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I have an integral of the form

$$I=\int_0^{\ell_p}dp_1\int_0^{\ell_p}dp_2\theta\left(-\lambda+\cos\left(\frac{2\pi p_1}{\ell_p}\right)+\cos\left(\frac{2\pi p_2}{B\ell_p}\right)\right),$$ Where $B,\ell_p$ are real constants and $\lambda\in(-2,2)$.

My EFFORT:

I know that $$\int_{\mathbb{R}^2}\theta(u(x,y))dxdy=\int_{\Omega}1dxdy$$ where $\Omega=\{(x,y) :u(x,y)>0\}$ Am I right in assuming that for my case we have $$I=\int_{\tilde{\Omega}}1dp_1dp_2,$$ where $\tilde{\Omega}=\{(p_1,p_2):0<u(p_1,p_2)<u(\ell_p,\ell_p)\},$ and $u(p_1,p_2)=-\lambda+\cos\left(\frac{2\pi p_1}{\ell_p}\right)+\cos\left(\frac{2\pi p_2}{B\ell_p}\right)$?

Now from this:

The condition $0<u(p_1,p_2)$ gives $$p_2>\pm\frac{\ell_pB}{2\pi}\arccos\left(\lambda-\cos\left(\frac{2\pi p_1}{\ell_p}\right)\right)+2\pi k,$$ where $k\in\mathbb{Z}$.

The condition $u(p_1,p_2)<u(\ell_p,\ell_p)$ gives $$p_2<\pm\frac{\ell_pB}{2\pi}\arccos\left(1+\cos\left(\frac{2\pi}{B}\right)-2\cos\left(\frac{2\pi p_1}{\ell_p}\right)\right)+2\pi m$$ where $m\in\mathbb{Z}$.

Taking $m,k=0$ and taking the positive branch we have $$I=\int_0^{\ell_p}\int_{\frac{\ell_pB}{2\pi}\arccos\left(\lambda-\cos\left(\frac{2\pi p_1}{\ell_p}\right)\right)}^{\frac{\ell_pB}{2\pi}\arccos\left(1+\cos\left(\frac{2\pi}{B}\right)-2\cos\left(\frac{2\pi p_1}{\ell_p}\right)\right)}dp_2dp_1\\ =\int_0^{\ell_p}\frac{\ell_pB}{2\pi}\arccos\left(1+\cos\left(\frac{2\pi}{B}\right)-2\cos\left(\frac{2\pi p_1}{\ell_p}\right)\right)-\frac{\ell_pB}{2\pi}\arccos\left(\lambda-\cos\left(\frac{2\pi p_1}{\ell_p}\right)\right)dp_1.$$

And so far I can't seem to solve this integral, explicitly at least...

I have tried letting $\cos(z)=\lambda-\cos\left(\frac{2\pi p_1}{\ell_p}\right)$ but this produces $\pm$ signs when looking the derivatives and when looking at the limits and I'm not sure on which branches to take etc., some help with this would be greatly appreciated also.

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  • $\begingroup$ $\theta$ is your name for the Heaviside function, $\theta(x)=0$ if $x<0$ and $=1$ if $x>0$, right? Are your $p_i$ integers? If so, you might as well take $p_1=p_2=1$, as the distribution of values of $\cos(kt)$ over a full period $0\le t<\pi$ does not depend on $k$ $\endgroup$ – kimchi lover May 24 at 12:54
  • $\begingroup$ Yes $\theta(x)$ is the Heaviside function, the $p_i$s are not integers they are integration variables and so are continuous in the interval $[0,\ell_p)$. $\endgroup$ – Lewis Proctor May 24 at 14:05
  • $\begingroup$ Careless reading on my part. Sorry. $\endgroup$ – kimchi lover May 24 at 14:28
  • $\begingroup$ no problem, I can't seem to find anything about the Heaviside step function restricted to a finite domain, I thing once I understand this I will be able to progress with my problem quickly. $\endgroup$ – Lewis Proctor May 24 at 14:33

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