# Dirac delta in spherical coordinates. What I'm doing wrong? [closed]

I must show that the integral $$\frac{1}{(2\pi)^{3}}\int_{\vec{k}}d^{3}k\frac{\cos(\vec{k}\cdot\vec{x})}{\left({\sqrt{|\vec{k}|^2+m^{2}}}\right)^{s}}=\delta^{3}(\vec{x})$$ when $$s=0$$ by using spherical coordinates. This should be true since $$\delta^{3}(\vec{x})=\frac{1}{(2\pi)^{3}}\int_{-\infty}^{\infty}e^{i\vec{k}\cdot\vec{x}}d^{3}k$$

My problem is just the lack of a factor $$\frac{1}{2}$$ at the final result.

Here is my procedure: (in spherical coordinates) \begin{aligned} \frac{1}{(2\pi)^{3}}\int_{-\infty}^{\infty}d^{3}k\frac{\cos(\vec{k}\cdot\vec{x})}{\left({\sqrt{|\vec{k}|^2+m^{2}}}\right)^{s}} &=\frac{1}{(2\pi)^{3}}\int_{0}^{\infty}dk\int_{0}^{2\pi}d\phi\int_{0}^{\pi}d\theta \sin(\theta) \frac{k^{2}}{(k^2+m^2)^{s/2}}\cos(kx \cos(\theta))\\ &=\frac{1}{(2\pi)^{2}}\int_{0}^{\infty}\frac{k^{2}}{(k^2+m^2)^{s/2}}\frac{2\sin(kx)}{kx}dk \\ &=\frac{1}{2\pi^{2}x}\int_{0}^{\infty}\frac{k\sin(kx)}{(k^{2}+m^{2})^{s/2}}dk \end{aligned} Now, if $$s=0$$, then \begin{aligned} &=\frac{1}{2\pi^{2} x} \int_{0}^{\infty} k\sin(kx)dk \\ &=\frac{1}{4 \pi^{2} x} \frac{1}{i}\int_{-\infty}^{\infty} ke^{ikx}dk \\ &=\frac{1}{2 \pi x} \left(\frac{1}{2 \pi i}\int_{-\infty}^{\infty} ke^{ikx}dk\right) \\ &=\frac{1}{2\pi x}\left(-\frac{\partial \delta(x)}{\partial x}\right) \\ &=\frac{1}{2\pi x^{2}} \delta(x) \end{aligned} where I used the identity $$x\delta'(x)=-\delta(x)$$.

This final result is almost the desired one since the Dirac delta in spherical coordinates in this case should be $$\delta^{3}(\vec{x})=\frac{1}{4 \pi x^{2}}\delta(x)$$ Where is the factor $$\frac{1}{2}$$ that I'm missing? Greetings and thank you a lot.

• Please note that check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual questions over those just asking for a specific computation. Jan 6, 2021 at 5:37
• Even when the delta function is on the boundary of the range of integration...? Jan 6, 2021 at 6:31
• Indeed. Maybe it's helpful to think of the gamma as one of its resolved versions, like a narrow gaussian. Try the calculation and then take the limit. Jan 6, 2021 at 6:48
• I meant delta, not gamma. This is what happens when you comment on here when you're in the middle of another calculation. Jan 6, 2021 at 7:06
• Would Mathematics be a better home for this question? Jan 7, 2021 at 5:38

The general delta function in 3D curved space is $$\frac{\delta^{3}({\bf x}-{\bf x}_c)}{\sqrt{ g }}$$ where $$g$$ is determinant of metric. For spherical coordinates, you should have $$\sqrt{g}=4\pi r^2$$, if you have spherical symmetry. Thus you should get $$\frac{\delta( r- r_c)}{4\pi r^2}$$
Edit: Delt function in spherical coordinates. Since $$\sqrt{g}=r^2\sin\theta$$, the full delta function without any additional symmetries should be $$\frac{1}{r^2\sin\theta} \delta( r- r_c) \delta( \theta- \theta_c) \delta( \phi- \phi_c)$$ if you have axial symmetry, the delta function becomes $$\frac{1}{2\pi r^2\sin\theta} \delta( r- r_c) \delta( \theta- \theta_c)$$
• Doesn't answer the question and doesn't address the case where there is a coordinate singularity at $\mathbf x_c$ which is precisely the case here. Jan 6, 2021 at 6:56
• @NiharKarve, kaylimekay, I just want to show that delta function in spherical coordinates is not $\delta(r)/(4\pi r^2)$. $\delta(r)/(4\pi r^2)$ is a delta function in spherical coordinates with an additional symmetry. Jan 6, 2021 at 7:02