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I believe the answer might involve some periodicity argument, but I am unable to make one work. Any insights are appreciated!

This integral is motivated by the second part of the question from this post: Conformal map from unit disk to polygon

Following User's advice, I have $$\int_0^{2\pi}\sin^2(n\theta/2)^{-1/n}\ d\theta=n\int_0^{2\pi/n}\sin^2(n\theta/2)^{-1/n}\ d\theta$$ Then I make the substitution $\varphi=n\theta/2$ with $d\varphi=n/2\ d\theta$ to get $$\int_0^{2\pi}\sin^2(n\theta/2)^{-1/n}\ d\theta=n\cdot\frac{2}{n}\int_0^{\pi}\sin^2(\varphi)^{-1/n}\ d\varphi=2\int_0^{\pi}\sin^2(\varphi)^{-1/n}\ d\varphi.$$ Thus, $$\int_0^{\pi}\sin^2(n\theta/2)^{-1/n}\ d\theta=\frac{1}{2}\left(2\int_0^{\pi}\sin^2(\varphi)^{-1/n}\ d\varphi\right)=\int_0^{\pi}\sin^2(\varphi)^{-1/n}\ d\varphi$$ as I wished to show. Thanks for the help!

Bruce Broderberg
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1 Answers1

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Hint:

Your guess is completely correct, if $n$ is assumed to be an integer number. Indeed $\sin^2\left(\frac\theta2\right)$ is an even $2\pi$-periodic function. Therefore $$ \int_0^\pi F(\sin^2(\theta/2))d\theta=\int_{m\pi}^{(m+1)\pi} F(\sin^2(\theta/2))d\theta $$ for any function $F(x)$ and any integer number $m$. It follows: $$ \int_0^{n\pi} F(\sin^2(\theta/2))d\theta=n\int_0^\pi F(\sin^2(\theta/2))d\theta. $$

Can you take it from here?

user
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