The equality $$\int_{0}^{2\pi}\left(-\frac{1+e^{-i\theta}}{1+\rho^{2}e^{i\theta}}\right)^{\!n}d\theta=\int_{\phi_{\rho}}^{2\pi-\phi_{\rho}}e^{in t}\frac{\sin(t/2)}{\sqrt{\rho^{2}-\cos^{2}(t/2)}}\,dt, $$ where $\phi_{\rho}=2\arccos(\rho)$, should hold for any $\rho\in(0,1)$ and $n\in\mathbb{N}$. Can anybody see a connection between the two integrals? Can one be transformed to the other? Thanks.
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The identity seems to hold. The following Wolfram Language code produces some interesting results:
FirstI[rho_,n_]=Integrate[(-(1+Exp[-I Theta])/(1+rho^2 Exp[I Theta]))^n,{Theta,0,2 Pi}]
SecondI[rho_,n_]=Integrate[Exp[I n t]Sin[t/2]/Sqrt[rho^2-(Cos[t/2])^2],{t,2 ArcCos[rho],2 Pi-2 ArcCos[rho]}]
FirstI[0.5,3]//N
SecondI[0.5,3]//N
The results are $1.76715-1.33227\times 10^{-15}i$ and $1.76454-0.000624587i$, which are quite close.
[EDIT] The following analysis is incorrect, as shown in the comments. However, I will not delete it, so that the comments still have context.
As for getting the link between the two integrals, it appears, via trig-substitution-on-steroids, that the substitution you want is the following: \begin{align*} \cos\left(\frac{t}{2}\right)&=\underbrace{\cos\left(\frac{\phi_{\rho}}{2}\right)}_{=\,\rho}\cos\left(\frac{\theta}{2}\right),\\ \sin\left(\frac{t}{2}\right)\,dt&=\cos\left(\frac{\phi_{\rho}}{2}\right)\sin\left(\frac{\theta}{2}\right)\,d\theta. \end{align*} You can see that $\theta=0$ corresponds to $t=\phi_{\rho}$, and $\theta=2\pi$ corresponds to $t=2\pi-\phi_{\rho}$. So that accounts for your interval of integration. When you substitute this into the RHS integral, you get some spectacular cancellations: $$\int_{\phi_{\rho}}^{2\pi-\phi_{\rho}}e^{in t}\,\frac{\sin(t/2)}{\sqrt{\rho^{2}-\cos^{2}(t/2)}}\,dt=\int_{0}^{2\pi}e^{int}\,\frac{\cos(\phi_{\rho}/2)\sin(\theta/2)\,d\theta}{\cos(\phi_{\rho}/2)\sin(\theta/2)}=\int_{0}^{2\pi}e^{int}\,d\theta. $$ It remains to show that $$e^{int}=\left(-\frac{1+e^{-i\theta}}{1+\rho^2e^{i\theta}}\right)^{\!n},$$ or equivalently that $$e^{it}=-\frac{1+e^{-i\theta}}{1+\rho^2e^{i\theta}}.$$ This doesn't seem beyond the pale. Perhaps you can finish?
Adrian Keister
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1Thanks for the hint but I don't think that the suggested substitution transforms the RHS integral to the LHS integral. The very last equation seems not to hold for $t$ and $\theta$ real. It can be seen from the fact that the LHS has absolute value 1 but the RHS does not. More specifically, $t=\pi$ corresponds to $\theta=\pi$ according to the substitution. For this choice, LHS=-1 but RHS=0. Am I mistaken somwhere? – Twi Apr 13 '18 at 07:24
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Yes, that unfortunately looks fatal to my plan. On the other hand, maybe I need to invert how I'm looking at the whole thing. Why not use that last equation as the substitution, and see where it leads? It's a fractional linear transformation (Mobius), so that might prove valuable. Calculating... – Adrian Keister Apr 13 '18 at 13:15
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That equation can't work, no-how. If you try $\theta=0$, you get $e^{it}=-\dfrac{2}{1+\rho^2}=\cos(t)+i\sin(t)$, which forces $\cos(t)=-\dfrac{2}{1+\rho^2}$ and $\sin(t)=0$. This, in turn, would force $\rho^2=1$, because otherwise you're outside the range of the cosine function. That would correspond to $t=\pi$, which, while it satisfies both those equations, does not work with the limits of the $t$ integral. – Adrian Keister Apr 13 '18 at 13:25
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I'm a bit mystified by this identity; it seems to work numerically, but finding out a relation between $t$ and $\theta$ is difficult. – Adrian Keister Apr 13 '18 at 13:27
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Actually, wait a minute. All is not lost with $e^{it}=-\dfrac{1+e^{-i\theta}}{1+\rho^2e^{i\theta}}$. $\rho^2=1$ would mean that $\phi_{\rho}=0$, which does work with the limits of the integral. – Adrian Keister Apr 13 '18 at 13:38
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NIntegrate. – Adrian Keister Apr 15 '18 at 00:08