Two conjectured identities of sums involving $\sin(x\sin x)/x^2$

Question

While tackling this question, I came up with the conjecture that the following identities hold:

Conjecture. Let $f(x)=\dfrac{\sin(x\sin x)}{x^2}$. Then

$$ \sum_{k=-\infty}^{\infty} f(k\pi+x) = 1 \qquad\text{and}\qquad \sum_{k=-\infty}^{\infty} (-1)^k f(k\pi + x) = \cos x. $$

Unfortunately, I have absolutely no idea how to tackle this problems. The only evidence I have for this one is a numerical simulation:

(Figure. Left: A comparison between the partial sum $\sum_{|k|\leq n} (-1)^k f(k\pi + x)$ and $\cos x$ / Right: A comparison between the partial sum $\sum_{|k|\leq n} f(k\pi + x)$ and $1$)

If the conjecture turns out to be true, then we can prove some fun identities, including the original posting that motivated this question. For example, we obtain the following identity as a by-product:

$$ \int_{-\infty}^{\infty} \frac{\sin(x\sin x)}{x^2} \, \mathrm{d}x = \int_{0}^{\pi} \sum_{k=-\infty}^{\infty} f(k\pi + x) \, \mathrm{d}x = \int_{0}^{\pi} \mathrm{d}x = \pi $$

Answer 1

This boils down to evaluating the sum $$ S(a,z)=\sum_{n\in\mathbb{Z}}\frac{e^{in\pi a}}{(n\pi+z)^2}\qquad(a\in\mathbb{R},z\in\mathbb{C}) $$ since clearly $$ \begin{gathered} \sum_{n\in\mathbb{Z}}f(n\pi+x)=\Im\big(e^{ix\sin x}S(1+\sin x,x)\big);\\ \sum_{n\in\mathbb{Z}}(-1)^n f(n\pi+x)=\Im\big(e^{ix\sin x}S(\sin x,x)\big). \end{gathered}\tag{*}\label{finalstep} $$

A possible approach to get $S(a,z)$ is via Poisson summation. Assume $0\leqslant a<2$ (w.l.o.g. by periodicity) and $\Im z>0$ (w.l.o.g. by analytic continuation). Let $f(t)=e^{ia\pi t}/(z+\pi t)^2$ and compute $$ \hat{f}(\omega)=\int_{-\infty}^\infty e^{-2\pi i\omega t}f(t)\,dt=g(a-2\omega),\quad g(\omega)=\frac1\pi\int_{-\infty}^\infty\frac{e^{i\omega t}\,dt}{(z+t)^2} $$ using, say, contour integration: if $\omega\geqslant 0$ then $g(\omega)=0$, otherwise $$ g(\omega)=-2i\operatorname*{Res}_{t=-z}\frac{e^{i\omega t}}{(z+t)^2}=2\omega e^{-i\omega z}. $$ Thus $$ S(a,z)=\sum_{n\in\mathbb{Z}}\hat{f}(n)=2e^{-iaz}\sum_{n=1}^\infty(a-2n)e^{2inz}=\frac{e^{-iaz}}{\sin^2 z}(1+iae^{iz}\sin z) $$ (the result holds for $0\leqslant a\leqslant 2$). It remains to do the substitutions in \eqref{finalstep}.

Answer 2

The following is an alternative way to show that $$S(a,z) = \sum_{n\in\mathbb{Z}}\frac{e^{i n \pi a}}{(n\pi+z)^2} = \frac{e^{-iaz}}{\sin^2 (z)}\left(1+iae^{iz}\sin (z)\right), \quad 0 \le a \le 2.$$

This series is evaluated in metamorphy's answer using the Poisson summation formula.

Assume that $f(w)$ is a meromorphic function and $f(w) = O\left(\frac{1}{w^{2}} \right)$ as $|w| \to \infty$.

A residue calculation using the cotangent trick can be used to evaluate $$\sum_{n \in \mathbb{Z}} e^{i n \pi a} f(n), \quad 0 \le a \le 2. $$

But instead of integrating $\pi \cot(\pi w) e^{i \pi a w} f(w)$ around a square contour with vertices at $ \pm \left(N+ \frac{1}{2} \right) \pm i\left(N+ \frac{1}{2} \right)$, we integrate $$\pi \left( \cot (\pi w) -i \right) e^{i \pi a w } f(w) =\pi e^{-i \pi w} \csc(\pi w) e^{i \pi a w}f(w).$$

The reason for doing this, as I explained in an answer here, is to counter the fact that the magnitude of $e^{i a w}$, $a >0$, grows exponentially as $\Im(w) \to - \infty$.

The restriction $0 \le a \le 2$ ensures that the integral vanishes as $N \to \infty$.

For $$f(w) = \frac{1}{(\pi w +z)^{2}},$$ we get $$ \begin{align} S(a,z) = \sum_{n\in\mathbb{Z}}\frac{e^{i n \pi a}}{(n\pi+z)^2} &= - \operatorname{Res} \left[\frac{ \pi e^{-i \pi w}\csc(\pi w) \, e^{i \pi a w}}{(\pi w +z)^{2}}, w= - \frac{z}{\pi} \right] \\ & = -\operatorname{Res} \left[\frac{\csc(\pi w) \, e^{i\pi (a-1)w }}{\pi (w +\frac{z}{\pi})^{2}}, w= - \frac{z}{\pi} \right] \\ &= -\lim_{w \to - \frac{z}{\pi}}\frac{1}{\pi} \frac{\mathrm d}{\mathrm dw}\csc(\pi w) \, e^{i \pi (a-1)w} \\ &= - \lim_{w \to - \frac{z}{\pi}}\frac{1}{\pi} \left(- \pi \cot(\pi w) \csc(\pi w) \, e^{i \pi (a-1)w} + i \pi (a-1) \csc(\pi w) \, e^{i \pi (a-1)w}\right) \\ &= e^{- i (a-1)z} \left(\cot(z)\csc(z)+ i (a-1) \csc(z) \right) \\ &= \frac{e^{-iaz} e^{iz}}{\sin^{2}(z)} \left(\cos(z) +i(a-1) \sin(z) \right) \\ &= \frac{e^{-iaz} e^{iz}}{\sin^{2}(z)} \left(e^{-i z} +i a \sin(z) \right) \\ &= \frac{e^{-i az}}{\sin^{2}(z)} \left(1 + i a e^{i z} \sin(z) \right). \end{align}$$

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To show that $$\sum_{n\in\mathbb{Z}}\frac{(-1)^n e^{in\pi a}}{(n\pi+z)^2}=\frac{e^{-iaz}}{\sin^2 (z)}\left( \cos (z)+ia\sin (z) \right), \quad -1 \le a \le 1, $$ we can use the typical cosecant trick.

Similar to before, the restriction $-1 \le a \le 1$ ensures that the integral vanishes as $N \to \infty$.

We get

$$ \begin{align} \sum_{n\in\mathbb{Z}}\frac{(-1)^n e^{in\pi a}}{(n\pi+z)^2} &= - \operatorname{Res} \left[\frac{ \pi \csc(\pi w) \, e^{i \pi a w}}{(\pi w +z)^{2}}, w= - \frac{z}{\pi} \right] \\ &= -\operatorname{Res} \left[\frac{\csc(\pi w) \, e^{i\pi a w }}{\pi (w +\frac{z}{\pi})^{2}}, w= - \frac{z}{\pi} \right] \\ &= -\lim_{w \to - \frac{z}{\pi}}\frac{1}{\pi} \frac{\mathrm d}{\mathrm dw}\csc(\pi w) \, e^{i \pi a w} \\ &= - \lim_{w \to - \frac{z}{\pi}}\frac{1}{\pi} \left(- \pi \cot(\pi w) \csc(\pi w) \, e^{i \pi a w} + i \pi a\csc(\pi w) \, e^{i \pi a w}\right) \\ &=e^{- i a z} \left(\cot(z)\csc(z)+ i a \csc(z) \right) \\ &= \frac{e^{-ia z}}{\sin^{2}(z)} \left(\cos(z) +ia \sin(z) \right). \end{align}$$