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How can I show that, if $0 < x < 1$, then $$\sum_{n=-\infty}^{\infty} \frac{e(n \alpha)}{x^2 - n^2} = \frac{\pi}{x} \frac{\sin(2 \pi \alpha x) - \sin(2 \pi (\alpha - 1)x)}{1 - \cos(2 \pi x)},$$ where $e(n \alpha) = e^{2 \pi i n \alpha}$?

This seems to work with Poisson summation given the limits of summation, but I'm not sure how in this particular case. If I haven't done anything wrong, the Fourier transform of $e(t \alpha)/(x^2 - t^2)$ with respect to $t$ should be $$\frac{1}{x} \sqrt{\pi/2} \operatorname{sgn}(2 \pi \alpha + \omega) \sin(x(2 \pi \alpha + \omega)).$$ Then, we could consider $e^{i x(2 \pi \alpha + \omega)}$, evaluate the corresponding geometric sum, and then take the imaginary part. However, the presence of the $\operatorname{sgn}$ function makes this a little difficult.

River Li
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huh
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3 Answers3

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Too long for a comment

The evaluation is basically the same as in this post.

Let's consider $$S(\alpha, x)=\sum_{n=-\infty}^\infty\frac{e^{2\pi i\alpha n}}{x^2-n^2}; \,\alpha\in[0;1];\,x\neq0,\pm1,\pm2...$$

To evaluate $S_0$ we integrate the function $\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2\pi i\alpha z}}{x^2-z^2}$ along the rectangular contour $\,C:\,-N-\frac12-iN\to N+\frac12-iN\to N+\frac12+iN\to-N-\frac12+iN\to-N-\frac12-iN$.

Integral along this contour tends to zero as $N\to \infty$. On the other hand, $$\oint_C\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2\pi i\alpha z}}{x^2-z^2}dz=2\pi i\sum \operatorname{Res}\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2\pi i\alpha z}}{x^2-z^2}$$

The residues at the poles of the function $\frac{2\pi i}{e^{2\pi iz}-1}$ give us $S$; we also have two poles at $z=\pm x$. Hence, $$S+\underset{z=\pm x}{\operatorname{Res}}\frac{2\pi i}{e^{2\pi iz}-1}\,\frac{e^{2\pi i\alpha z}}{x^2-z^2}=0$$ $$\Rightarrow\,S=\underset{z=\pm x}{\operatorname{Res}}\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{2\pi i\alpha z}}{(z-x)(z+x)}=\frac{2\pi i}{2x}\left(\frac {e^{2\pi i\alpha x-\pi ix}}{e^{\pi ix}-e^{-\pi ix}}-\frac {e^{-2\pi i\alpha x+\pi ix}}{e^{-\pi ix}-e^{\pi ix}}\right)$$ $$\boxed{\,\,S=\frac\pi x\frac{\cos\big(\pi x(2\alpha-1)\big)}{\sin(\pi x)}\,\,}$$ what is the same as the original form of the answer, because $$\frac{\sin(2 \pi \alpha x) - \sin(2 \pi (\alpha - 1)x)}{1 - \cos(2 \pi x)}=2\frac{\sin(\pi x)\cos( 2\pi x\frac{2\alpha-1}2)}{2\sin^2(\pi x)}$$

Svyatoslav
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  • It's not immediate to me why the integral along this contour tends to zero as $N\to\infty$, could you, what am I missing? – Zima Oct 13 '24 at 08:08
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    @Zima I think you'll need to parameterize each segment, bound the integrand on each one, and then apply the M-L inequality. Each integral should then be $O(n^{-1})$ (or potentially $O(n^{-2})$, though I’m leaning towards $O(n^{-1})$—I haven’t worked through all the details). – huh Oct 13 '24 at 08:49
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    @Zima On the contour, at $\alpha\in[0;1]$ $$|\frac{e^{2\pi i\alpha z}}{e^{2\pi iz}-1}\frac1{z^2-x^2}|<|\frac{e^{2\pi i\alpha z}}{e^{2\pi iz}-1}||\frac1{z^2-x^2}|<|\frac{const}{z^2-x^2}|$$ (we avoid "dangerous points $e^{2\pi iz}-1=0$) – Svyatoslav Oct 13 '24 at 13:07
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Here is an alternative method for evaluating the sum. I proved here that if $k < x < k+1$ where $k \in \mathbb{Z}$ that

\begin{align*} \sum_{n=-\infty}^\infty \frac{e^{2 \pi i \alpha n}}{x + n} & = e^{-x i 2 \pi \{ \alpha \} + x \pi i} \dfrac{\pi}{\sin x \pi} \end{align*}

where $\{ \alpha \}$ is the fraction function. Using this, we have

\begin{align} \sum_{n=-\infty}^\infty \frac{e^{2 \pi i \alpha n}}{x^2-n^2} & = \frac{1}{2x} \sum_{n=-\infty}^\infty \frac{e^{2 \pi i \alpha n}}{x + n} - \frac{1}{2x} \sum_{n=-\infty}^\infty \frac{e^{2 \pi i \alpha n}}{-x + n} \nonumber \\ & = \frac{\pi}{2x} e^{-x i 2 \pi \{ \alpha \} + x \pi i} \dfrac{1}{\sin x \pi} + \frac{\pi}{2x} e^{x i 2 \pi \{ \alpha \} - x \pi i} \dfrac{1}{\sin x \pi} \nonumber \\ & = \frac{\pi}{x} \dfrac{1}{2\sin x \pi} (e^{-x i 2 \pi \{ \alpha \} + x \pi i} + e^{x i 2 \pi \{ \alpha \} - x \pi i}) \nonumber \\ & = \frac{\pi}{x} \dfrac{\cos (x \pi (2 \{ \alpha \} - 1))}{\sin x \pi} \end{align}

Dave77
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2

Here, I answer the OP's question directly. While Poisson's summation formula yields the correct result, it does not seem to provide a rigorous derivation.

Poisson's summation formula:

\begin{align*} \sum_{n=-\infty}^\infty F(n+t) & = \sum_{m=-\infty}^\infty \tilde{F} (2 \pi m) e^{2 \pi i m t} \end{align*}

where

\begin{align*} \tilde{F} (y) = \int_{-\infty}^\infty F(s) e^{-i y s} ds \end{align*}

So

\begin{align*} F (s) = \frac{e^{2 \pi i \alpha s}}{x^2-s^2} \end{align*}

and

\begin{align*} \tilde{F} (y) = \int_{-\infty}^\infty \frac{e^{2 \pi i \alpha s - iy s}}{x^2-s^2} ds \qquad (1) \end{align*}

We have the Fourier transform:

\begin{align*} \int_{-\infty}^\infty \frac{e^{-i ys}}{x^2-s^2} ds & = \frac{\pi}{x} \sin (xy) sgn (y) \end{align*}

Using this in (1):

\begin{align*} \tilde{F} (y) = \frac{\pi}{x} \sin (x (2 \pi \alpha - y)) sgn (2 \pi \alpha - y) \end{align*}

and putting $y=2 \pi m$:

\begin{align*} \tilde{F} (2 \pi m) = \frac{\pi}{x} \sin (x 2 \pi (\alpha - m)) sgn (2 \pi (\alpha - m)) \end{align*}

then

\begin{align*} \sum_{n=-\infty}^\infty F(n+t) & =\sum_{m=-\infty}^\infty \tilde{F} (2 \pi m) e^{2 \pi i m t} \nonumber \\ & = \frac{\pi}{x} \sum_{m=-\infty}^\infty \sin (x 2 \pi (\alpha - m)) sgn (2 \pi (\alpha - m)) e^{2 \pi i m t} \nonumber \\ & = \frac{\pi}{x} \sum_{m=-\infty}^\infty \sin (x 2 \pi (\alpha + m)) sgn (2 \pi (\alpha + m)) e^{-2 \pi i m t} \end{align*}

Presence of sgn:

Put $\alpha = r + \{ \alpha \}$ where $\{ \alpha \}$ is the fractional part of $\alpha$, then

\begin{align*} \sum_{n=-\infty}^\infty F(n+t) & = \frac{\pi}{x} \sum_{m=-\infty}^\infty \sin (2 \pi x (r+m + \{ \alpha \})) sgn (2 \pi (r+m + \{ \alpha \})) e^{-2 \pi i m t} \nonumber \\ & = \frac{\pi}{x} e^{2 \pi i t r} \sum_{m=-\infty}^\infty \sin (2 \pi x (r+m + \{ \alpha \})) sgn (2 \pi (r+m + \{ \alpha \})) e^{-2 \pi i (m+r) t} \nonumber \\ & = \frac{\pi}{x} e^{2 \pi i t r} \sum_{m=-\infty}^\infty \sin (2 \pi x (m + \{ \alpha \})) sgn (2 \pi (m + \{ \alpha \})) e^{-2 \pi i m t} \nonumber \\ & = \frac{\pi}{x} e^{2 \pi i t r} [- \sum_{m=-\infty}^{-1} \sin (2 \pi x (m + \{ \alpha \})) e^{-2 \pi i m t} + \sum_{m=0}^\infty \sin (2 \pi x (m + \{ \alpha \})) e^{-2 \pi i m t}] \end{align*}

Summing geometric series and taking imaginary part:

Then $t=0$,

\begin{align*} \sum_{n=-\infty}^\infty \frac{e^{2 \pi i \alpha n}}{x^2-n^2} & = \frac{\pi}{x} [- \sum_{m=-\infty}^{-1} \sin (2 \pi x (m+\{ \alpha \})) + \sum_{m=0}^\infty \sin (2 \pi x (m + \{ \alpha \}))] \nonumber \\ & = \frac{\pi}{x} [\sum_{m=1}^\infty \sin (2 \pi x (m-\{ \alpha \})) + \sum_{m=0}^\infty \sin (2 \pi x (m + \{ \alpha \}))] \nonumber \\ & = \frac{\pi}{x} Im [\sum_{m=1}^\infty e^{i2 \pi x (m-\{ \alpha \})} + \sum_{m=0}^\infty e^{i2 \pi x (m + \{ \alpha \})})] \nonumber \\ & = \frac{\pi}{x} Im [ \frac{e^{-i2 \pi x \{ \alpha \}}}{e^{- i2 \pi x} - 1} + \frac{e^{i2 \pi x \{ \alpha \}}}{1-e^{i2 \pi x}}] \nonumber \\ & = \frac{\pi}{x} Im [-\frac{e^{-i2 \pi x \{ \alpha \} +i \pi x}}{e^{i \pi x} - e^{-i \pi x}} - \frac{e^{i2 \pi x \{ \alpha \} - i \pi x}}{e^{i \pi x} - e^{-i \pi x}}] \nonumber \\ & = \frac{\pi}{x} Im [\frac{i}{2} \frac{e^{-i2 \pi x \{ \alpha \} +i \pi x}}{\sin (\pi x)} + \frac{i}{2} \frac{e^{i2 \pi x \{ \alpha \} - i \pi x}}{\sin (\pi x)}] \nonumber \\ & = \frac{\pi}{x} \frac{\cos (\pi x (2\{ \alpha \} - 1))}{\sin (\pi x)} \end{align*}

giving the correct answer.

How to make it legitimate?

Doing $\sum_{n=0}^\infty e^{2 \pi i xm} = \frac{1}{1-e^{2 \pi i x}}$ is not legitimate as this

\begin{align*} (1 - e^{2 \pi i x}) \sum_{n=0}^N e^{2 \pi i xm} = 1 - e^{2\pi i (N+1)x} = 1 - \cos (2 \pi (N+1)x) - i \sin (2 \pi (N+1)x) \end{align*}

is ill-defined in the limit $N \rightarrow \infty$. How do you adjust the above calculation to make it legitimate?

Dave77
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