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In this paper (https://arxiv.org/pdf/hep-th/9506067.pdf), it is claimed that: \begin{equation} \widetilde{a}_n(\omega)=\frac{1}{2 \pi} \int_{-\pi}^\pi e^{i \omega \lambda} \frac{\sinh (n \gamma)}{\cosh (n \gamma)-\cos \lambda} d \lambda=e^{-n \gamma|\omega|} \end{equation}

which I can verify numerically. Here $\omega\in \mathbb{Z}$.

How do I perform this integral?

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    $\begingroup$ Ummm.... which equation in that paper? $\endgroup$ May 14, 2023 at 0:01
  • $\begingroup$ First, take out the constant factors.... $\frac{1}{2 \pi} \sinh(n \gamma)$ and replace needlessly complex constants, e.g., $\cosh(n \gamma) \rightarrow a$. Then use Euler's formula. $\endgroup$ May 14, 2023 at 1:26
  • $\begingroup$ @DavidG.Stork It is equation 20 in the paper (the middle equation). They claim Fourier transform of $\phi'(n)$ is $e^{-2\gamma|n|}$ and from Eq.16 in the paper, you see that $\phi'(n)$ is $\frac{\sinh (2 \gamma )}{\cosh (2 \gamma )-\cos (\alpha )}$. Which is the subcase of $n=2$ in my question. $\endgroup$
    – user824530
    May 14, 2023 at 1:30
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    $\begingroup$ Oh geez.... that is SO non-obvious!! $\endgroup$ May 14, 2023 at 1:31

2 Answers 2

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Let $a=e^{-n\gamma}$ and note that $$\frac{1-a^2}{1-2a\cos \lambda+a^2}=\sum_{k=-\infty}^{\infty} a^{|k|}e^{i k \lambda} $$

\begin{align} &\ \frac{1}{2 \pi} \int_{-\pi}^\pi \frac{e^{i \omega \lambda} \sinh (n \gamma)}{\cosh (n \gamma)-\cos \lambda} \ d \lambda\\ = &\ \frac{1}{2 \pi} \int_{-\pi}^\pi \frac{e^{i \omega \lambda} (1-a^2)}{1-2a\cos \lambda+a^2}\ d \lambda\\ =&\ \frac{1}{2 \pi}\sum_{k=-\infty}^{\infty} a^{|k|}\int_{-\pi}^\pi e^{i (\omega +k)\lambda} \ d \lambda = a^{|-w|}= e^{-n \gamma|\omega|} \end{align} (Implicit is that $\omega$ is an integer.)

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    $\begingroup$ It is not consistent with the numerical result: Let $a=0.1, \omega=0.5$ and write the integral $\int_{-\pi}^{\pi}e^{i(\omega +k)\lambda}d\lambda=\frac{2\sin(\pi(\omega+k))}{\omega+k}$. Wolfram gives 0.6782, but $a^\omega=\sqrt{0.1}=0.3162$ wolframalpha.com/input?i=1%2F%282pi%29*sum%5B0.1%5Eabs%5Bk%5D*%282+sin%5Bpi*%28k+%2B+0.5%29%5D%29%2F%28k+%2B0.5%29%2C%7Bk%2C-infinity%2C+infinity%7D%5D $\endgroup$
    – user1026811
    May 14, 2023 at 17:45
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Note: The OP's result holds only for $\omega\in\mathbb{Z}$. If $\omega\notin\mathbb{Z}$, then $z=0$ is a branch point, and we need to introduce a branch cut (i.e. along the negative x-axis), hence, the current unit circle contour cannot be used. Instead, we can use a key-hole contour, which will give an extra term, as posted in the comment below.

From now on, we assume $\omega\in\mathbb{Z}$.

$$I(\omega)=\frac{1}{2 \pi} \int_{-\pi}^\pi e^{i \omega \theta} \frac{\sinh (n \gamma)}{\cosh (n \gamma)-\cos \theta} d \theta=\frac{\sinh (n \gamma)}{2 \pi} \int_{-\pi}^\pi \frac{e^{i \omega \theta}}{\cosh (n \gamma)-\cos \theta} d \theta$$

Since $e^{i \omega \theta}=\cos(\omega\theta)+i\sin(\omega\theta)$, the integrand on the $\sin(\omega\theta)$ part is odd, hence vanishes. For cosine part, note that: $\cos(\omega\theta)=\cos(-\omega\theta)$, hence $I(-\omega)=I(\omega)$. Without loss of generality, we let $\omega\in\mathbb{N}$, and denote: $I=I(-\omega)=I(\omega)$.

Let $z=e^{i\theta}$, then we have,

$$e^{i\omega\theta}=z^\omega,~~~ \cos\theta=\frac{z+\frac{1}z}{2},~~~d\theta=\frac{1}{iz}dz$$

Plug in substitution, we get

$$I=-\frac{\sinh (n \gamma)}{\pi i} \oint_C \frac{z^{\omega}}{z^2-2\cosh (n \gamma)z+1} d z=-\frac{\sinh (n \gamma)}{\pi i} \oint_C \frac{z^{\omega}}{(z-z_1)(z-z_2)} d z$$ where $C$ is the unit circle, and $z_1=e^{n\gamma}, z_2=e^{-n\gamma}$. Note that only $z_2$ is inside the unit circle,

$$\oint_C \frac{z^{\omega}}{(z-z_1)(z-z_2)} d z=2\pi i\cdot \text {Res}(z_2)=2\pi i\cdot \frac{z_2^{\omega}}{z_2-z_1}=-\pi i\frac{e^{-n\gamma\omega}}{\sinh(n\gamma)}$$

Therefore,

$$I=e^{-n\gamma\omega}=e^{-n\gamma|\omega|}$$

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    $\begingroup$ You cannot apply residue theorem to general cases since $z^\omega$ has a branch cut and the contour is not closed. $\endgroup$
    – Po1ynomial
    May 14, 2023 at 5:26
  • $\begingroup$ Yes, in that case, we can apply a key-hole contour, with the branch cut on the negative x-axis. There will be an extra term $$\int_0^1 \frac{r^\omega}{(r-z_1)(r-z_2)}(e^{i\omega\pi}-e^{-i\omega\pi})dr$$ @Po1ynomial $\endgroup$
    – MathFail
    May 14, 2023 at 6:24
  • $\begingroup$ Your answer should address the general case more carefully than in a brief comment, this post will mislead many people otherwise. As written, your answer is not correct since $z\mapsto z^{\omega}$ need not be holomorphic on $C$ and its interior $\endgroup$
    – FShrike
    May 14, 2023 at 17:18
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    $\begingroup$ It is not consistent with the numerical result: Let $a=0.1, \omega=0.5$. Wolfram gives 0.6782, but $a^\omega=\sqrt{0.1}=0.3162$ wolframalpha.com/input?i=1%2F%282pi%29*sum%5B0.1%5Eabs%5Bk%5D*%282+sin%5Bpi*%28k+%2B+0.5%29%5D%29%2F%28k+%2B0.5%29%2C%7Bk%2C-infinity%2C+infinity%7D%5D $\endgroup$
    – user1026811
    May 14, 2023 at 17:48
  • $\begingroup$ The OP's equation holds only for integer cases, I added this note in my answer. Hope this won't mislead audience. @FShrike $\endgroup$
    – MathFail
    May 14, 2023 at 19:12

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