# 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


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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