I am trying to understand the equations from a paper. basically the paper defines:
$$\rho = r_m(\varphi) = \frac{1}{2}\lim_{n\to+\infty}\left(|\cos{\frac{m\varphi}{4}|^n}+|\sin{\frac{m\varphi}{4}}|^n\right)^{-\frac{1}{n}}=\frac{1}{2}\min{\left(\left|\sec{\frac{m\varphi}{4}}\right|,\left|\csc{\frac{m\varphi}{4}}\right|\right)}\label{1}(1)$$ $$A_m = \frac{1}{2}\int_{0}^{2\pi} r_m^2(\varphi) \,d\varphi ~(2)$$
The authors conclude that $A_m$ is equal to 1. They replace (1) in (2) to do so. To my understanding I would need to divide the integral in intervals where sec > csc and where sec < csc. However, it seems that I lack some trigonometry simplification to achieve the same result, as my development of the equations is getting quite big for the result given from the authors. Can anyone point me the realization I must have been missing?
If you set $r_m(\varphi)=\frac{1}{2}\min{\left(\left|\sec{\frac{m\varphi}{4}}\right|,\left|\csc{\frac{m\varphi}{4}}\right|\right)}$, notice that this function is periodic with the period $\frac{2\pi}{m}$ because $\min{\left(|\sec\varphi|,|\csc\varphi|\right)}$ is periodic with the period $\frac{\pi}{2}$.
This means that
$$2A_m=\int_{0}^{2\pi}r^2_m(\varphi)d\varphi=m\int_0^{2\pi/m}r^2_m(\varphi)d\varphi=m\int_0^{2\pi}r^2_1(\psi)\frac{1}{m}d\psi=\int_0^{2\pi}r^2_1(\psi)d\psi$$
(taking substitution $\psi=m\varphi$ in the second step, and taking the integral of a periodic function over $m$ periods in the first step).
So the value of the integral does not depend on $m$, and you can just calculate it for the case you find the easiest (e.g. $m=1$). I guess this simplifies the problem enough for you to complete it, if not, please let me know.