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Hi I have the following integral and something strange happens with the coefficient. I do not see what I am missing.

It is claimed that $\frac{1}{(2\pi)^2}\int_{[-\pi,\pi]^2}{\frac{1-\cos{n(\theta_1+\theta_2)}}{1-1/2\left[\cos{\theta_1}+\cos\theta_2\right]}d\theta}=\frac{1}{\pi}\int_{[-\pi,\pi]}{\frac{\sin^2(2n\alpha)}{\mid\sin\alpha\mid}}$

I transformed using that the determinant of the jacobian is $1/2$ and that $\int_0^\pi{\frac{1}{1-\cos(x)\cos(y)}dy}=\frac{\pi}{\mid\sin(x)\mid}$

$\frac{1}{(2\pi)^2}\int_{[-\pi,\pi]^2}{\frac{1-\cos{n(\theta_1+\theta_2)}}{1-1/2\left[\cos{\theta_1}+\cos\theta_2\right]}d\theta}=\frac{1}{8\pi^2}\int_{[-\pi,\pi]^2}{\frac{1-\cos{(2n\alpha)}}{1-\cos{\alpha}\cos{\beta}}d\alpha d\beta} =\frac{1}{4\pi}\int_{[-\pi,\pi]}{\frac{1-\cos(2n\alpha)}{\mid\sin\alpha\mid}}=\frac{1}{2\pi}\int_{[-\pi,\pi]}{\frac{\sin^2(2n\alpha)}{\mid\sin\alpha\mid}}$

Can anyone tell me why am I getting the $1/2$?

Thanks

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Since $\alpha=\frac{\theta_1+\theta_2}2$ and $\beta=\frac{\theta_1-\theta_2}2$, $\theta_1=\alpha+\beta$ and $\theta_2=\alpha-\beta$, your Jacobian is $$d\theta_1d\theta_2=\left|\det\begin{bmatrix}\frac{\partial\theta_1}{\partial\alpha}&\frac{\partial\theta_1}{\partial\beta}\\\frac{\partial\theta_2}{\partial\alpha}&\frac{\partial\theta_2}{\partial\beta}\end{bmatrix}\right|d\alpha\,d\beta=\left|\det\begin{bmatrix}1&1\\1&-1\end{bmatrix}\right|d\alpha\,d\beta=2\,d\alpha\,d\beta$$ So now we have $$\begin{align}&\frac1{(2\pi)^2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{1-\cos n(\theta_1+\theta_2)}{1-\frac12(\cos\theta_1+\cos\theta_2)}d\theta_1\,d\theta_2\\ &=\frac1{2\pi^2}\int_0^{\pi}\int_0^{\pi-\alpha}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha+\frac1{2\pi^2}\int_{-\pi}^0\int_0^{\pi+\alpha}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha\\ &+\frac1{2\pi^2}\int_{-\pi}^0\int_{-\pi-\alpha}^0\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha+\frac1{2\pi^2}\int_0^{\pi}\int_{\alpha-\pi}^0\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha\\ &=\frac1{2\pi^2}\int_0^{\pi}\int_0^{\pi-\alpha}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha+\frac1{2\pi^2}\int_{-\pi}^0\int_0^{\pi+\alpha}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha\\ &+\frac1{2\pi^2}\int_0^{\pi}\int_{-\phi}^0\frac{1-\cos2n\phi}{1+\cos\phi\cos\beta}d\beta\,d\phi+\frac1{2\pi^2}\int_{-\pi}^0\int_{\phi}^0\frac{1-\cos2n\phi}{1+\cos\phi\cos\beta}d\beta\,d\phi\\ &=\frac1{2\pi^2}\int_0^{\pi}\int_0^{\pi-\alpha}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha+\frac1{2\pi^2}\int_{-\pi}^0\int_0^{\pi+\alpha}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha\\ &+\frac1{2\pi^2}\int_0^{\pi}\int_{\pi-\phi}^{\pi}\frac{1-\cos2n\phi}{1-\cos\phi\cos\psi}d\psi\,d\phi+\frac1{2\pi^2}\int_{-\pi}^0\int_{\pi+\phi}^{\pi}\frac{1-\cos2n\phi}{1-\cos\phi\cos\psi}d\psi\,d\phi\\ &=\frac1{2\pi^2}\int_0^{\pi}\int_0^{\pi}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha+\frac1{2\pi^2}\int_{-\pi}^0\int_0^{\pi}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha\\ &=\frac1{2\pi^2}\int_{-\pi}^{\pi}\int_0^{\pi}\frac{1-\cos2n\alpha}{1-\cos\alpha\cos\beta}d\beta\,d\alpha\\ &=\frac1{2\pi}\int_{-\pi}^{\pi}\frac{1-\cos2n\alpha}{|\sin\alpha|}d\alpha=\frac1{\pi}\int_{-\pi}^{\pi}\frac{\sin^2n\alpha}{|\sin\alpha|}d\alpha\end{align}$$ So you had the Jacobian upside down and you didn't properly take into account the ratio of areas between the square original region and the diamond-shaped transformed region of integration. Note: you might want to copy some of my MathJax up into your original question to make the site elders happier with your formatting.

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