Evaluating the integral $int^{infty}_{-infty} frac{dx}{x^4-2cos(2theta)x^2 +1}$

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:
$$frac{1}{z^4-2cos(2theta)z^2 +1}=frac{1}{(z-e^{itheta})(z+e^{itheta})(z-e^{-itheta})(z+e^{-itheta})}$$
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!

Mathematics Asked by user793781 on November 12, 2021

2 Answers

2 Answers

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$ $ds{underline{underline{mbox{With} theta notin pimathbb{Z}}}}$: begin{align} &bbox[10px,#ffe]{int_{-infty}^{infty} {dd x over x^{4} - 2cospars{2theta}x^{2} + 1}} = 2int_{0}^{infty} {x^{-2},dd x over x^{2} - 2cospars{2theta} + x^{-2}} \[5mm] = & 2int_{0}^{infty} {x^{-2},dd x over pars{x - x^{-1}}^{2} + 2bracks{1 -cospars{2theta}}} \[5mm] = & int_{0}^{infty} {x^{-2},dd x over pars{x - x^{-1}}^{2} + 4sin^{2}pars{theta}} + int_{infty}^{0} {-,dd x over pars{x^{-1} - x}^{2} + 4sin^{2}pars{theta}} \[5mm] = & int_{0}^{infty} {pars{1 + x^{-2}},dd x over pars{x - x^{-1}}^{2} + 4sin^{2}pars{theta}} = int_{-infty}^{infty} {dd x over x^{2} + 4sin^{2}pars{theta}} \[5mm] = & {1 over 4sin^{2}pars{theta}}, 2verts{sinpars{theta}}int_{-infty}^{infty} {dd x/bracks{2verts{sinpars{theta}}} over braces{x/bracks{2verts{sinpars{theta}}}}^{2} + 1} \[5mm] = & {1 over 2verts{sinpars{theta}}}int_{-infty}^{infty} {dd x over x^{2} + 1} = bbx{pi over 2verts{sinpars{theta}}},,qquad theta notin pimathbb{Z} end{align}

Answered by Felix Marin on November 12, 2021

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 $z=e^{itheta}$, and the other at $z=-e^{-itheta}$. The integral over the semi-circle will vanish as $Rtoinfty$. The sum of your residues is $$frac{1}{2e^{itheta}}frac{1}{2isintheta}frac{1}{2costheta}+frac{1}{-2e^{-itheta}}frac{1}{-2isintheta}frac{1}{2costheta} = frac{1}{4isintheta}.$$ Hence, multiplying by $i2pi $, $$int_{-infty}^infty frac{dx}{x^4 - 2 cos(2theta)x^2+1}= frac{pi}{2sintheta}.$$

Answered by Zachary on November 12, 2021

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