# Heaviside multivariable integral on a finite domain

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 $$\tilde{\Omega}=\{(p_1,p_2):0 and $$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 $$0 gives $$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 $$u(p_1,p_2) gives $$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.

• $\theta$ 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 $k$ – kimchi lover May 24 at 12:54
• 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 $[0,\ell_p)$. – Lewis Proctor May 24 at 14:05
• Careless reading on my part. Sorry. – kimchi lover May 24 at 14:28
• 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. – Lewis Proctor May 24 at 14:33