Evaluate $$A=\lim_{r\to 1}\int_{0}^{2\pi}\frac{d\theta}{(1+r^2\cos2\theta)^{\frac{1}{3}}}$$
Question How to calculate A?
Attempt
Write, $z=e^{i\theta}$
then , $d\theta=\frac{dz}{iz}$
$$\cos2\theta= 2cos^2\theta-1$$
$$\cos2\theta=\frac{1}{2}\left(z+\frac{1}{z}\right)^2-1$$
$$\cos2\theta=\frac{z^2+\frac{1}{z^2}}{2}$$
So, A becomes ,
$$A=\lim_{r\to 1}\frac{1}{i} \oint_{|z|=1}\frac{dz}{z\left[1+\frac{r^2}{2}\left(z^2+\frac{1}{z^2}\right)\right]^{\frac{1}{3}}}$$
By Cauchy's Residue theorem, $$A=2\pi \sum_{k} Res[f(z_k)]$$
$$f(z)=\frac{1}{z\left[1+\frac{r^2}{2}\left(z^2+\frac{1}{z^2}\right) \right]^{\frac{1}{3}}}$$ So $f(z)$ has a pole at $z=0$.
Also, $f$ has singularities at the points where
$$1+\frac{r^2}{2}\left(z^2+\frac{1}{z^2}\right)=0$$
Which gives $$z^2=\frac{-1\pm\sqrt{1-r^4}}{r^2} $$
$$z=\frac{\pm\sqrt{-1\pm\sqrt{1-r^4}}}{r} $$
All these 4 singularities of $f(z)$ lie inside $|z|=1$. Next how to apply residue theorem or otherwise to calculate A?
Best Answer
By symmetry considerations we can integrate only over $[0,\pi/2]$ – the other three parts are congruent. Then for any sequence of integrands $f_n(\theta)=(1+r_n^2\cos2\theta)^{-1/3}$ with $r_n\to1^-$ (pointwise convergence) and for all $\theta\in[0,\pi/2)$ $$0\le f_n(\theta)\le(1+\cos2\theta)^{-1/3}=\cos^{-2/3}\theta\le2(\pi/2-\theta)^{-2/3}$$ Then $\int_0^{\pi/2}2(\pi/2-\theta)^{-2/3}$ converges. Hence, by the dominated convergence theorem, the limit can be moved inside the integral.
But the integral is not elementary. $$A=\int_0^{2\pi}(1+\cos2\theta)^{-1/3}\,d\theta=2^{5/3}\int_0^{\pi/2}\cos^{-2/3}\theta\,d\theta$$ Substitute $u=\cos\theta$ followed by $v=u^2$ to get $$A=2^{2/3}\int_0^1v^{-5/6}(1-v)^{-1/2}\,dv=2^{2/3}\mathrm B(1/6,1/2)$$ where $\mathrm B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is the beta function.