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

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

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 $$i2\pi$$, $$\int_{-\infty}^\infty \frac{dx}{x^4 - 2 \cos(2\theta)x^2+1}= \frac{\pi}{2\sin\theta}.$$
• 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? – user793781 May 29 at 17:32
• 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. – Zachary May 29 at 17:44
• Shouldn't this be $\pi/(2\sin|\theta|)$? The integrand is strictly positive, so the integral can't be negative. – eyeballfrog Jul 21 at 20:22