This integral comes from equation (3.15) in an older paper I've been reading: $$ \int \mathrm{d} \Omega_k \, \delta\left(|\vec{k}|^2 - |\vec{k}+\vec{q}_1|^2\right) \delta\left(|\vec{k}|^2 - |\vec{k}-\vec{q}_2|^2\right) . $$ The integration over $\Omega_k$ refers to the angular coordinates of the vector $\vec{k}$. I'm most interested in the case of three spatial dimensions, for which $$ \int \mathrm{d} \Omega_k \to \int_0^{2\pi} d\theta \int_0^\pi d\varphi \sin \varphi, $$ where $\theta$ and $\varphi$ are the azimuthal and polar angles associated with $\vec{k}$. The result should depend on the lengths $|\vec{k}|$, $|\vec{q}_1|$, and $|\vec{q}_2|$, as well as the dot product $\vec{q}_1\cdot\vec{q}_2$.

The authors write that the integral is straightforward and don't give an explicit result, so I suspect that I'm missing something simple. I've tried a few strategies, such as using integral representations for the delta functions, but I haven't yet found an elegant approach.

  • $\begingroup$ Using both $k$ and $\vec{k}$ makes the notation a little unclear. What is the $k$? Also what do you mean by $(\vec{k}+\vec{q}_1)^2$? It is equal to $\|\vec{k}+\vec{q}_1\|^2$? $\endgroup$
    – mucciolo
    Commented Jan 29, 2018 at 22:01
  • $\begingroup$ Being used to physicists notation, I assume that $u^2 = \|\vec{u}\|^2$ and $\endgroup$
    – md2perpe
    Commented Jan 29, 2018 at 22:03
  • $\begingroup$ I made some tweaks to the notation for clarity. Thanks for the comment. $\endgroup$
    – wcw
    Commented Jan 29, 2018 at 22:10
  • $\begingroup$ Does $\text d\Omega_k$ indicate integration over some sphere $\{ |k| = c \}$? $\endgroup$ Commented Jan 29, 2018 at 22:14
  • $\begingroup$ @CalvinKhor: yes, that's right. $\endgroup$
    – wcw
    Commented Jan 29, 2018 at 22:21

1 Answer 1


This isn't a full solution (yet), but it's the closest I've come so far. I still think there is likely a more elegant approach. I will use the convention that $k = |\vec{k}|$.

Coordinate system (three dimensions)

Choose a spherical coordinate system for $\vec{k}$ such that $\vec{q}_1$ coincides with the positive $z$-axis, and let $\theta$ and $\varphi$ represent the azimuthal and polar angles associated with $\vec{k}$. Then, $$ \vec{k} \cdot \vec{q}_1 = k \, q_1 \cos \varphi . $$

Rotate the coordinate system about the z-axis until $\vec{q}_2$ lies on the $x>0$ portion of the $x$-$z$ plane. Let $\alpha$ represent the angle between $\vec{q}_2$ and the $z$-axis. Then, using the formula for a dot product in spherical coordinates,

$$ \vec{k} \cdot \vec{q}_2 = k \, q_2 \left[ \sin \varphi \sin \alpha \cos \theta + \cos \varphi \cos \alpha \right]. $$

The original integral may now be written $$ \int_0^{2\pi} d\theta \int_0^\pi d\varphi \, \sin \varphi \,\, \delta\left(f(\varphi)\right) \delta\left(g(\theta,\varphi)\right) $$


$$ f(\varphi) = 2k \, q_1\cos\varphi + q_1^2 $$


$$ g(\theta,\varphi) = 2kq_2 \left[ \sin \varphi \sin \alpha \cos \theta + \cos \varphi \cos \alpha \right] -q_2^2. $$

Evaluating the integral over $\varphi$

Using the formula suggested by Calvin Khor, we can say that

$$ \delta\left(f(\varphi)\right) \to \frac{\delta(\varphi-\varphi_0)}{2k q_1 |\sin \varphi_0|} $$

with $\varphi_0 = \arccos \left(-q_1 / 2k\right)$, which is valid for $q_1 \le 2k$. We can now evaluate the integral over $\varphi$, obtaining $$ \frac{1}{2kq_1} \int_0^{2\pi} d\theta \, g(\theta,\varphi_0) $$

If $q_1 > 2k$, the integral evaluates to $0$.

Evaluating the integral over $\theta$

Work in progress...

  • $\begingroup$ Hey I think you're basically done, just that you have to integrate in $\theta $ first $\endgroup$ Commented Jan 31, 2018 at 12:53

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