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I get into trouble in evaluating this integral: $$ C(a)=\frac{1}{i\beta}\int_\Gamma \cot\frac{\pi z}{\beta}\frac{1}{\sin^2\frac{z}{2}}dz $$ where the contour $\Gamma$ consists of two vertical lines, (−π − i∞, −π + i∞) and (π + i∞,π − i∞).The result is: $$ -\frac{2}{3}((\frac{2\pi}{\beta})^2-1) $$ The integral can be evaluated via residues. Could show me how to do that integral? Thanks a lot!

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  • $\begingroup$ What are the restrictions on $\beta$? $\endgroup$ – Mark Viola Sep 18 '15 at 3:55
  • $\begingroup$ @Dr.MV $\beta$ is positive real number. $\endgroup$ – JQ Skywalker Sep 18 '15 at 4:14
  • $\begingroup$ @Dr.MV It just tells me this can be evaluated via residues and gives me the result above. I do not know how to obtain it...There are many papers which use this result such as (26) in this paper: arxiv.org/pdf/1104.3712v1.pdf $\endgroup$ – JQ Skywalker Sep 18 '15 at 4:33
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We begin with the contour integral

$$C(a)=\frac{1}{i\beta}\oint_\Gamma \frac{\cot\frac{\pi z}{\beta}}{\sin^2\frac{z}{2}}dz$$

where for $|\beta|>\pi$, the only singularity within $\Gamma$ is at $z=0$ (a pole of order three). In order to calculate the residue at the origin, the following expansions are useful.

$$\begin{align} \cos (\pi z/\beta)&=1-\frac12 (\pi z/\beta)^2+O(z^4) \tag 1\\\\ \sin^2(z/2)&=(z/2)^2\left(1-\frac{1}{12}z^2+O(z^4)\right) \tag 2\\\\ \sin(\pi z/\beta)&=(\pi z/\beta)\left(1-\frac16 (\pi z/\beta)^2+O(z^4)\right)\tag 3 \end{align}$$

Using $(1)-(3)$, we have the following expansion

$$\begin{align} \frac{\cot\frac{\pi z}{\beta}}{\sin^2\frac{z}{2}}&=\frac{\cos (\pi z/\beta)}{\sin^2(z/2)\sin(\pi z/\beta)}\\\\ &=\frac{1-\frac12 (\pi z/\beta)^2+O(z^4)}{(z/2)^2(\pi z/\beta)\left(1-\frac{1}{12}z^2+O(z^4)\right)\left(1-\frac16 (\pi z/\beta)^2+O(z^4)\right)}\\\\ &=\frac{1-\frac12 (\pi z/\beta)^2+O(z^4)}{(z/2)^2(\pi z/\beta)\left(1-\left(\frac{1}{12}+\frac{\pi^2}{6\beta^2}\right)+O(z^4)\right)}\\\\ &=\frac{\left(1-\frac12 (\pi z/\beta)^2+O(z^4)\right)\left(1+\left(\frac{1}{12}+\frac{\pi^2}{6\beta^2}\right)+O(z^4)\right)}{(z/2)^2(\pi z/\beta)}\\\\ &=\frac{1+\left(\frac1{12}-\frac{\pi^2}{3\beta^2}\right)z^2+O(z^4)}{(z/2)^2(\pi z/\beta)}\\\\ &=\frac{4\beta/\pi}{z^3}-\frac{(4\beta/\pi)\left(\frac1{12}-\frac{\pi^2}{3\beta^2}\right)}{z^1}+O(z) \end{align}$$

Therefore, the residue of $\frac{\cot\frac{\pi z}{\beta}}{\sin^2\frac{z}{2}}$ at $z=0$ is given by

$$\text{Res}\left(\frac{\cot(\pi z/\beta)}{\sin^2(z/2)},z=0\right)=(4\beta/\pi)\left(\frac1{12}-\frac{\pi^2}{3\beta^2}\right)$$

Finally, using the Residue Theorem, we obtain

$$\bbox[5px,border:2px solid #C0A000]{\frac{1}{i\beta}\oint_\Gamma \frac{\cot\frac{\pi z}{\beta}}{\sin^2\frac{z}{2}}dz=2\pi i \frac{1}{i\beta}(4\beta/\pi)\left(\frac1{12}-\frac{\pi^2}{3\beta^2}\right)=-\frac23\left(\left(\frac{2\pi}{\beta}\right)^2-1\right)}$$

as was to be shown!


NOTE:

While the use of expansions can often facilitate analyses such as the one herein, we can also have used Cauchy's Integral Formula or here to calculate the residue as

$$\text{Res}\left(\frac{\cot(\pi z/\beta)}{\sin^2(z/2)},z=0\right)=\frac1{2!} \lim_{z\to 0}\frac{d^2}{dz^2}\left(z^3\frac{\cot(\pi z/\beta)}{\sin^2(z/2)}\right)$$

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  • $\begingroup$ I See! It is stupid that I almost forget the power of Taylor expansion...It's very kind of you and thank you very much~ $\endgroup$ – JQ Skywalker Sep 18 '15 at 5:57
  • $\begingroup$ You're welcome. My pleasure. And it wasn't stupid at all. There is another way to go which uses the Cauchy Integral Formula. You can find the residue at $z=0$ as $$\frac1{2!} \lim_{z\to 0}\frac{d^2}{dz^2}\left(z^3\frac{\cot(\pi z/\beta)}{\sin^2(z/2)}\right)$$ $\endgroup$ – Mark Viola Sep 18 '15 at 6:05

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