Evaluating $\int_{\mathbb{R}^n} \frac{|x|^{2}}{|x|^{8} + 2|x|^{4}\cos \left( \theta \right) + 1} \, dx $

Question

Evaluate the integral over $\mathbb{R}^n$

$$\int_{\mathbb{R}^n} \frac{|x|^{2}}{|x|^{8} + 2|x|^{4}\cos \left( \theta \right) + 1} \, dx \quad \textrm{where } \theta \in \left( 0, \frac{\pi}{2} \right)$$

We start with the integral over $\mathbb{R}^n$

$$ I_n(\theta) = \int_{\mathbb{R}^n} \frac{|x|^{2}}{|x|^{8} + 2|x|^{4} \cos(\theta) + 1} \, dx \quad \theta \in \left(0, \frac{\pi}{2}\right) $$

Since the integrand depends only on the norm $r = |x|$, we change to spherical coordinates in $\mathbb{R}^n$

$$ dx = \omega_n \, r^{n-1} dr $$

where $\omega_n = \frac{2 \pi^{\frac{n}{2}}}{\Gamma\left(\frac{n}{2}\right)}$ is the surface area of the unit sphere in $\mathbb{R}^n$. Then, the integral becomes

$$ I_n(\theta) = \omega_n \int_0^\infty \frac{r^{2} \cdot r^{n-1}}{r^{8} + 2 r^{4} \cos(\theta) + 1} \, dr = \omega_n \int_0^\infty \frac{r^{n+1}}{r^{8} + 2 r^{4} \cos(\theta) + 1} \, dr $$

The denominator becomes

$$ r^{8} + 2 r^{4} \cos(\theta) + 1 = (r^{4})^{2} + 2 (r^{4}) \cos(\theta) + 1 $$

which factors as $(r^{4} + e^{i \theta})(r^{4} + e^{-i \theta})$. We then make the substitution $t = r^{4}$

$$ t = r^{4} \implies dt = 4 r^{3} dr \implies dr = \frac{dt}{4 r^{3}} = \frac{dt}{4 t^{3/4}} $$

Rewriting the numerator gives $r^{n+1} = (r^{4})^{\frac{n+1}{4}} = t^{\frac{n+1}{4}}$, then the integral becomes

$$ I_n(\theta) = \omega_n \int_0^\infty \frac{t^{\frac{n+1}{4}}}{t^{2} + 2 t \cos(\theta) + 1} \cdot \frac{dt}{4 t^{3/4}} = \frac{\omega_n}{4} \int_0^\infty \frac{t^{\frac{n+1}{4} - \frac{3}{4}}}{t^{2} + 2 t \cos(\theta) + 1} \, dt $$

Simplifying the exponent $\frac{n+1}{4} - \frac{3}{4} = \frac{n - 2}{4}$, so we have

$$ I_n(\theta) = \frac{\omega_n}{4} \int_0^\infty \frac{t^{\frac{n-2}{4}}}{t^{2} + 2 t \cos(\theta) + 1} \, dt $$

I'm not sure of how to proceed from here.

Answer 1

The integral $$ I=\int_0^\infty\frac{t^\alpha\,dt}{1+2t\cos\theta+t^2}=\frac{\pi\sin\alpha\theta}{\sin\alpha\pi\,\sin\theta}\qquad(0<|\alpha|<1,0<|\theta|<\pi) $$ can be evaluated in many ways (e.g. here).

One of the simplest for me is via contour integration. Consider, as the contour, the edge of $|t|<R$ slit along the positive real axis (as the branch cut of $t^\alpha$, taking real values on the upper edge of the cut); as $R>1$, the integral along that contour equals the sum of the residues of the integrand at $t=-e^{\pm i\theta}$, which are easy to compute; at the same time, as $R\to\infty$, this integral tends to $(1-e^{2i\pi\alpha})I$.