{a^{2}}{2} +i(0),$$
where, integrating by parts, we get $$ \begin{align}\int_{0}^{\infty}\left(\frac{\operatorname{Si}(ax)}{x} \right)^{2} \, dx  &= 2 \int_{0}^{\infty}\frac{\sin (ax)}{x^{2}} \operatorname{Si}(ax) \, dx \\ &= 2 \int_{0}^{\infty} \frac{\sin(ax)}{x^{2}} \int_{0}^{1} \frac{\sin(axt)}{t} \, dt \, dx \\ &=2 \int_{0}^{1} \frac{1}{t} \int_{0}^{\infty} \frac{\sin(ax) \sin(atx)}{x^{2}} \, dx \, dt \tag{1} \\ &= \pi \int_{0}^{1} \frac{\min \left(|a|, |a|t\right)}{t} \, dt \\ &=\pi \int_{0}^{1} |a| \, dt \\ &= \pi |a|. \end{align}$$
Therefore, $$\sum_{n=1}^{\infty} \left(\frac{\operatorname{Si}(an)}{n} \right)^{2} = \pi |a|- \frac{a^{2}}{2}, \quad -\pi \le a \le \pi. $$
The convergence is indeed quite slow.

The asymptotic expansion of the sine integral as $|z| \to \infty$
$(1)$ How do I show that $\int_0^\infty \frac{\sin(ax) \sin(bx)}{x^{2}} \, \mathrm dx = \pi \min(a,b)/2$

The alternating version of your series, namely, $$\sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{\operatorname{Si(an)}}{n}\right)^{2}  $$ also has a simple closed-form expression for $- \frac{\pi}{2} \le a \le \frac{\pi}{2}$.
Specifically, $$\sum_{n=1}^{\infty} (-1)^{n-1}\left(\frac{\operatorname{Si(an)}}{n}\right)^{2} = \frac{a^{2}}{2} ,  \quad  -\frac{\pi}{2} \le a \le \frac{\pi}{2}. \tag{1}$$
What's particularly interesting about $(1)$ is that it remains true if $\operatorname{Si}(an)$ is replaced by $\sin(an)$.
To prove $(1)$, we can integrate the complex function $$f(z) = \pi \csc(\pi z) \left(\frac{\operatorname{Si}(az)}{z}\right)^{2} $$ around a rectangular contour (call it $C_{N}$) with vertices at $\pm \left(N+ \frac{1}{2} \right)\pm i \left(N+ \frac{1}{2} \right)$, where $N$ is some positive integer.
Notice that except for a removable singularity at the origin, the function $\left( \frac{\operatorname{Si}(az)}{z} \right)^{2}$ is an analytic function on the entire complex plane.
So by integrating around the contour, and using the fact that $\pi \csc(\pi z)$ has simple poles at the integers with residues that alternate between $1$ and $-1$, we get $$\int_{C_{N}} f(z) \, dz = 2 \pi i \left(2 \sum_{n=1}^{N} (-1)^{n} \left(\frac{\operatorname{Si}(az)}{z}\right)^{2}  + \operatorname{Res}[f(z), 0]\right), $$ where $$\operatorname{Res}[f(z),0] = \pi \lim_{z \to 0} \frac{z}{\sin(\pi z)} \left(\frac{\operatorname{Si}(az)}{z}\right)^{2} = \pi \left(\frac{1}{\pi}\right) \left(a^{2} \right) = a^{2}.$$
If we can  argue that $\csc(\pi z) \operatorname{Si}^{2}(ax)$ remains bounded on the contour as $N \to \infty$ through the positive integers, then it follows from the estimation lemma that $\int_{C_{N}} f(z) \, dz$ vanishes as $N \to \infty$.
The asymptotic expansion of the sine integral  as $|z| \to \infty$ tells us that the magnitude of $\operatorname{Si}^{2}(az)$ eventually starts growing exponentially as $\operatorname{Im}(z) \to \pm \infty$.
But if $- \frac{\pi}{2} \le a \le \frac{\pi}{2}$, the exponential decay of the magnitude of $\csc(\pi z)$ as $\operatorname{Im}(z) \to \pm \infty$ counters this growth.
Therefore, if $\frac{\pi}{2} \le a \le \frac{\pi}{2}$, then $$ \lim_{N \to \infty} \int_{C_{N}} f(z) \, dz = 0 = 2 \pi i  \left(2 \sum_{n=1}^{\infty} (-1)^{n}\left(\frac{\operatorname{Si}(az)}{z}\right)^{2}  + a^{2}\right),$$ and the result follows.
Alternatively, you could use the alternating version of the Abel-Plana formula.
