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I would like to solve that problem: $$ \int d^2 \mathbf{k} \, \delta(f(k,\phi)) = \int_{k_1}^{k_2} dk \, k \int_0^{2\pi} d\phi \, \delta(f(k,\phi)) \,, $$ where $f(k,\phi) = a - bk^2 - ck^3 |\sin(2\phi)|$ ($a$, $b$ and $c$ are positive constants, $k=\sqrt{k_x^2+k_y^2}$, $k_x=k\cos(\phi)$, $k_y=k\sin(\phi)$).

I know that $$\delta(g(x))=\sum_i \frac{\delta(x-x_i)}{|g'(x)|_{x=x_i}} \,$$ and that for 2D Dirac delta function in polar coordinates we have $$\delta(\mathbf{k}-\mathbf{k}_0) = \frac{1}{k_0}\delta(k-k_0)\delta(\phi-\phi_0) \,.$$ But I have doubt that I can write (if so what about $\frac{1}{k_0}$ from above formula): $$\delta(f(k,\phi))=\sum_i \frac{\delta(\phi-\phi_i)}{|g'(\phi)|_{\phi=\phi_i}} \sum_j \frac{\delta(k-k_j)}{|g'(k)|_{k=k_j}}\,.$$

1. First try

I begin with the first part of above formula (with $\delta(\phi-\phi_i)$). I can calculate that

$$|\frac{\partial f(k,\phi)}{\partial \phi}|_{\phi=\phi_i}= ck^3|\cos(2\phi_i)|\,,$$

in the region $\phi \in [0,2\pi]$ function $f(k,\phi)$ has 8 roots ($|\sin(2\phi_0)|=\frac{a-bk^2}{ck^3}$), so: $$\delta(f(\phi))=\frac{8\delta(\phi-\phi_0)}{ck^3|\cos(2\phi_0)|}= \frac{8\delta(\phi-\phi_0)}{\sqrt{c^2 k^6 - (a-bk^2)^2}} \,.$$ And now should I do the same for the second part (with $\delta(k-k_j)$), multiply and integrate it?

3. Second try

If I consider, for beginning, only integral over $\phi$ of function $f(k,\phi)$ and substitute it to the primarly equation: $$ \int_{k_1}^{k_2} dk \, k \int_0^{2\pi} d\phi \, \delta(f(k,\phi))= \int_{k_1}^{k_2} dk \, k \int_0^{2\pi} d\phi \, \frac{8\delta(\phi-\phi_0)}{\sqrt{c^2 k^6 - (a-bk^2)^2}}= \int_{k_1}^{k_2} dk \, \frac{8k}{\sqrt{c^2 k^6 - (a-bk^2)^2}} \,, $$

Based on above formula, I can try to define the limits of integration over $k$ (from requirement $c^2 k^6 - (a-bk^2)^2>0$) and just integrate it over $k$. But I think that is also not appropriate approach to solve this problem.

I do not understand it well and perhaps I do not see something obvious.

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  • $\begingroup$ You need a concrete definition of integrals with $\delta$. For example $ \lim_{n \to \infty} \int_{\mathbb{R}^2} n e^{-\pi n^2 |f(k)|^2}d^2 k$. Here $f(k)$ is non-zero for $\|k\| \not \in [k_1,k_2]$ and $ \to \infty$ fast enough as $\|k\| \to \infty$ to state it is $=\lim_{n \to \infty} \int_{\mathbb{R}^2, \|k \| \in [k_1,k_2]} n e^{-\pi n^2 |f(k)|^2}d^2 k$. Then apply the radial change of variable $k = (r \cos \phi, r \sin \phi), d^2 k = d\phi r dr$. Conclude by looking for each $r$ at the finitely many zeros of $\phi \mapsto f(r \cos \phi, r \sin \phi)$ $\endgroup$ – reuns Mar 8 '19 at 16:42
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You need to use the relationship: $$\int_{\Omega}{g(\omega)\delta(f(\omega))\,d\omega} = \int_{f^{-1}(0)}{g(x)/|\nabla f| \,d\sigma(x)}$$

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