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The first part of this question required me to find out the zeroes of the denominator, and to treat the equation as that of a complex number, which allows us to write: $$\frac{1}{z^4-2\cos(2\theta)z^2 +1}=\frac{1}{(z-e^{i\theta})(z+e^{i\theta})(z-e^{-i\theta})(z+e^{-i\theta})}$$ As far as I can understand these zeros have the effect of giving a circle of singularities in the complex plane with radius 1.

I assume to do the integral given in the question, I would have to do some kind of contour integration over the complex plane, but I am stumped as to which contour I should use to do this. I assume a branch cut will also be needed due to the transcendental nature of the zeroes in the denominator.

As always and help is appreciated - thanks!

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For the contour, you can simply choose a semi-circle of radius $R>1$in the upper half plane, as usual oriented counter-clockwise. I don't think you need to consider any branch cuts. You'll have two singularities in your contour, one at $z=e^{i\theta}$, and the other at $z=-e^{-i\theta}$. The integral over the semi-circle will vanish as $R\to\infty$. The sum of your residues is $$\frac{1}{2e^{i\theta}}\frac{1}{2i\sin\theta}\frac{1}{2\cos\theta}+\frac{1}{-2e^{-i\theta}}\frac{1}{-2i\sin\theta}\frac{1}{2\cos\theta} = \frac{1}{4i\sin\theta}.$$ Hence, multiplying by $i2\pi $, $$\int_{-\infty}^\infty \frac{dx}{x^4 - 2 \cos(2\theta)x^2+1}= \frac{\pi}{2\sin\theta}.$$

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  • $\begingroup$ Thank you for your answer - I am still slightly confused though, as it seems as though $z=e^{i\theta}$ and $z=-e^{-i\theta}$ are loci in the complex plane as opposed to a fixed numbers. Also wouldn't the contour run through these singularities as opposed to enclosing them? $\endgroup$ – user793781 May 29 at 17:32
  • $\begingroup$ EDIT - sorry I just saw that you said a semicircle with radius > 1 - in that case the contour would enclose the singularities. I am still a bit of stuck on the first point though. $\endgroup$ – user793781 May 29 at 17:36
  • $\begingroup$ I see what you're saying, but the integrand is done for a fixed value of $\theta$ so that e.g. $\exp(i\theta)$ is just a complex number. $\endgroup$ – Zachary May 29 at 17:44
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    $\begingroup$ Shouldn't this be $\pi/(2\sin|\theta|)$? The integrand is strictly positive, so the integral can't be negative. $\endgroup$ – eyeballfrog Jul 21 at 20:22
  • $\begingroup$ @eyeballfrog See my answer. $\endgroup$ – Felix Marin Jul 21 at 20:28
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\underline{\underline{\mbox{With}\ \theta \not\in \pi\mathbb{Z}}}}$: \begin{align} &\bbox[10px,#ffe]{\int_{-\infty}^{\infty} {\dd x \over x^{4} - 2\cos\pars{2\theta}x^{2} + 1}} = 2\int_{0}^{\infty} {x^{-2}\,\dd x \over x^{2} - 2\cos\pars{2\theta} + x^{-2}} \\[5mm] = &\ 2\int_{0}^{\infty} {x^{-2}\,\dd x \over \pars{x - x^{-1}}^{2} + 2\bracks{1 -\cos\pars{2\theta}}} \\[5mm] = &\ \int_{0}^{\infty} {x^{-2}\,\dd x \over \pars{x - x^{-1}}^{2} + 4\sin^{2}\pars{\theta}} + \int_{\infty}^{0} {-\,\dd x \over \pars{x^{-1} - x}^{2} + 4\sin^{2}\pars{\theta}} \\[5mm] = &\ \int_{0}^{\infty} {\pars{1 + x^{-2}}\,\dd x \over \pars{x - x^{-1}}^{2} + 4\sin^{2}\pars{\theta}} = \int_{-\infty}^{\infty} {\dd x \over x^{2} + 4\sin^{2}\pars{\theta}} \\[5mm] = &\ {1 \over 4\sin^{2}\pars{\theta}}\, 2\verts{\sin\pars{\theta}}\int_{-\infty}^{\infty} {\dd x/\bracks{2\verts{\sin\pars{\theta}}} \over \braces{x/\bracks{2\verts{\sin\pars{\theta}}}}^{2} + 1} \\[5mm] = & {1 \over 2\verts{\sin\pars{\theta}}}\int_{-\infty}^{\infty} {\dd x \over x^{2} + 1} = \bbx{\pi \over 2\verts{\sin\pars{\theta}}}\,,\qquad \theta \not\in \pi\mathbb{Z} \end{align}

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