# Integrating a Dirac delta with multidimensional periodic argument

I am trying to solve the following $$\int_0^{2\pi}dx_1\int_0^{\frac{2\pi}{B}}dx_2\delta\left(\lambda+\cos(x_1)+\cos(x_2)\right)$$ where $$B,\lambda$$ are constants.

Since this is not a vector function, I can't use the multidimensional simplification to get the delta function in a simple form. I am write in saying that I use the one dimensional version? But this then leaves me with a simpler but still a 2-dimensional scalar function in the argument.

Do I infact use the multidimensional analogue of $$\int_\mathbb{R}dx\delta(f(x))=\sum_{x_0}\frac{\delta(x-x_0)}{|f'(x_0)|}$$ where $$f(x_0)=0$$?

I have tried changing variables to get it into polar coordinates but then in my final answer the delta function zeros were no longer in the integration limits and thus was zero.

Does anybody know anything that could help me?

EDIT: Finding the roots of the argument for $$x_1$$, i.e., $$f(x_1,x_2,\lambda)=\lambda+\cos(x_1)+\cos(x_2)$$ and therefore $$f(x_1^0,x_2,\lambda)=0$$, I get $$x_1^0=\arccos(-\lambda-\cos(x_2))+2\pi k,$$ $$\forall k\in\mathbb{Z}$$. Therefore our integral can be written $$\int_0^{2\pi}dx_1\int_0^{\frac{2\pi}{B}}dx_2\sum_{k}\frac{\delta(\frac{x_1}{2\pi}-\frac{\arccos(-\lambda-\cos(x_2))}{2\pi}- k)}{2\pi|\sqrt{1-(\lambda-\cos(x_2))^2}|}.$$

Using the Poisson summation formula $$\sum_{k}\delta(y-m)=\sum_{k}\exp(2\pi iym),$$

we have $$\int_0^{2\pi}dx_1\int_0^{\frac{2\pi}{B}}dx_2\sum_k\frac{\exp(ikx_1)\exp(-ik(\arccos(-\lambda-\cos(x_2))))}{2\pi|\sqrt{1-(\lambda-\cos(x_2))^2}|}$$ thus using the integral representation of the delta function, i.e., $$\frac{1}{2\pi}\int_0^{2\pi}\exp(ikx_1)dx_1=\delta(k),$$ with this the integral is

$$\int_0^{2\pi/B}\frac{1}{|\sqrt{1-(\lambda-\cos(x_2))^2}|}dx_2.$$

I am now trying to solve this. Can anybody comment on the validity of my calculations? Am I right to do what I did?

EDIT2: I will add that $$\lambda\in(-1,3)$$

## 1 Answer

The coarea formula is used to tell you how to integrate $$\delta(f(\mathbf{x}))$$ by changing to a (perhaps local) coordinate system in which $$f(\mathbf{x})$$ itself is one of the coordinates, and correctly take into account how the integration of the other coordinates will be changed.

You don't really need it here. The point is that you can evaluate the inner integral using the 1D version of the coarea formula, which is what you stated. That is:

$$\int_0^l \delta(\lambda + \cos(x_1) + \cos(x_2)) dx_2 = \sum_{x_2 \in [0,l] : \cos(x_2)=-\lambda-\cos(x_1)} \frac{1}{|\sin(x_2)|} \\ = \sum_{x_2 \in [0,l] : \cos(x_2)=-\lambda-\cos(x_1)} \frac{1}{\sqrt{1-(\lambda+\cos(x_1))^2}} \\ = \frac{\# \{ x \in [0,l] : \cos(x) = -\lambda - \cos(x_1) \}}{\sqrt{1-(\lambda+\cos(x_1))^2}}.$$

Naturally if $$l=2\pi$$ (which seems to be an interesting case for you) then the cardinality in the numerator has the behavior:

• If $$|\lambda+\cos(x_1)|<1$$ then the cardinality in the numerator is $$2$$
• If $$|\lambda+\cos(x_1)|=1$$ then the cardinality in the numerator is $$1$$
• If $$|\lambda+\cos(x_1)|>1$$ then the cardinality in the numerator is $$0$$.

The middle case will have measure zero, so if $$l=2\pi$$ then the numerator will just serve to restrict the $$x_1$$ integral to $$\{ x_1 : \lambda+\cos(x_1) \in (-1,1) \}$$, and then to multiply it by $$2$$ at the end.

Anyway, this is a function of just $$x_1$$, so you can integrate it in principle. It does, however, have some singularities (wherever $$|\lambda+\cos(x_1)|=1$$), but it looks like they should be integrable.

• Yeah, I have made an edit to my answer that $\lambda\in(-1,3)$, and of course one would have to look out for the singularities, I think I'll have to numerically calculate this integral... – Lewis Proctor May 16 at 16:29
• this is great thanks! – Lewis Proctor May 16 at 19:37