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I have a really convoluted proof of the following: $$ (1) \quad \quad - \int_0^\pi \frac{\sin(2 y) \log(\sin (y/2)) } {r + 1/r + 2 \cos{(2 y)} } dy = \big(\arctan(\sqrt{r})\big)^2 $$ I proved it for $0<r<1.$ To my surprise it holds for complex $r$ if $-1<\Re{(r)}<1.$ Numerically it holds that $$ - \int_0^\pi \frac{\sin(2 y) \log(\sin (y/2)) } {r + 1/r + 2 \cos{(2 y)} } dy = \big(\text{arccot}(\sqrt{r})\big)^2 $$ if $\Re{(r)}>1$ or $\Re{(r)}<-1.$ This is not surprising given that $\text{arccot}(x) =\arctan(1/x)$ for $x>0.$ However, there is a $\pi$ to be added if $x<0.$

Given the symmetry within the integral, (1) is sufficient, so long as the user makes sure that the $r$ put in on the right-hand side has a real part $-1<\Re{(r)}<1$

I would like a proof (or if the formula exists, a reference) of (1), and with special attention paid to the complexity of the parameter $r.$

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  • $\begingroup$ May be complexification and contour integration could help. Those integral looks like it was initially a complex integration. The denominator is looks like $z+z^{-1}$ etc. If so would you care to formulate it? $\endgroup$ – Andrew Jun 3 at 17:55
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From the generating function of the Chebyshev polynomials of the second kind: \begin{equation} \frac{1}{1-2xz+z^{2}}=\sum_{n=0}^{\infty}U_{n}\left(x\right)z^{n} \end{equation} valid for $\left|z\right|<1$, we chose $z=-r, x=\cos\left( 2y \right)$ to express \begin{equation} \frac{\sin(2 y) } {r + 1/r + 2 \cos{(2 y)} } =\sum_{n=0}^{\infty}(-1)^n \sin\left( 2(n+1)y \right)r^{n+1} \end{equation} as $U_n(\cos z)=\sin\left( (n+1)z \right)/\sin z$.

The Fourier series for $\ln\left( \sin(y/2) \right)$ reads \begin{equation} \log\sin\frac{y}{2}=-\log 2-\sum_{p=1}^\infty\frac{1}{p}\cos(py)\qquad\forall y\in(0,2\pi) \end{equation} and thus \begin{align} I&=- \int_0^\pi \frac{\sin(2 y) \log(\sin (y/2)) } {r + 1/r + 2 \cos{(2 y)} } \,dy\\ &=\int_0^\pi \left( \log 2+\sum_{p= 1}^\infty\frac{1}{p}\cos(py) \right)\left( \sum_{n=0}^{\infty}(-1)^n \sin\left( 2(n+1)y \right)r^{n+1} \right)\,dy \end{align} Using symmetry properties, the contributions of the $\log2$ term and of the terms with even values of $p$ vanish: \begin{align} I&=\sum_{n=0}^\infty\sum_{s=0}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\int_0^\pi\cos\left( (2s+1)y \right)\sin\left( 2(n+1)y \right)\,dy\\ &=\sum_{n=0}^\infty\sum_{s=0}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\left[\frac{1}{2n-2s+1}+\frac{1}{2n+2s+3}\right] \end{align} A decomposition of the double sum can be written as \begin{align} I&=\sum_{n=0}^\infty\sum_{s=0}^n\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n-2s+1}+ \sum_{n=0}^\infty\sum_{s=n+1}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n-2s+1}+ \sum_{n=0}^\infty\sum_{s=0}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n+2s+3} \end{align} By changing $k=s+n+1$ in the last double sum, we obtain \begin{equation} \sum_{n=0}^\infty\sum_{s=0}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n+2s+3}=-\sum_{n=0}^\infty\sum_{k=n+1}^\infty\frac{(-1)^nr^{n+1}}{2n-2k+1}\frac{1}{2k+1} \end{equation} which is the opposite of the second double sum. Now, if $r^{1/2}$ is defined with its principal value, the integral can be expressed as \begin{align} I&=\sum_{n=0}^\infty\sum_{s=0}^n\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n-2s+1}\\ &=\sum_{n=0}^\infty\sum_{s=0}^n \frac{(-1)^{s}r^{s+1/2}}{2s+1} \frac{(-1)^{n-s}r^{n-s+1/2}}{2(n-s)+1}\\ &=\left( \sum_{t=0}^\infty \frac{(-1)^{t}r^{t+1/2}}{2t+1} \right)^2\\ &=\left( \arctan\left(r^{1/2}\right) \right)^2 \end{align} as proposed.

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    $\begingroup$ Very nice double series solution, and much more elegant than the way I discovered it. $\endgroup$ – skbmoore Jun 11 at 20:33

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