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!

  • $\begingroup$ 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

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!


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

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

  • $\begingroup$ 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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