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

Question

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!

Felix Marin · Answer

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
 newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
 newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
 newcommand{dd}{mathrm{d}}
 newcommand{ds}[1]{displaystyle{#1}}
 newcommand{expo}[1]{,mathrm{e}^{#1},}
 newcommand{ic}{mathrm{i}}
 newcommand{mc}[1]{mathcal{#1}}
 newcommand{mrm}[1]{mathrm{#1}}
 newcommand{pars}[1]{left(,{#1},right)}
 newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}}
 newcommand{root}[2][]{,sqrt[#1]{,{#2},},}
 newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
 newcommand{verts}[1]{leftvert,{#1},rightvert}$
$ds{underline{underline{mbox{With} theta notin pimathbb{Z}}}}$:
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}

Zachary · Answer

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 $, 
$$int_{-infty}^infty frac{dx}{x^4 - 2 cos(2theta)x^2+1}= frac{pi}{2sintheta}.$$

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

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